{
"author": "Jiří Lebl",
"title": "Basic Analysis I–II",
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"id": "L2.3.2",
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\nProposition2.3.2.\n
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Let \\(\\{ x_n \\}_{n=1}^\\infty\\) be a bounded sequence. Let \\(a_n\\) and \\(b_n\\) be as in the definition above.\n
\n\n
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The sequence \\(\\{ a_n \\}_{n=1}^\\infty\\) is bounded monotone decreasing and \\(\\{ b_n \\}_{n=1}^\\infty\\) is bounded monotone increasing. In particular, \\(\\liminf\\limits_{n\\to\\infty} x_n\\) and \\(\\limsup\\limits_{n\\to\\infty} x_n\\) exist.\n
\n\n
\n
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\n\\(\\displaystyle \\limsup_{n \\to \\infty} x_n = \\inf \\{ a_n : n \\in \\N \\}\\) and \\(\\displaystyle \\liminf_{n \\to \\infty} x_n = \\sup \\{ b_n : n \\in \\N \\}\\text{.}\\)\n
Let us see why \\(\\{ a_n \\}_{n=1}^\\infty\\) is a decreasing sequence. As \\(a_n\\) is the least upper bound for \\(\\{ x_k : k \\geq n \\}\\text{,}\\) it is also an upper bound for the subset \\(\\{ x_k : k \\geq n+1 \\}\\text{.}\\) Therefore, \\(a_{n+1}\\text{,}\\) the least upper bound for \\(\\{ x_k : k \\geq n+1 \\}\\text{,}\\) has to be less than or equal to \\(a_n\\text{,}\\) the least upper bound for \\(\\{ x_k : k \\geq n \\}\\text{.}\\) That is, \\(a_n \\geq a_{n+1}\\) for all \\(n\\text{.}\\) Similarly (an exercise), \\(\\{ b_n \\}_{n=1}^\\infty\\) is an increasing sequence. It is left as an exercise to show that if \\(\\{ x_n \\}_{n=1}^\\infty\\) is bounded, then \\(\\{ a_n \\}_{n=1}^\\infty\\) and \\(\\{ b_n \\}_{n=1}^\\infty\\) must be bounded.\n
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The second item follows as the sequences \\(\\{ a_n \\}_{n=1}^\\infty\\) and \\(\\{ b_n \\}_{n=1}^\\infty\\) are monotone and bounded.\n
\n
\n
For the third item, note that \\(b_n \\leq a_n\\text{,}\\) as the \\(\\inf\\) of a nonempty set is less than or equal to its \\(\\sup\\text{.}\\) The sequences \\(\\{ a_n \\}_{n=1}^\\infty\\) and \\(\\{ b_n \\}_{n=1}^\\infty\\) converge to the limsup and the liminf, respectively. Apply Lemma 2.2.3 to obtain
Define \\(a_n \\coloneqq \\sup \\{ x_k : k \\geq n \\}\\text{.}\\) Write \\(x \\coloneqq \\limsup_{n\\to\\infty} x_n = \\lim_{n\\to\\infty} a_n\\text{.}\\) We define the subsequence inductively. Let \\(n_1 \\coloneqq 1\\text{,}\\) and suppose \\(n_1,n_2,\\ldots,n_{k-1}\\) are already defined for some \\(k \\geq 2\\text{.}\\) Pick an \\(m \\geq n_{k-1} + 1\\) such that
Such an \\(m\\) exists as \\(a_{(n_{k-1}+1)}\\) is a supremum of the set \\(\\{ x_\\ell : \\ell \\geq n_{k-1} + 1 \\}\\) and hence there are elements of the sequence arbitrarily close (or even possibly equal) to the supremum. Set \\(n_{k} \\coloneqq m\\text{.}\\) The subsequence \\(\\{ x_{n_k} \\}_{k=1}^\\infty\\) is defined. Next, we must prove that it converges to \\(x\\text{.}\\)\n
\n\n
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For all \\(k \\geq 2\\text{,}\\) we have \\(a_{(n_{k-1}+1)} \\geq a_{n_k}\\) (why?) and \\(a_{n_{k}} \\geq x_{n_k}\\text{.}\\) Therefore, for every \\(k \\geq 2\\text{,}\\)\n
Let us show that \\(\\{ x_{n_k} \\}_{k=1}^\\infty\\) converges to \\(x\\text{.}\\) Note that the subsequence need not be monotone. Let \\(\\epsilon > 0\\) be given. As \\(\\{ a_n \\}_{n=1}^\\infty\\) converges to \\(x\\text{,}\\) the subsequence \\(\\{ a_{n_k} \\}_{k=1}^\\infty\\) converges to \\(x\\text{.}\\) Thus there exists an \\(M_1 \\in \\N\\) such that for all \\(k \\geq M_1\\text{,}\\) we have
First suppose \\(\\liminf_{n\\to\\infty} x_n = \\limsup_{n\\to\\infty} x_n\\text{.}\\) Then \\(\\{ a_n \\}_{n=1}^\\infty\\) and \\(\\{ b_n \\}_{n=1}^\\infty\\) both converge to the same limit. By the squeeze lemma (Lemma 2.2.1), \\(\\{ x_n \\}_{n=1}^\\infty\\) converges and
Now suppose \\(\\{ x_n \\}_{n=1}^\\infty\\) converges to \\(x\\text{.}\\) By Theorem 2.3.4, there exists a subsequence \\(\\{ x_{n_k} \\}_{k=1}^\\infty\\) converging to \\(\\limsup_{n\\to\\infty} x_n\\text{.}\\) As \\(\\{ x_n \\}_{n=1}^\\infty\\) converges to \\(x\\text{,}\\) every subsequence converges to \\(x\\) and so \\(\\limsup_{n\\to\\infty} x_n = \\lim_{k\\to\\infty} x_{n_k} = x\\text{.}\\) Similarly, \\(\\liminf_{n\\to\\infty} x_n = x\\text{.}\\)\n
Suppose a sequence \\(\\{ x_n \\}_{n=1}^\\infty\\) of real numbers is bounded. Then there exists a convergent subsequence \\(\\{ x_{n_i} \\}_{i=1}^\\infty\\text{.}\\)\n
\n\n\n
Proof.
\n
\nTheorem 2.3.4 says that there exists a subsequence whose limit is \\(\\limsup_{n\\to\\infty} x_n\\text{.}\\)\n
\n\n",
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"proof_scope": "Bolzano–Weierstrass via the first complete proof. Alternate bisection proof also read, but not needed by this graph."
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\nProposition2.4.4.\n
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If a sequence is Cauchy, then it is bounded.\n
\n\n\n
Proof.
\n
\n
Suppose \\(\\{ x_n \\}_{n=1}^\\infty\\) is Cauchy. Pick an \\(M\\) such that for all \\(n,k \\geq M\\text{,}\\) we have \\(\\sabs{x_n-x_k} < 1\\text{.}\\) In particular, for all \\(n \\geq M\\text{,}\\)\n
Then \\(\\sabs{x_n} \\leq B\\) for all \\(n \\in \\N\\text{.}\\)\n
\n\n
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\nTheorem2.4.5.\n
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A sequence of real numbers is Cauchy if and only if it converges.\n
\n\n\n
Proof.
\n
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Suppose \\(\\{ x_n \\}_{n=1}^\\infty\\) converges to \\(x\\text{,}\\) and let \\(\\epsilon > 0\\) be given. Then there exists an \\(M\\) such that for \\(n \\geq M\\text{,}\\)\n
Hence, for \\(n \\geq M\\) and \\(k \\geq M\\text{,}\\)\n
\n
\r\n\\begin{equation}\r\n\\sabs{x_n - x_k} = \r\n\\sabs{x_n - x + x - x_k}\r\n\\leq \\sabs{x_n-x} + \\sabs{x-x_k} < \\frac{\\epsilon}{2} + \\frac{\\epsilon}{2} =\r\n\\epsilon .\r\n\\end{equation}\r\n
\n\n
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Alright, that direction was easy. Now suppose \\(\\{ x_n \\}_{n=1}^\\infty\\) is Cauchy. We have shown that \\(\\{ x_n \\}_{n=1}^\\infty\\) is bounded. For a bounded sequence, liminf and limsup exist, and this is where we use the least-upper-bound property. If we show that
then \\(\\{ x_n \\}_{n=1}^\\infty\\) must be convergent by Proposition 2.3.5.
\n\n
\n
\n
Define \\(a \\coloneqq \\limsup_{n\\to\\infty} x_n\\) and \\(b \\coloneqq \\liminf_{n\\to\\infty} x_n\\text{.}\\) By Theorem 2.3.4, there exist subsequences \\(\\{ x_{n_i} \\}_{i=1}^\\infty\\) and \\(\\{ x_{m_i} \\}_{i=1}^\\infty\\) such that
Given an \\(\\epsilon > 0\\text{,}\\) there exists an \\(M_1\\) such that \\(\\sabs{x_{n_i} - a} < \\nicefrac{\\epsilon}{3}\\) for all \\(i \\geq M_1\\) and an \\(M_2\\) such that \\(\\sabs{x_{m_i} - b} < \\nicefrac{\\epsilon}{3}\\) for all \\(i \\geq M_2\\text{.}\\) There also exists an \\(M_3\\) such that \\(\\sabs{x_n-x_k} < \\nicefrac{\\epsilon}{3}\\) for all \\(n,k \\geq M_3\\text{.}\\) Let \\(M \\coloneqq \\max \\{ M_1, M_2, M_3 \\}\\text{.}\\) If \\(i \\geq M\\text{,}\\) then \\(n_i \\geq M\\) and \\(m_i \\geq M\\text{.}\\) Hence,
Sine and cosine are \\(2\\pi\\)-periodic and not periodic with any smaller period. That is, \\(2\\pi\\) is the smallest number such that for all \\(z \\in \\C\\text{,}\\)\n
The function \\(x \\mapsto e^{ix}\\) is a bijective map from \\([0,2\\pi)\\) onto the set of \\(z \\in \\C\\) such that \\(\\sabs{z} = 1\\text{.}\\)\n
\n\n
\n\n\n\n
Proof.
\n
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The first three items follow directly from the definition. The computation of the power series for both is left as an exercise. As the complex conjugate is a continuous function, the definition of \\(e^z\\) implies \\(\\overline{e^z} = e^{\\bar{z}}\\text{.}\\) If \\(x\\) is real,
Thus for real \\(x\\text{,}\\)\\(\\cos(x) =\r\n\\frac{e^{ix}+e^{-ix}}{2} =\r\n\\frac{e^{ix}+\\overline{e^{ix}}}{2} =\r\n\\Re (e^{ix})\\) and similarly \\(\\sin(x) = \\Im (e^{ix})\\text{.}\\)\n
A slightly more complicated computation shows this fact for complex numbers, see Exercise 11.4.6. In particular, \\(e^{ix}\\) is unimodular for real \\(x\\text{;}\\) the values lie on the unit circle. A square of a real number is always nonnegative:
So \\(\\babs{\\sin(x)} \\leq 1\\) and similarly \\(\\babs{\\cos(x)} \\leq 1\\text{.}\\)\n
\n\n
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We leave the computation of the derivatives to the reader as exercises. Let us prove that \\(\\sin(x) \\leq x\\) for \\(x \\geq 0\\text{.}\\) Consider \\(f(x) \\coloneqq x-\\sin(x)\\) and differentiate:
for all \\(x \\in \\R\\) as \\(\\babs{\\cos(x)} \\leq 1\\text{.}\\) In other words, \\(f\\) is increasing and \\(f(0) = 0\\text{.}\\) So \\(f\\) must be nonnegative when \\(x \\geq 0\\) and hence, \\(\\sin(x) \\leq x\\text{.}\\)\n
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Next, we claim there exists a positive \\(x\\) such that \\(\\cos(x) = 0\\text{.}\\) As \\(\\cos(0) = 1 > 0\\text{,}\\)\\(\\cos(x) > 0\\) for \\(x\\) near \\(0\\text{.}\\) Namely, there is some \\(y > 0\\) such that \\(\\cos(x) > 0\\) on \\([0,y)\\text{.}\\) Then \\(\\sin(x)\\) is strictly increasing on \\([0,y)\\text{.}\\) As \\(\\sin(0) = 0\\text{,}\\) we have \\(\\sin(x) > 0\\) for \\(x \\in (0,y)\\text{.}\\) Take \\(a \\in (0,y)\\text{.}\\) By the mean value theorem, there is a \\(c \\in (a,y)\\) such that
As \\(a \\in (0,y)\\text{,}\\) we have \\(\\sin(a) > 0\\) and so
\n
\r\n\\begin{equation}\r\ny \\leq \\frac{2}{\\sin(a)} + a .\r\n\\end{equation}\r\n
\n
Hence there is some largest \\(y\\) such that \\(\\cos(x) > 0\\) in \\([0,y)\\text{,}\\) and let \\(y\\) be the largest such number. By continuity, \\(\\cos(y) = 0\\text{.}\\) In fact, \\(y\\) is the smallest positive \\(y\\) such that \\(\\cos(y) = 0\\text{.}\\) As mentioned, \\(\\pi\\) is defined to be \\(2y\\text{.}\\)\n
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As \\(\\cos(\\nicefrac{\\pi}{2}) = 0\\text{,}\\) we find \\({\\bigl(\\sin(\\nicefrac{\\pi}{2})\\bigr)}^2 = 1\\text{.}\\) As \\(\\sin\\) is positive on \\((0,\\nicefrac{\\pi}{2})\\text{,}\\) we have \\(\\sin(\\nicefrac{\\pi}{2}) = 1\\text{.}\\) Hence,
\n
\r\n\\begin{equation}\r\ne^{i \\pi /2} = i ,\r\n\\end{equation}\r\n
for all \\(z \\in \\C\\text{.}\\) Immediately, we also obtain \\(\\cos(z+2\\pi) = \\cos(z)\\) and \\(\\sin(z+2\\pi) = \\sin(z)\\text{.}\\) So \\(\\sin\\) and \\(\\cos\\) are \\(2\\pi\\)-periodic.
\n\n
\n
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We claim that \\(\\sin\\) and \\(\\cos\\) are not periodic with a smaller period. It suffices to show that if \\(e^{ix} = 1\\) for the smallest positive \\(x\\text{,}\\) then \\(x = 2\\pi\\text{.}\\) Let \\(x\\) be the smallest positive \\(x\\) such that \\(e^{ix} = 1\\text{.}\\) Of course, \\(x \\leq 2\\pi\\text{.}\\) By the law of exponents,
Then either \\(a = 0\\) or \\(a^2 = b^2\\text{.}\\) As \\(\\nicefrac{x}{4} \\leq \\nicefrac{\\pi}{2}\\text{,}\\) we have \\(a = \\cos(\\nicefrac{x}{4}) \\geq 0\\) and \\(b = \\sin(\\nicefrac{x}{4}) > 0\\text{.}\\) If \\(a^2=b^2\\text{,}\\) then \\(a^4-6a^2b^2+b^4 = -4a^4 < 0\\) and in particular not equal to 1. Therefore \\(a=0\\text{,}\\) in which case \\(\\nicefrac{x}{4} = \\nicefrac{\\pi}{2}\\text{.}\\) Hence \\(2\\pi\\) is the smallest period we could choose for \\(e^{ix}\\) and so also for \\(\\cos\\) and \\(\\sin\\text{.}\\)\n
\n\n
\n
Finally, we wish to show that \\(e^{ix}\\) is one-to-one and onto from the set \\([0,2\\pi)\\) to the set of \\(z \\in \\C\\) such that \\(\\sabs{z} = 1\\text{.}\\) Suppose \\(e^{ix} = e^{iy}\\) and \\(x > y\\text{.}\\) Then \\(e^{i(x-y)} = 1\\text{,}\\) meaning \\(x-y\\) is a multiple of \\(2\\pi\\) and hence only one of them can live in \\([0,2\\pi)\\text{.}\\) To show \\(e^{ix}\\) is onto, pick \\((a,b) \\in \\R^2\\) such that \\(a^2+b^2 = 1\\text{.}\\) Suppose first that \\(a,b \\geq 0\\text{.}\\) By the intermediate value theorem, there must exist an \\(x \\in [0,\\nicefrac{\\pi}{2}]\\) such that \\(\\cos(x) = a\\text{,}\\) and hence \\(b^2 = \\bigl(\\sin(x)\\bigr)^2\\text{.}\\) As \\(b\\) and \\(\\sin(x)\\) are nonnegative, \\(b = \\sin(x)\\text{.}\\) Since \\(-\\sin(x)\\) is the derivative of \\(\\cos(x)\\) and \\(\\cos(-x) = \\cos(x)\\text{,}\\) we have that \\(\\sin(x) < 0\\) for \\(x \\in [\\nicefrac{-\\pi}{2},0)\\text{.}\\) Using the same reasoning, we obtain that if \\(a > 0\\) and \\(b \\leq 0\\text{,}\\) we can find an \\(x\\) in \\([\\nicefrac{-\\pi}{2},0)\\text{,}\\) and by periodicity, \\(x \\in [\\nicefrac{3\\pi}{2},2\\pi)\\) such that \\(\\cos(x) = a\\) and \\(\\sin(x)=b\\text{.}\\) Multiplying by \\(-1\\) is the same as multiplying by \\(e^{i\\pi}\\) or \\(e^{-i\\pi}\\text{.}\\) So we can always assume that \\(a \\geq 0\\) (details are left as an exercise).\n
\n\n",
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"proof_scope": "Proposition and retained argument completed by P16.1–P16.3, including its actual exercises, first-zero supremum, return-time existence issue and omitted axes. The following arc-length assertion is excluded."
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\nTheorem9.1.1.Leibniz integral rule.\n
\n
\n
Suppose \\(f \\colon [a,b] \\times [c,d] \\to \\R\\) is a continuous function, such that \\(\\frac{\\partial f}{\\partial y}\\) exists for all \\((x,y) \\in [a,b]\r\n\\times [c,d]\\) and is continuous. Define \\(g \\colon [c,d] \\to \\R\\) by
Fix \\(y \\in [c,d]\\) and let \\(\\epsilon > 0\\) be given. As \\(\\frac{\\partial f}{\\partial y}\\) is continuous on \\([a,b] \\times [c,d]\\) it is uniformly continuous. In particular, there exists \\(\\delta > 0\\) such that whenever \\(y_1 \\in [c,d]\\) with \\(\\sabs{y_1-y} < \\delta\\) and all \\(x \\in [a,b]\\text{,}\\) we have
Suppose \\(h\\) is such that \\(y+h \\in [c,d]\\) and \\(\\sabs{h} < \\delta\\text{.}\\) Fix \\(x\\) for a moment and apply the mean value theorem to find a \\(y_1\\) between \\(y\\) and \\(y+h\\) such that
We defined uniform convergence for sequences although the idea is the same. You may replace \\(h\\) with a sequence of nonzero numbers \\(\\{ h_n \\}_{n=1}^\\infty\\) converging to \\(0\\) such that \\(y+h_n \\in [c,d]\\) and let \\(n \\to \\infty\\text{.}\\)\n
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Consider the difference quotient of \\(g\\text{,}\\)\n
Let \\(x \\coloneqq \\lim_{n\\to\\infty} a_n = \\lim_{n\\to\\infty} b_n\\text{.}\\) Let \\(\\epsilon > 0\\) be given. Find an \\(M_1\\) such that for all \\(n \\geq M_1\\text{,}\\) we have \\(\\sabs{a_n-x} < \\epsilon\\text{,}\\) and an \\(M_2\\) such that for all \\(n \\geq M_2\\text{,}\\) we have \\(\\sabs{b_n-x} < \\epsilon\\text{.}\\) Set \\(M \\coloneqq \\max \\{M_1, M_2 \\}\\text{.}\\) Suppose \\(n \\geq M\\text{.}\\) In particular, \\(x - a_n < \\epsilon\\text{,}\\) or \\(x - \\epsilon < a_n\\text{.}\\) Similarly, \\(b_n < x + \\epsilon\\text{.}\\) Putting everything together, we find
In other words, \\(-\\epsilon < x_n-x < \\epsilon\\) or \\(\\sabs{x_n-x} < \\epsilon\\text{.}\\) So \\(\\{x_n\\}_{n=1}^\\infty\\) converges to \\(x\\text{.}\\) See Figure 2.3.
Let \\(x \\coloneqq \\lim_{n\\to\\infty} x_n\\) and \\(y \\coloneqq \\lim_{n\\to\\infty} y_n\\text{.}\\) Let \\(\\epsilon > 0\\) be given. Find an \\(M_1\\) such that for all \\(n \\geq M_1\\text{,}\\) we have \\(\\sabs{x_n-x} < \\nicefrac{\\epsilon}{2}\\text{.}\\) Find an \\(M_2\\) such that for all \\(n \\geq M_2\\text{,}\\) we have \\(\\sabs{y_n-y} < \\nicefrac{\\epsilon}{2}\\text{.}\\) In particular, for any \\(n \\geq \\max\\{ M_1, M_2 \\}\\text{,}\\) we have \\(x-x_n < \\nicefrac{\\epsilon}{2}\\) and \\(y_n-y < \\nicefrac{\\epsilon}{2}\\text{.}\\) We add these inequalities to obtain
If \\(\\lim_{n\\to\\infty} y_n \\neq 0\\) and \\(y_n \\neq 0\\) for all \\(n \\in \\N\\text{,}\\) then the sequence \\(\\{ z_n \\}_{n=1}^\\infty\\text{,}\\) where \\(z_n \\coloneqq \\dfrac{x_n}{y_n}\\text{,}\\) converges and
We start with i. Suppose \\(\\{ x_n \\}_{n=1}^\\infty\\) and \\(\\{ y_n \\}_{n=1}^\\infty\\) are convergent sequences and write \\(z_n \\coloneqq x_n + y_n\\text{.}\\) Let \\(x \\coloneqq \\lim_{n\\to\\infty} x_n\\text{,}\\)\\(y \\coloneqq \\lim_{n\\to\\infty} y_n\\text{,}\\) and \\(z \\coloneqq x+y\\text{.}\\) Let \\(\\epsilon > 0\\) be given. Find an \\(M_1\\) such that for all \\(n \\geq M_1\\text{,}\\) we have \\(\\sabs{x_n - x} < \\nicefrac{\\epsilon}{2}\\text{.}\\) Find an \\(M_2\\) such that for all \\(n \\geq M_2\\text{,}\\) we have \\(\\sabs{y_n - y} < \\nicefrac{\\epsilon}{2}\\text{.}\\) Take \\(M \\coloneqq \\max \\{ M_1, M_2 \\}\\text{.}\\) For all \\(n \\geq M\\text{,}\\) we have
Therefore, i is proved. The proof of ii is almost identical and is left as an exercise.
\n\n
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Let us tackle iii. Suppose again that \\(\\{ x_n \\}_{n=1}^\\infty\\) and \\(\\{ y_n \\}_{n=1}^\\infty\\) are convergent sequences and write \\(z_n \\coloneqq x_n y_n\\text{.}\\) Let \\(x \\coloneqq \\lim_{n\\to\\infty} x_n\\text{,}\\)\\(y \\coloneqq \\lim_{n\\to\\infty} y_n\\text{,}\\) and \\(z \\coloneqq xy\\text{.}\\) Let \\(\\epsilon > 0\\) be given. Let \\(K \\coloneqq \\max\\{ \\sabs{x}, \\sabs{y}, \\nicefrac{\\epsilon}{3} , 1 \\}\\text{.}\\) Find an \\(M_1\\) such that for all \\(n \\geq M_1\\text{,}\\) we have \\(\\sabs{x_n - x} < \\frac{\\epsilon}{3K}\\text{.}\\) Find an \\(M_2\\) such that for all \\(n \\geq M_2\\text{,}\\) we have \\(\\sabs{y_n - y} < \\frac{\\epsilon}{3K}\\text{.}\\) Take \\(M \\coloneqq \\max \\{ M_1, M_2 \\}\\text{.}\\) For all \\(n \\geq M\\text{,}\\) we have
Finally, we examine iv. We prove the following simpler claim:\n
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Claim: If \\(\\{ y_n \\}_{n=1}^\\infty\\) is a convergent sequence such that \\(\\lim_{n\\to\\infty} y_n \\neq 0\\) and \\(y_n \\neq 0\\) for all \\(n \\in \\N\\text{,}\\) then \\(\\{ \\nicefrac{1}{y_n} \\}_{n=1}^\\infty\\) converges and\n
Once the claim is proved, we take the sequence \\(\\{ \\nicefrac{1}{y_n} \\}_{n=1}^\\infty\\text{,}\\) multiply it by the sequence \\(\\{ x_n \\}_{n=1}^\\infty\\text{,}\\) and apply item iii.\n
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Proof of claim: Let \\(\\epsilon > 0\\) be given. Let \\(y \\coloneqq \\lim_{n\\to\\infty} y_n\\text{.}\\) As \\(\\sabs{y} \\neq 0\\text{,}\\) we have \\(\\min \\left\\{ \\sabs{y}^2\\frac{\\epsilon}{2}, \\, \\frac{\\sabs{y}}{2} \\right\\} > 0\\text{.}\\) Find an \\(M\\) such that for all \\(n \\geq M\\text{,}\\) we have
If \\(c > 1\\text{,}\\) then \\(\\{ c^n \\}_{n=1}^\\infty\\) is unbounded.\n
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Proof.
\n
First consider \\(c < 1\\text{.}\\) As \\(c > 0\\text{,}\\) we get \\(c^n > 0\\) for all \\(n \\in \\N\\) by induction. Then \\(c < 1\\) implies that \\(c^{n+1} < c^n\\) for all \\(n\\text{.}\\) The sequence \\(\\{ c^n \\}_{n=1}^\\infty\\) is thus bounded below and decreasing. Hence, it is convergent. Let \\(x \\coloneqq \\lim_{n\\to\\infty} c^n\\text{.}\\) The 1-tail \\(\\{ c^{n+1} \\}_{n=1}^\\infty\\) also converges to \\(x\\text{.}\\) Taking the limit of both sides of \\(c^{n+1} = c \\cdot\r\nc^n\\text{,}\\) we obtain \\(x = cx\\text{,}\\) or \\((1-c)x=0\\text{.}\\) It follows that \\(x=0\\) as \\(c \\neq 1\\text{.}\\)\n
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Now consider \\(c > 1\\text{.}\\) Let \\(B > 0\\) be arbitrary. As \\(\\nicefrac{1}{c} < 1\\text{,}\\) the sequence \\(\\bigl\\{ {(\\nicefrac{1}{c})}^n \\bigr\\}_{n=1}^\\infty\\) converges to \\(0\\text{.}\\) Hence, for some large enough \\(n\\text{,}\\) we get
In other words, \\(c^n > B\\text{,}\\) and \\(B\\) is not an upper bound for \\(\\{ c^n \\}_{n=1}^\\infty\\text{.}\\) As \\(B\\) was arbitrary, \\(\\{ c^n \\}_{n=1}^\\infty\\) is unbounded.
\n\n
\n\n",
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"proof_scope": "For 01 part is not needed"
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\nTheorem5.3.1.\n
\n
\n
Let \\(F \\colon [a,b] \\to \\R\\) be a continuous function, differentiable on \\((a,b)\\text{.}\\) Let \\(f \\in \\sR\\bigl([a,b]\\bigr)\\) be such that \\(f(x) = F'(x)\\) for \\(x \\in\r\n(a,b)\\text{.}\\) Then
\n
\r\n\\begin{equation}\r\n\\int_a^b f = F(b)-F(a) .\r\n\\end{equation}\r\n
\n\n
\n\n\n
Proof.
\n
\n
Let \\(P = \\{ x_0, x_1, \\ldots, x_n \\}\\) be a partition of \\([a,b]\\text{.}\\) For each interval \\([x_{i-1},x_i]\\text{,}\\) use the mean value theorem to find a \\(c_i \\in (x_{i-1},x_i)\\) such that
See Figure 5.5, and note that the area of the \\(i\\)th rectangle is \\(F(x_{i})-F(x_{i-1})\\text{,}\\) and the total area of all three rectangles pictured is \\(F(x_{i+1})-F(x_{i-2})\\text{.}\\) The idea is that taking smaller and smaller subintervals, the total area of all these rectangles converges to the integral of \\(f\\text{.}\\)\n
\n\n
\n\nFigure5.5.Mean value theorem on subintervals of a partition approximating the area under the curve.\n\n
\n
Using the notation from the definition of the integral, \\(m_i \\leq f(c_i) \\leq M_i\\text{,}\\) and multiplying by \\(\\Delta x_i\\) gets
In the middle sum, all the terms except the first and last cancel and we end up with \\(F(x_n)-F(x_0) = F(b)-F(a)\\text{.}\\) The sums on the left and on the right are the lower and the upper sums, respectively. So
We take the supremum of \\(L(P,f)\\) over all partitions \\(P\\) and the left inequality yields
\n
\r\n\\begin{equation}\r\n\\underline{\\int_a^b} f \\leq F(b)-F(a) .\r\n\\end{equation}\r\n
\n
Similarly, taking the infimum of \\(U(P,f)\\) over all partitions \\(P\\) yields
\n
\r\n\\begin{equation}\r\nF(b)-F(a) \\leq \\overline{\\int_a^b} f .\r\n\\end{equation}\r\n
\n
As \\(f\\) is Riemann integrable, we have
\n
\r\n\\begin{equation}\r\n\\int_a^b f =\r\n\\underline{\\int_a^b} f \\leq F(b)-F(a) \\leq \\overline{\\int_a^b} f\r\n= \\int_a^b f .\r\n\\end{equation}\r\n
\n
The inequalities must be equalities and we are done.
\n\n
\n\n",
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\nTheorem5.3.3.\n
\n
\n
Let \\(f \\colon [a,b] \\to \\R\\) be a Riemann integrable function. Define
\n
\r\n\\begin{equation}\r\nF(x) \\coloneqq \\int_a^x f .\r\n\\end{equation}\r\n
\n
First, \\(F\\) is continuous on \\([a,b]\\text{.}\\) Second, if \\(f\\) is continuous at \\(c \\in [a,b]\\text{,}\\) then \\(F\\) is differentiable at \\(c\\) and \\(F'(c) = f(c)\\text{.}\\)\n
\n\n
\n\n\n
Proof.
\n
\n
As \\(f\\) is bounded, there is an \\(M > 0\\) such that \\(\\babs{f(x)} \\leq M\\) for all \\(x \\in [a,b]\\text{.}\\) Suppose \\(x,y \\in [a,b]\\) with \\(x > y\\text{.}\\) Then
\n
\r\n\\begin{equation}\r\n\\babs{F(x)-F(y)} =\r\n\\abs{\\int_a^x f - \\int_a^y f}\r\n=\r\n\\abs{\\int_y^x f}\r\n\\leq\r\nM\\sabs{x-y} .\r\n\\end{equation}\r\n
\n
By symmetry, the same also holds if \\(x < y\\text{.}\\) So \\(F\\) is Lipschitz continuous and hence continuous.
\n\n
\n
\n
Now suppose \\(f\\) is continuous at \\(c\\text{.}\\) Let \\(\\epsilon > 0\\) be given. Let \\(\\delta > 0\\) be such that for \\(x \\in [a,b]\\text{,}\\)\\(\\sabs{x-c} < \\delta\\) implies \\(\\babs{f(x)-f(c)} < \\epsilon\\text{.}\\) In particular, for such \\(x\\text{,}\\) we have
The result follows. It is left to the reader to see why is it OK that we just have a non-strict inequality.
\n\n
\n\n",
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"id": "L5.3.5",
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\nTheorem5.3.5.Change of variables.\n
\n
\n
Let \\(g \\colon [a,b] \\to \\R\\) be a continuously differentiable function, let \\(f \\colon [c,d] \\to \\R\\) be continuous, and suppose \\(g\\bigl([a,b]\\bigr) \\subset [c,d]\\text{.}\\) Then
\n
\r\n\\begin{equation}\r\n\\int_a^b f\\bigl(g(x)\\bigr)\\, g'(x)\\, dx =\r\n\\int_{g(a)}^{g(b)} f(u)\\, du .\r\n\\end{equation}\r\n
\n\n
\n\n\n
Proof.
\n
As \\(g\\text{,}\\)\\(g'\\text{,}\\) and \\(f\\) are continuous, \\(f\\bigl(g(x)\\bigr)\\,g'(x)\\) is a continuous function of \\([a,b]\\text{,}\\) therefore it is Riemann integrable. Similarly, \\(f\\) is integrable on every subinterval of \\([c,d]\\text{.}\\)\n
By the second form of the fundamental theorem of calculus (see Remark 5.3.4 and Exercise 5.3.4), \\(F\\) is a differentiable function and \\(F'(y) = f(y)\\text{.}\\) Apply the chain rule,
\n
\r\n\\begin{equation}\r\n\\bigl( F \\circ g \\bigr)' (x) =\r\nF'\\bigl(g(x)\\bigr) g'(x)\r\n=\r\nf\\bigl(g(x)\\bigr) g'(x) .\r\n\\end{equation}\r\n
\n
Note that \\(F\\bigl(g(a)\\bigr) = 0\\) and use the first form of the fundamental theorem to obtain
\n
\r\n\\begin{multline*}\r\n\\int_{g(a)}^{g(b)} f(u)\\,du = F\\bigl(g(b)\\bigr) = F\\bigl(g(b)\\bigr)-F\\bigl(g(a)\\bigr)\r\n\\\\\r\n=\r\n\\int_a^b \r\n\\bigl( F \\circ g \\bigr)' (x) \\,dx\r\n=\r\n\\int_a^b \r\nf\\bigl(g(x)\\bigr) g'(x)\r\n\\,dx .\r\n\\qedhere\r\n\\end{multline*}\r\n
\n\n
\n\n",
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"proof_scope": "Oriented one-variable substitution, including noninjective g and endpoint range values."
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\nProposition5.5.3.\n
\n
\n
Let \\(f \\colon [a,\\infty) \\to \\R\\) be a function that is Riemann integrable on \\([a,b]\\) for all \\(b > a\\text{.}\\) For every \\(b > a\\text{,}\\) the integral \\(\\int_b^\\infty f\\) converges if and only if \\(\\int_a^\\infty f\\) converges, in which case
\n
\r\n\\begin{equation}\r\n\\int_a^\\infty f\r\n=\r\n\\int_a^b f +\r\n\\int_b^\\infty f .\r\n\\end{equation}\r\n
\n\n
\n\n\n
Proof.
\n
\n
Let \\(c > b\\text{.}\\) Then
\n
\r\n\\begin{equation}\r\n\\int_a^c f\r\n=\r\n\\int_a^b f +\r\n\\int_b^c f .\r\n\\end{equation}\r\n
\n
Taking the limit \\(c \\to \\infty\\) finishes the proof.
\n\n
\n\n",
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\nProposition5.5.4.\n
\n
Suppose \\(f \\colon [a,\\infty) \\to \\R\\) is nonnegative (\\(f(x)\r\n\\geq 0\\) for all \\(x\\)) and \\(f\\) is Riemann integrable on \\([a,b]\\) for all \\(b > a\\text{.}\\)\n
\n\n
\n
\r\n\\begin{equation}\r\n\\int_a^\\infty f = \\sup \\left\\{ \\int_a^x f : x \\geq a \\right\\} .\r\n\\end{equation}\r\n
\n\n
\n
\n
\n
Suppose \\(\\{ x_n \\}_{n=1}^\\infty\\) is a sequence with \\(\\lim_{n\\to\\infty} x_n = \\infty\\text{.}\\) Then \\(\\int_a^\\infty f\\) converges if and only if \\(\\lim_{n\\to\\infty} \\int_a^{x_n} f\\) exists, in which case
\n
\r\n\\begin{equation}\r\n\\int_a^\\infty f = \\lim_{n\\to\\infty} \\int_a^{x_n} f .\r\n\\end{equation}\r\n
\n\n
\n\n
\n\n\n\n
Proof.
\n
We start with the first item. As \\(f\\) is nonnegative, \\(\\int_a^x f\\) is increasing as a function of \\(x\\text{.}\\) If the supremum is infinite, then for every \\(M \\in \\R\\) we find \\(N\\) such that \\(\\int_a^N f \\geq M\\text{.}\\) As \\(\\int_a^x f\\) is increasing, \\(\\int_a^x f \\geq M\\) for all \\(x \\geq N\\text{.}\\) So \\(\\int_a^\\infty f\\) diverges to infinity.\n
\n
Next suppose the supremum is finite, say \\(A \\coloneqq \\sup \\left\\{ \\int_a^x f : x \\geq a \\right\\}\\text{.}\\) For every \\(\\epsilon > 0\\text{,}\\) we find an \\(N\\) such that \\(A - \\int_a^N f < \\epsilon\\text{.}\\) As \\(\\int_a^x f\\) is increasing, then \\(A - \\int_a^x f < \\epsilon\\) for all \\(x \\geq N\\) and hence \\(\\int_a^\\infty f\\) converges to \\(A\\text{.}\\)\n
\n
\n
Let us look at the second item. If \\(\\int_a^\\infty f\\) converges, then every sequence \\(\\{ x_n \\}_{n=1}^\\infty\\) going to infinity works. The trick is proving the other direction. Suppose \\(\\{ x_n \\}_{n=1}^\\infty\\) is such that \\(\\lim_{n\\to\\infty} x_n = \\infty\\) and
\n
\r\n\\begin{equation}\r\n\\lim_{n\\to\\infty} \\int_a^{x_n} f = A\r\n\\end{equation}\r\n
\n
converges. Given \\(\\epsilon > 0\\text{,}\\) pick \\(N\\) such that for all \\(n \\geq N\\text{,}\\) we have \\(A - \\epsilon < \\int_a^{x_n} f < A + \\epsilon\\text{.}\\) Because \\(\\int_a^x f\\) is increasing as a function of \\(x\\text{,}\\) we have that for all \\(x \\geq x_N\\)\n
\n
\r\n\\begin{equation}\r\nA - \\epsilon < \\int_a^{x_N} f \\leq \\int_a^x f .\r\n\\end{equation}\r\n
\n
As \\(\\{ x_n \\}_{n=1}^\\infty\\) goes to \\(\\infty\\text{,}\\) we have that for any \\(x\\text{,}\\) there is an \\(x_m\\) such that \\(m \\geq N\\) and \\(x \\leq x_m\\text{.}\\) Then
\n
\r\n\\begin{equation}\r\n\\int_a^{x} f \\leq \\int_a^{x_m} f < A + \\epsilon .\r\n\\end{equation}\r\n
\n
In particular, for all \\(x \\geq x_N\\text{,}\\) we have \\(\\abs{\\int_a^{x} f - A} < \\epsilon\\text{.}\\)\n
\nProposition5.5.5.Comparison test for improper integrals.\n
\n
\n
Let \\(f \\colon [a,\\infty) \\to \\R\\) and \\(g \\colon [a,\\infty) \\to \\R\\) be functions that are Riemann integrable on \\([a,b]\\) for all \\(b > a\\text{.}\\) Suppose that for all \\(x \\geq a\\text{,}\\)\n
If \\(\\int_a^\\infty g\\) converges, then \\(\\int_a^\\infty f\\) converges, and in this case \\(\\abs{\\int_a^\\infty f} \\leq \\int_a^\\infty g\\text{.}\\)\n
\n\n
\n
\n
If \\(\\int_a^\\infty f\\) diverges, then \\(\\int_a^\\infty g\\) diverges.\n
\n\n
\n\n\n\n
Proof.
\n
\n
We start with the first item. For every \\(b\\) and \\(c\\text{,}\\) such that \\(a \\leq b \\leq c\\text{,}\\) we have \\(-g(x) \\leq f(x) \\leq g(x)\\text{,}\\) and so
\n
\r\n\\begin{equation}\r\n\\int_b^c -g \\leq \\int_b^c f \\leq \\int_b^c g .\r\n\\end{equation}\r\n
\n
In other words, \\(\\abs{\\int_b^c f} \\leq \\int_b^c g\\text{.}\\)\n
\r\n\\begin{equation}\r\n\\int_a^\\infty g =\r\n\\int_a^b g +\r\n\\int_b^\\infty g .\r\n\\end{equation}\r\n
\n
As \\(\\int_a^b g\\) goes to \\(\\int_a^\\infty g\\) as \\(b\\) goes to infinity, \\(\\int_b^\\infty g\\) goes to 0 as \\(b\\) goes to infinity. Choose \\(B\\) such that
\n
\r\n\\begin{equation}\r\n\\int_B^\\infty g < \\epsilon .\r\n\\end{equation}\r\n
\n
As \\(g\\) is nonnegative, if \\(B \\leq b < c\\text{,}\\) then \\(\\int_b^c g < \\epsilon\\) as well. Let \\(\\{ x_n \\}_{n=1}^\\infty\\) be a sequence going to infinity. Let \\(M\\) be such that \\(x_n \\geq B\\) for all \\(n \\geq M\\text{.}\\) Take \\(n, m \\geq M\\text{,}\\) with \\(x_n \\leq x_m\\text{,}\\)\n
\n
\r\n\\begin{equation}\r\n\\abs{\\int_a^{x_m} f - \\int_a^{x_n} f} \r\n=\r\n\\abs{\\int_{x_n}^{x_m} f} \r\n\\leq \\int_{x_n}^{x_m} g < \\epsilon .\r\n\\end{equation}\r\n
\n
Therefore, the sequence \\(\\bigl\\{ \\int_a^{x_n} f \\bigr\\}_{n=1}^\\infty\\) is Cauchy and hence converges.
\n\n
\n
\n
We need to show that the limit is unique. Suppose \\(\\{ x_n \\}_{n=1}^\\infty\\) is a sequence converging to infinity such that \\(\\bigl\\{ \\int_a^{x_n} f \\bigr\\}_{n=1}^\\infty\\) converges to \\(L_1\\text{,}\\) and \\(\\{\r\ny_n \\}_{n=1}^\\infty\\) is a sequence converging to infinity such that \\(\\bigl\\{ \\int_a^{y_n} f \\bigr\\}_{n=1}^\\infty\\) converges to \\(L_2\\text{.}\\) Then there must be some \\(n\\) such that \\(\\babs{\\int_a^{x_n} f - L_1} < \\epsilon\\) and \\(\\babs{\\int_a^{y_n} f - L_2} < \\epsilon\\text{.}\\) We can also suppose \\(x_n \\geq B\\) and \\(y_n \\geq B\\text{.}\\) Then
As \\(\\epsilon > 0\\) was arbitrary, \\(L_1 = L_2\\text{,}\\) and hence \\(\\int_a^\\infty f\\) converges. Above we have shown that \\(\\abs{\\int_a^c f} \\leq \\int_a^c g\\) for all \\(c > a\\text{.}\\) By taking the limit \\(c \\to \\infty\\text{,}\\) the first item is proved.
\n\n
\n
The second item is simply a contrapositive of the first item.\n
converges to \\(\\frac{1}{1-p}\\) if \\(0 < p < 1\\) and diverges if \\(p \\geq 1\\text{.}\\)\n
\n\n
\n\n\n
Proof.
\n
The proof follows by application of the fundamental theorem of calculus. Let us do the proof for \\(p > 1\\) for the infinite right endpoint and leave the rest to the reader. Hint: You should handle \\(p=1\\) separately.\n
\n
\n
Suppose \\(p > 1\\text{.}\\) Then using the fundamental theorem,
As \\(p > 1\\text{,}\\) we have \\(p-1 > 0\\text{.}\\) Take the limit as \\(b \\to \\infty\\) to obtain that \\(\\frac{1}{b^{p-1}}\\) goes to 0. The result follows.
\n\n
\n\n",
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\nTheorem10.2.2.Fubini version A.\n
\n
\n
\n Let \\(R \\times S \\subset \\R^n \\times \\R^m\\) be a closed rectangle and \\(f \\colon R \\times S \\to \\R\\) be integrable. The functions \\(g \\colon R \\to \\R\\) and \\(h \\colon R \\to \\R\\) defined by
\r\n\\begin{equation}\r\n\\int_R g = \\int_R h = \\int_{R \\times S} f .\r\n\\end{equation}\r\n
\n\n
\n\n\n
Proof.
\n
A partition of \\(R \\times S\\) can be written as \\((P,P') = (P_1,P_2,\\ldots,P_n,P'_1,P'_2,\\ldots,P'_m)\\text{,}\\) where \\(P = (P_1,P_2,\\ldots,P_n)\\) and \\(P' = (P'_1,P'_2,\\ldots,P'_m)\\) are partitions of \\(R\\) and \\(S\\) respectively. Let \\(R_1,R_2,\\ldots,R_N\\) be the subrectangles of \\(P\\) and \\(R'_1,R'_2,\\ldots,R'_K\\) be the subrectangles of \\(P'\\text{.}\\) The subrectangles of \\((P,P')\\) are \\(R_i \\times R'_j\\) where \\(1 \\leq i \\leq N\\) and \\(1 \\leq j \\leq K\\text{.}\\)\n
Similarly, \\(U\\bigl((P,P'),f\\bigr) \\geq U(P,h)\\text{,}\\) and the proof of this inequality is left as an exercise. Putting the two inequalities together with the fact that \\(g(x) \\leq h(x)\\) for all \\(x\\text{,}\\)\n
and we can make the right-hand side arbitrarily small. As for any partition we have \\(L\\bigl((P,P'),f\\bigr) \\leq L(P,g) \\leq U\\bigl((P,P'),f\\bigr)\\text{,}\\) we have \\(\\int_R g = \\int_{R \\times S} f\\text{.}\\)\n
As \\(f\\) is integrable, so is \\(h\\text{.}\\) Moreover, \\(L\\bigl((P,P'),f\\bigr) \\leq L(P,h) \\leq U\\bigl((P,P'),f\\bigr)\\) implies \\(\\int_R h = \\int_{R \\times S} f\\text{.}\\)\n
\n\n
\n\n",
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\nProposition10.5.1.\n
\n
A bounded set \\(S \\subset \\R^n\\) is Jordan measurable if and only if the boundary \\(\\partial S\\) is a measure zero set.\n
\n\n\n
Proof.
\n
Suppose \\(R\\) is a closed rectangle such that \\(S\\) is contained in the interior of \\(R\\text{.}\\) If \\(x \\in \\partial S\\text{,}\\) then for every \\(\\delta > 0\\text{,}\\) the sets \\(S \\cap B(x,\\delta)\\) (where \\(\\chi_S\\) is 1) and the sets \\((R \\setminus S) \\cap B(x,\\delta)\\) (where \\(\\chi_S\\) is 0) are both nonempty. So \\(\\chi_S\\) is not continuous at \\(x\\text{.}\\) If \\(x\\) is either in the interior of \\(S\\) or in the complement of the closure \\(\\widebar{S}\\text{,}\\) then \\(\\chi_S\\) is either identically 1 or identically 0 in a whole neighborhood of \\(x\\) and hence \\(\\chi_S\\) is continuous at \\(x\\text{.}\\) Therefore, the set of discontinuities of \\(\\chi_S\\) is precisely the boundary \\(\\partial S\\text{.}\\) The proposition follows.\n
Let \\(R_1,R_2,\\ldots,R_k\\) be all the subrectangles of \\(P\\) such that \\(\\chi_S\\) is not identically zero on each \\(R_j\\text{.}\\) That is, there is some point \\(x \\in R_j\\) such that \\(x \\in S\\) (i.e. \\(\\chi_S(x)=1\\)). Let \\(O_j\\) be an open rectangle such that \\(R_j \\subset O_j\\) and \\(V(O_j) < V(R_j) + \\nicefrac{\\epsilon}{k}\\text{.}\\) Notice that \\(S \\subset\r\n\\bigcup_j O_j\\text{.}\\) Then
As \\(U(P,\\chi_S) \\leq V(S) + \\epsilon\\text{,}\\) then \\(m^*(S) - \\epsilon \\leq V(S) + \\epsilon\\text{,}\\) or in other words \\(m^*(S) \\leq V(S)\\text{.}\\)\n
\n\n
\n
\n
Let \\(R'_1,R'_2,\\ldots,R'_\\ell\\) be all the subrectangles of \\(P\\) such that \\(\\chi_S\\) is identically one on each \\(R'_j\\text{.}\\) In other words, these are the subrectangles contained in \\(S\\text{.}\\) The interiors of the subrectangles \\(R'^\\circ_j\\) are disjoint and \\(V(R'^\\circ_j) = V(R'_j)\\text{.}\\) Via Exercise 10.3.16,
\n\n",
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"proof_scope": "Full equality of Jordan volume and outer measure, with the finite-disjoint-rectangle and empty-family steps supplied."
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"reader": "sec_jordansets.html",
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\nProposition10.5.5.\n
\n
If \\(S \\subset \\R^n\\) is a bounded Jordan measurable set and \\(f \\colon S \\to \\R\\) is a bounded continuous function, then \\(f\\) is integrable on \\(S\\text{.}\\)\n
\n\n\n
Proof.
\n
Define the function \\(\\widetilde{f}\\) as above for some closed rectangle \\(R\\) with \\(S\r\n\\subset R\\text{.}\\) If \\(x \\in R \\setminus \\widebar{S}\\text{,}\\) then \\(\\widetilde{f}\\) is identically zero in a neighborhood of \\(x\\text{.}\\) Similarly, if \\(x\\) is in the interior of \\(S\\text{,}\\) then \\(\\widetilde{f} = f\\) on a neighborhood of \\(x\\) and \\(f\\) is continuous at \\(x\\text{.}\\) Therefore, \\(\\widetilde{f}\\) is only ever possibly discontinuous at \\(\\partial S\\text{,}\\) which is a set of measure zero, and we are finished.\n
\n\n",
"licence": "CC-BY-SA-4.0",
"proof_scope": "Every bounded continuous function on a bounded Jordan set is Riemann integrable; full proof."
},
{
"id": "L10.5.9",
"reader": "sec_jordansets.html",
"author_url": "https://www.jirka.org/ra/html/sec_jordansets.html",
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\nProposition10.5.9.\n
\n
Suppose \\(U \\subset \\R^n\\) is open and \\(S \\subset U\\) is a compact Jordan measurable set. Suppose \\(g \\colon U \\to \\R^n\\) is a one-to-one continuously differentiable mapping such that the Jacobian determinant \\(J_g\\) is never zero on \\(S\\text{.}\\) Then \\(g(S)\\) is bounded and Jordan measurable.\n
\n\n\n
Proof.
\n
Let \\(T \\coloneqq g(S)\\text{.}\\) By Lemma 7.5.5, the set \\(T\\) is also compact and so closed and bounded. We claim \\(\\partial T \\subset g(\\partial S)\\text{.}\\) Suppose the claim is proved. As \\(S\\) is Jordan measurable, \\(\\partial S\\) is measure zero. Then \\(g(\\partial S)\\) is measure zero by Proposition 10.3.10. As \\(\\partial T \\subset g(\\partial\r\nS)\\text{,}\\) then \\(T\\) is Jordan measurable.\n
\n
It is therefore left to prove the claim. As \\(T\\) is closed, \\(\\partial T \\subset T\\text{.}\\) Suppose \\(y \\in \\partial T\\text{.}\\) There must exist an \\(x \\in S\\) such that \\(g(x) = y\\text{,}\\) and by hypothesis \\(J_g(x) \\neq 0\\text{.}\\) We use the inverse function theorem (Theorem 8.5.1). We find a neighborhood \\(V \\subset U\\) of \\(x\\) and an open set \\(W\\) such that the restriction \\(f|_V\\) is a one-to-one and onto function from \\(V\\) to \\(W\\) with a continuously differentiable inverse. In particular, \\(g(x) = y \\in W\\text{.}\\) As \\(y \\in \\partial T\\text{,}\\) there exists a sequence \\(\\{ y_k \\}_{k=1}^\\infty\\) in \\(W\\) with \\(\\lim_{k\\to\\infty} y_k = y\\) and \\(y_k \\notin T\\text{.}\\) As \\(g|_V\\) is invertible and in particular has a continuous inverse, there exists a sequence \\(\\{ x_k \\}_{k=1}^\\infty\\) in \\(V\\) such that \\(g(x_k) = y_k\\) and \\(\\lim_{k\\to\\infty} x_k = x\\text{.}\\) Since \\(y_k \\notin T = g(S)\\text{,}\\) clearly \\(x_k \\notin S\\text{.}\\) Since \\(x \\in S\\text{,}\\) we conclude that \\(x \\in \\partial S\\text{.}\\) The claim is proved: \\(\\partial T \\subset\r\ng(\\partial S)\\text{.}\\)\n
\n\n",
"licence": "CC-BY-SA-4.0",
"proof_scope": "Full compact Jordan image theorem, with the local inverse theorem already closed; source f|V is the map g|V."
},
{
"id": "L3.1.7",
"reader": "sec_limoffunc.html",
"author_url": "https://www.jirka.org/ra/html/sec_limoffunc.html",
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\nLemma3.1.7.\n
\n
Let \\(S \\subset \\R\\text{,}\\) let \\(c\\) be a cluster point of \\(S\\text{,}\\) let \\(f \\colon S \\to\r\n\\R\\) be a function, and let \\(L \\in \\R\\text{.}\\)\n
\n
Then \\(f(x) \\to L\\) as \\(x \\to c\\) if and only if for every sequence \\(\\{ x_n \\}_{n=1}^\\infty\\) such that \\(x_n \\in S \\setminus \\{c\\}\\) for all \\(n\\text{,}\\) and such that \\(\\lim_{n\\to\\infty} x_n = c\\text{,}\\) we have that the sequence \\(\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty\\) converges to \\(L\\text{.}\\)\n
\n\n\n
Proof.
\n
\n
Suppose \\(f(x) \\to L\\) as \\(x \\to c\\text{,}\\) and \\(\\{ x_n \\}_{n=1}^\\infty\\) is a sequence such that \\(x_n \\in S \\setminus \\{c\\}\\) and \\(\\lim_{n\\to\\infty} x_n = c\\text{.}\\) We wish to show that \\(\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty\\) converges to \\(L\\text{.}\\) Let \\(\\epsilon > 0\\) be given. Find a \\(\\delta > 0\\) such that if \\(x \\in S \\setminus \\{c\\}\\) and \\(\\sabs{x-c} < \\delta\\text{,}\\) then \\(\\babs{f(x) - L} < \\epsilon\\text{.}\\) As \\(\\{ x_n \\}_{n=1}^\\infty\\) converges to \\(c\\text{,}\\) find an \\(M\\) such that for \\(n \\geq M\\text{,}\\) we have that \\(\\sabs{x_n - c} < \\delta\\text{.}\\) Therefore, for \\(n \\geq M\\text{,}\\)\n
Thus \\(\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty\\) converges to \\(L\\text{.}\\)\n
\n\n
\n
For the other direction, we use proof by contrapositive. Suppose it is not true that \\(f(x) \\to L\\) as \\(x \\to c\\text{.}\\) The negation of the definition is that there exists an \\(\\epsilon > 0\\) such that for every \\(\\delta > 0\\) there exists an \\(x \\in S \\setminus \\{c\\}\\text{,}\\) where \\(\\sabs{x-c} < \\delta\\) and \\(\\babs{f(x)-L} \\geq \\epsilon\\text{.}\\)\n
\n
Let us use \\(\\nicefrac{1}{n}\\) for \\(\\delta\\) in the statement above to construct a sequence \\(\\{ x_n \\}_{n=1}^\\infty\\text{.}\\) We have that there exists an \\(\\epsilon > 0\\) such that for every \\(n\\text{,}\\) there exists a point \\(x_n \\in S \\setminus \\{c\\}\\text{,}\\) where \\(\\sabs{x_n-c} < \\nicefrac{1}{n}\\) and \\(\\babs{f(x_n)-L} \\geq \\epsilon\\text{.}\\) The sequence \\(\\{ x_n \\}_{n=1}^\\infty\\) just constructed converges to \\(c\\text{,}\\) but the sequence \\(\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty\\) does not converge to \\(L\\text{.}\\) And we are done.\n
Let us prove iii. Property ii together with the fact that \\(L'(x) = \\nicefrac{1}{x} > 0\\) for \\(x > 0\\text{,}\\) implies that \\(L\\) is strictly increasing and hence one-to-one. Let us show \\(L\\) is onto. As \\(\\nicefrac{1}{t} \\geq \\nicefrac{1}{2}\\) when \\(t \\in [1,2]\\text{,}\\)\n
Given \\(y > 0\\text{,}\\) by the Archimedean property of the real numbers (notice \\(L(2) > 0\\)), there is an \\(n \\in \\N\\) such that \\(L(2^n) > y\\text{.}\\) The intermediate value theorem gives an \\(x_1 \\in (1,2^n)\\) such that \\(L(x_1) = y\\text{.}\\) Thus \\((0,\\infty)\\) is in the image of \\(L\\text{.}\\) As \\(L\\) is increasing, \\(L(x) > y\\) for all \\(x > 2^n\\text{,}\\) and so
Next \\(0 = L(\\nicefrac{x}{x}) = L(x) + L(\\nicefrac{1}{x})\\text{,}\\) and so \\(L(x) = - L(\\nicefrac{1}{x})\\text{.}\\) Using \\(x=2^{-n}\\text{,}\\) we obtain as above that \\(L\\) achieves all negative numbers. And
In the limits, note that only \\(x > 0\\) are in the domain of \\(L\\text{.}\\)\n
\n\n
\n
\n
Let us prove v. Fix \\(x > 0\\text{.}\\) As above, iv implies \\(L(x^n) = n L(x)\\) for all \\(n \\in \\N\\text{.}\\) We already found that \\(L(x) = - L(\\nicefrac{1}{x})\\text{,}\\) so \\(L(x^{-n}) = - L(x^n) = -n L(x)\\text{.}\\) Then for \\(m \\in \\N\\)\n
\n
\r\n\\begin{equation}\r\nL(x) = L\\Bigl({(x^{1/m})}^m\\Bigr) = m L\\bigl(x^{1/m}\\bigr) .\r\n\\end{equation}\r\n
\n
Putting everything together for \\(n \\in \\Z\\) and \\(m \\in \\N\\text{,}\\) we have \\(L(x^{n/m}) = n L(x^{1/m}) = (\\nicefrac{n}{m}) L(x)\\text{.}\\)\n
\n\n
\n
\n
Uniqueness follows using properties i and ii. Via the first form of the fundamental theorem of calculus (Theorem 5.3.1),
\n\\(E(x+y) = E(x)E(y)\\) for all \\(x,y \\in \\R\\text{.}\\)\n
\n\n
\n
\n
If \\(q \\in \\Q\\text{,}\\) then \\(E(qx) = {E(x)}^q\\text{.}\\)\n
\n\n
\n\n\n\n
Proof.
\n
Again, we prove existence of such a function by defining a candidate and proving that it satisfies all the properties. The \\(L = \\ln\\) defined above is invertible. Let \\(E\\) be the inverse function of \\(L\\text{.}\\) Property i is immediate.\n
\n
\n
Property ii follows via the inverse function theorem, in particular via Lemma 4.4.1: \\(L\\) satisfies all the hypotheses of the lemma, and hence
Let us look at property iii. The function \\(E\\) is strictly increasing since \\(E'(x) = E(x) > 0\\text{.}\\) As \\(E\\) is the inverse of \\(L\\text{,}\\) it must also be bijective. To find the limits, we use that \\(E\\) is strictly increasing and onto \\((0,\\infty)\\text{.}\\) For every \\(M > 0\\text{,}\\) there is an \\(x_0\\) such that \\(E(x_0) = M\\) and \\(E(x) \\geq M\\) for all \\(x \\geq x_0\\text{.}\\) Similarly, for every \\(\\epsilon > 0\\text{,}\\) there is an \\(x_0\\) such that \\(E(x_0) = \\epsilon\\) and \\(E(x) < \\epsilon\\) for all \\(x < x_0\\text{.}\\) Therefore,
To prove property iv, we use the corresponding property for the logarithm. Take \\(x, y \\in \\R\\text{.}\\) As \\(L\\) is bijective, find \\(a\\) and \\(b\\) such that \\(x = L(a)\\) and \\(y = L(b)\\text{.}\\) Then
\n
\r\n\\begin{equation}\r\nE(x+y) =\r\nE\\bigl(L(a)+L(b)\\bigr) = \r\nE\\bigl(L(ab)\\bigr) = ab = E(x)E(y) .\r\n\\end{equation}\r\n
\n\n
\n
\n
Property v also follows from the corresponding property of \\(L\\text{.}\\) Given \\(x \\in \\R\\text{,}\\) let \\(a\\) be such that \\(x = L(a)\\) and
Therefore, by Proposition 4.2.6, \\(F(x)E(-x) = F(0)E(-0) = 1\\) for all \\(x \\in \\R\\text{.}\\) Doing the computation with \\(F = E\\text{,}\\) we obtain \\(E(x)E(-x) = 1\\text{.}\\) Then
\n\\(E\\) is a function into \\((0,\\infty)\\) after all. However, \\(E(-x) \\neq 0\\) also follows from \\(E(x)E(-x) = 1\\text{.}\\) Therefore, we can prove uniqueness of \\(E\\) given i and ii, even for functions \\(E \\colon \\R\r\n\\to \\R\\text{.}\\)\n
for all \\(x \\in \\R\\text{.}\\) So \\(F(x)-E(x) = 0\\) for all \\(x\\text{,}\\) and we are done.
\n\n
\n\n",
"licence": "CC-BY-SA-4.0",
"proof_scope": "Parts (i)–(iv) and full uniqueness proof, with smooth inverse supplied by P3. Rational-power part (v) excluded."
},
{
"id": "L7.4.2",
"reader": "sec_metcompact.html",
"author_url": "https://www.jirka.org/ra/html/sec_metcompact.html",
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\nProposition7.4.2.\n
\n
A convergent sequence in a metric space is Cauchy.\n
\n\n\n
Proof.
\n
Suppose \\(\\{ x_n \\}_{n=1}^\\infty\\) converges to \\(p\\text{.}\\) Given \\(\\epsilon > 0\\text{,}\\) there is an \\(M\\) such that for all \\(n \\geq M\\text{,}\\) we have \\(d(p,x_n) < \\nicefrac{\\epsilon}{2}\\text{.}\\) Hence, for all \\(n,k \\geq M\\text{,}\\) we have \\(d(x_n,x_k) \\leq d(x_n,p) + d(p,x_k) < \\nicefrac{\\epsilon}{2} +\r\n\\nicefrac{\\epsilon}{2} = \\epsilon\\text{.}\\)\n
The space \\(\\R^n\\) with the standard metric is a complete metric space.\n
\n\n\n
Proof.
\n
\n
Let \\(\\{ x_m \\}_{m=1}^\\infty\\) be a Cauchy sequence in \\(\\R^n\\text{,}\\) where \\(x_m = \\bigl(x_{m,1},x_{m,2},\\ldots,x_{m,n}\\bigr) \\in \\R^n\\text{.}\\) As the sequence is Cauchy, given \\(\\epsilon > 0\\text{,}\\) there exists an \\(M\\) such that for all \\(i,j \\geq M\\text{,}\\)\n
Hence, the sequence \\(\\{ x_{m,k} \\}_{m=1}^\\infty\\) is Cauchy. As \\(\\R\\) is complete, the sequence converges; there exists a \\(y_k \\in \\R\\) such that \\(y_k = \\lim_{m\\to\\infty} x_{m,k}\\text{.}\\) Write \\(y = (y_1,y_2,\\ldots,y_n) \\in \\R^n\\text{.}\\) By Proposition 7.3.9, \\(\\{ x_m \\}_{m=1}^\\infty\\) converges to \\(y \\in \\R^n\\text{,}\\) and hence \\(\\R^n\\) is complete.
As \\(K\\) is contained in a ball, \\(K\\) is bounded. See the left-hand side of Figure 7.11.
\n\n
\n
\n
Next, we show that a set that is not closed is not compact. Suppose \\(\\widebar{K} \\neq K\\text{,}\\) that is, there is a point \\(x \\in \\widebar{K}\r\n\\setminus K\\text{.}\\) If \\(y \\neq x\\text{,}\\) then \\(y \\notin C(x,\\nicefrac{1}{n})\\) for \\(n \\in \\N\\) such that \\(\\nicefrac{1}{n} < d(x,y)\\text{.}\\) Furthermore, \\(x \\notin K\\text{,}\\) so
A closed ball is closed, so its complement \\({C(x,\\nicefrac{1}{n})}^c\\) is open, and we have an open cover. If we take any finite collection of indices \\(n_1 < n_2 < \\ldots < n_m\\text{,}\\) then
As \\(x\\) is in the closure of \\(K\\text{,}\\) then \\(C(x,\\nicefrac{1}{n_m}) \\cap K \\neq \\emptyset\\text{.}\\) So there is no finite subcover and \\(K\\) is not compact. See the right-hand side of Figure 7.11.
\n\n
\n\n",
"licence": "CC-BY-SA-4.0",
"proof_scope": "Compact sets are closed and bounded, with the empty-set case explicit locally"
},
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"id": "L7.4.10",
"reader": "sec_metcompact.html",
"author_url": "https://www.jirka.org/ra/html/sec_metcompact.html",
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\nLemma7.4.10.Lebesgue covering lemma.\n
\n
\n Let \\((X,d)\\) be a metric space and \\(K \\subset X\\text{.}\\) Suppose every sequence in \\(K\\) has a subsequence convergent in \\(K\\text{.}\\) Given an open cover \\(\\{ U_\\lambda \\}_{\\lambda \\in I}\\) of \\(K\\text{,}\\) there exists a \\(\\delta > 0\\) such that for every \\(x \\in K\\text{,}\\) there exists a \\(\\lambda \\in I\\) with \\(B(x,\\delta) \\subset U_\\lambda\\text{.}\\)\n
\n\n\n
Proof.
\n
\n
We prove the lemma by contrapositive. If the conclusion is not true, then there is an open cover \\(\\{ U_\\lambda \\}_{\\lambda \\in I}\\) of \\(K\\) with the following property. For every \\(n \\in \\N\\text{,}\\) there exists an \\(x_n \\in K\\) such that \\(B(x_n,\\nicefrac{1}{n})\\) is not a subset of any \\(U_\\lambda\\text{.}\\) Take any \\(x \\in K\\text{.}\\) There is a \\(\\lambda \\in I\\) such that \\(x \\in U_\\lambda\\text{.}\\) As \\(U_\\lambda\\) is open, there is an \\(\\epsilon > 0\\) such that \\(B(x,\\epsilon) \\subset U_\\lambda\\text{.}\\) Take \\(M\\) such that \\(\\nicefrac{1}{M} < \\nicefrac{\\epsilon}{2}\\text{.}\\) If \\(y \\in \r\nB(x,\\nicefrac{\\epsilon}{2})\\) and \\(n \\geq M\\text{,}\\) then
where \\(B(y,\\nicefrac{\\epsilon}{2}) \\subset B(x,\\epsilon)\\) follows by triangle inequality. See Figure 7.12. Thus \\(y \\neq x_n\\text{.}\\) In other words, for all \\(n \\geq M\\text{,}\\)\\(x_n \\notin B(x,\\nicefrac{\\epsilon}{2})\\text{.}\\) The sequence cannot have a subsequence converging to \\(x\\text{.}\\) As \\(x \\in K\\) was arbitrary, we are done.
Let \\((X,d)\\) be a metric space. Then \\(K \\subset X\\) is compact if and only if every sequence in \\(K\\) has a subsequence converging to a point in \\(K\\text{.}\\)\n
\n\n\n
Proof.
\n
Claim: Let \\(K \\subset X\\) be a subset of \\(X\\) and \\(\\{ x_n \\}_{n=1}^\\infty\\) a sequence in \\(K\\text{.}\\) Suppose that for each \\(x \\in K\\text{,}\\) there is a ball \\(B(x,\\alpha_x)\\) for some \\(\\alpha_x > 0\\) such that \\(x_n \\in B(x,\\alpha_x)\\) for only finitely many \\(n \\in \\N\\text{.}\\) Then \\(K\\) is not compact.\n
Any finite collection of these balls contains at most finitely many elements of \\(\\{ x_n \\}_{n=1}^\\infty\\text{,}\\) and so there must be an \\(x_n \\in K\\) not in their union. Hence, \\(K\\) is not compact and the claim is proved.
\n\n
\n
So suppose that \\(K\\) is compact and \\(\\{ x_n \\}_{n=1}^\\infty\\) is a sequence in \\(K\\text{.}\\) Then there exists an \\(x \\in K\\) such that for all \\(\\delta > 0\\text{,}\\)\\(B(x,\\delta)\\) contains \\(x_n\\) for infinitely many \\(n \\in \\N\\text{.}\\) We define the subsequence inductively. The ball \\(B(x,1)\\) contains some \\(x_k\\text{,}\\) so let \\(n_1 \\coloneqq k\\text{.}\\) Suppose \\(n_{j-1}\\) is defined. There must exist a \\(k > n_{j-1}\\) such that \\(x_k \\in B(x,\\nicefrac{1}{j})\\text{.}\\) Define \\(n_j \\coloneqq k\\text{.}\\) We now possess a subsequence \\(\\{ x_{n_j} \\}_{j=1}^\\infty\\text{.}\\) Since \\(d(x,x_{n_j}) < \\nicefrac{1}{j}\\text{,}\\)Proposition 7.3.5 says \\(\\lim_{j\\to\\infty} x_{n_j} = x\\text{.}\\)\n
\n
For the other direction, suppose every sequence in \\(K\\) has a subsequence converging in \\(K\\text{.}\\) Take an open cover \\(\\{ U_\\lambda \\}_{\\lambda \\in I}\\) of \\(K\\text{.}\\) Using the Lebesgue covering lemma above, find a \\(\\delta > 0\\) such that for every \\(x \\in K\\text{,}\\) there is a \\(\\lambda \\in I\\) with \\(B(x,\\delta) \\subset U_\\lambda\\text{.}\\)\n
\n
Pick \\(x_1 \\in K\\) and find \\(\\lambda_1 \\in I\\) such that \\(B(x_1,\\delta) \\subset\r\nU_{\\lambda_1}\\text{.}\\) If \\(K \\subset U_{\\lambda_1}\\text{,}\\) we stop as we have found a finite subcover. Otherwise, there must be a point \\(x_2 \\in K \\setminus U_{\\lambda_1}\\text{.}\\) Note that \\(d(x_2,x_1) \\geq \\delta\\text{.}\\) There must exist some \\(\\lambda_2 \\in I\\) such that \\(B(x_2,\\delta) \\subset U_{\\lambda_2}\\text{.}\\) We work inductively. Suppose \\(\\lambda_{n-1}\\) is defined. Either \\(U_{\\lambda_1} \\cup\r\nU_{\\lambda_2} \\cup \\cdots \\cup\r\nU_{\\lambda_{n-1}}\\) is a finite cover of \\(K\\text{,}\\) in which case we stop, or there must be a point \\(x_n \\in K \\setminus \\bigl( U_{\\lambda_1} \\cup\r\nU_{\\lambda_2} \\cup \\cdots \\cup\r\nU_{\\lambda_{n-1}}\\bigr)\\text{.}\\) Note that \\(d(x_n,x_j) \\geq \\delta\\) for all \\(j = 1,2,\\ldots,n-1\\text{.}\\) Next, there must be some \\(\\lambda_n \\in I\\) such that \\(B(x_n,\\delta) \\subset U_{\\lambda_n}\\text{.}\\) See Figure 7.13.\n
\n\nFigure7.13.Covering \\(K\\) by \\(U_{\\lambda}\\text{.}\\) The points \\(x_1,x_2,x_3,x_4\\text{,}\\) the three sets \\(U_{\\lambda_1}\\text{,}\\)\\(U_{\\lambda_2}\\text{,}\\)\\(U_{\\lambda_3}\\text{,}\\) and the first three balls of radius \\(\\delta\\) are drawn.\n\n
Either at some point we obtain a finite subcover of \\(K\\text{,}\\) or we obtain an infinite sequence \\(\\{ x_n \\}_{n=1}^\\infty\\) as above. For contradiction, suppose that there is no finite subcover and we have the sequence \\(\\{ x_n \\}_{n=1}^\\infty\\text{.}\\) For all \\(n\\) and \\(k\\text{,}\\)\\(n \\neq k\\text{,}\\) we have \\(d(x_n,x_k) \\geq \\delta\\text{.}\\) So no subsequence of \\(\\{ x_n \\}_{n=1}^\\infty\\) is Cauchy. Hence, no subsequence of \\(\\{ x_n \\}_{n=1}^\\infty\\) is convergent, which is a contradiction.\n
\n\n",
"licence": "CC-BY-SA-4.0",
"proof_scope": "Sequential compactness and the finite-subcover definition are equivalent"
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"id": "FIGURE-fig_compactbndclosed",
"reader": "sec_metcompact.html",
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"html": "Figure7.11.Proving compact set is bounded (left) and closed (right).\n",
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"id": "FIGURE-fig_lebesguedelta",
"reader": "sec_metcompact.html",
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"source_page_sha256": "eff46c68c5e77865104ac9c33ebdbf25da45b5d9c97dd91f88177665177eb95b",
"html": "Figure7.12.Proof of Lebesgue covering lemma. Note that \\(B(y,\\nicefrac{\\epsilon}{2}) \\subset\r\nB(x,\\epsilon)\\) by triangle inequality.\n",
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},
{
"id": "L7.5.2",
"reader": "sec_metcont.html",
"author_url": "https://www.jirka.org/ra/html/sec_metcont.html",
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\nProposition7.5.2.\n
\n
Let \\((X,d_X)\\) and \\((Y,d_Y)\\) be metric spaces. Then \\(f \\colon X \\to Y\\) is continuous at \\(c \\in X\\) if and only if for every sequence \\(\\{ x_n \\}_{n=1}^\\infty\\) in \\(X\\) converging to \\(c\\text{,}\\) the sequence \\(\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty\\) converges to \\(f(c)\\text{.}\\)\n
\n\n\n
Proof.
\n
Suppose \\(f\\) is continuous at \\(c\\text{.}\\) Let \\(\\{ x_n \\}_{n=1}^\\infty\\) be a sequence in \\(X\\) converging to \\(c\\text{.}\\) Given \\(\\epsilon > 0\\text{,}\\) there is a \\(\\delta > 0\\) such that \\(d_X(x,c) < \\delta\\) implies \\(d_Y\\bigl(f(x),f(c)\\bigr) < \\epsilon\\text{.}\\) So take \\(M\\) such that for all \\(n \\geq M\\text{,}\\) we have \\(d_X(x_n,c) < \\delta\\text{,}\\) and then \\(d_Y\\bigl(f(x_n),f(c)\\bigr) < \\epsilon\\text{.}\\) Hence, \\(\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty\\) converges to \\(f(c)\\text{.}\\)\n
\n
On the other hand, suppose \\(f\\) is not continuous at \\(c\\text{.}\\) Then there exists an \\(\\epsilon > 0\\) such that for every \\(n \\in \\N\\) there exists an \\(x_n \\in X\\text{,}\\) with \\(d_X(x_n,c) < \\nicefrac{1}{n}\\) such that \\(d_Y\\bigl(f(x_n),f(c)\\bigr) \\geq\r\n\\epsilon\\text{.}\\) Then \\(\\{ x_n \\}_{n=1}^\\infty\\) converges to \\(c\\text{,}\\) but \\(\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty\\) does not converge to \\(f(c)\\text{.}\\)\n
Let \\((X,d_X)\\) and \\((Y,d_Y)\\) be metric spaces and \\(f \\colon X \\to Y\\) a continuous function. If \\(K \\subset X\\) is a compact set, then \\(f(K)\\) is a compact set.\n
\n\n\n
Proof.
\n
\n
A sequence in \\(f(K)\\) can be written as \\(\\bigl\\{ f(x_n) \\bigr\\}_{n=1}^\\infty\\text{,}\\) where \\(\\{ x_n \\}_{n=1}^\\infty\\) is a sequence in \\(K\\text{.}\\) The set \\(K\\) is compact and therefore there is a subsequence \\(\\{ x_{n_j} \\}_{j=1}^\\infty\\) that converges to some \\(x \\in K\\text{.}\\) By continuity,
So every sequence in \\(f(K)\\) has a subsequence convergent to a point in \\(f(K)\\text{,}\\) and \\(f(K)\\) is compact by Theorem 7.4.11.
\n\n
\n\n",
"licence": "CC-BY-SA-4.0",
"proof_scope": "Continuous image of a compact set is compact"
},
{
"id": "L7.5.6",
"reader": "sec_metcont.html",
"author_url": "https://www.jirka.org/ra/html/sec_metcont.html",
"source_page_sha256": "30b3fcc05eec06699be2cb0764d6b2184b1efded82b88417a4e69c74100215c0",
"html": "
\nTheorem7.5.6.\n
\n
Let \\((X,d)\\) be a nonempty compact metric space and let \\(f \\colon X \\to \\R\\) be continuous. Then \\(f\\) is bounded and in fact \\(f\\) achieves an absolute minimum and an absolute maximum on \\(X\\text{.}\\)\n
\n\n\n
Proof.
\n
As \\(X\\) is compact and \\(f\\) is continuous, \\(f(X) \\subset \\R\\) is compact. Hence, \\(f(X)\\) is closed and bounded. In particular, \\(\\sup f(X) \\in f(X)\\) and \\(\\inf f(X) \\in f(X)\\text{,}\\) because both the sup and the inf can be achieved by sequences in \\(f(X)\\) and \\(f(X)\\) is closed. Therefore, there is some \\(x \\in X\\) such that \\(f(x) = \\sup f(X)\\) and some \\(y \\in X\\) such that \\(f(y) = \\inf f(X)\\text{.}\\)\n
\n\n",
"licence": "CC-BY-SA-4.0",
"proof_scope": "Continuous real functions on nonempty compact spaces attain both extrema"
},
{
"id": "L7.5.11",
"reader": "sec_metcont.html",
"author_url": "https://www.jirka.org/ra/html/sec_metcont.html",
"source_page_sha256": "30b3fcc05eec06699be2cb0764d6b2184b1efded82b88417a4e69c74100215c0",
"html": "
\nTheorem7.5.11.\n
\n
Let \\((X,d_X)\\) and \\((Y,d_Y)\\) be metric spaces. Suppose \\(f \\colon X \\to Y\\) is continuous and \\(X\\) is compact. Then \\(f\\) is uniformly continuous.\n
\n\n\n
Proof.
\n
Let \\(\\epsilon > 0\\) be given. For each \\(c \\in X\\text{,}\\) pick \\(\\delta_c > 0\\) such that \\(d_Y\\bigl(f(x),f(c)\\bigr) < \\nicefrac{\\epsilon}{2}\\) whenever \\(x \\in B(c,\\delta_c)\\text{.}\\) The balls \\(B(c,\\delta_c)\\) cover \\(X\\text{,}\\) and the space \\(X\\) is compact. Apply the Lebesgue covering lemma to obtain a \\(\\delta > 0\\) such that for every \\(x \\in X\\text{,}\\) there is a \\(c \\in X\\) for which \\(B(x,\\delta) \\subset B(c,\\delta_c)\\text{.}\\)\n
\n
\n
Suppose \\(p, q \\in X\\) where \\(d_X(p,q) < \\delta\\text{.}\\) Find a \\(c \\in X\\) such that \\(B(p,\\delta) \\subset B(c,\\delta_c)\\text{.}\\) Then \\(q \\in B(c,\\delta_c)\\text{.}\\) By the triangle inequality and the definition of \\(\\delta_c\\text{,}\\)\n
Fix \\(y \\in [c,d]\\) and let \\(\\epsilon > 0\\) be given. As \\(f\\) is continuous on \\([a,b] \\times [c,d]\\text{,}\\) which is compact, \\(f\\) is uniformly continuous. In particular, there exists a \\(\\delta > 0\\) such that whenever \\(z \\in [c,d]\\) and \\(\\sabs{z-y} < \\delta\\text{,}\\) we have \\(\\babs{f(x,z)-f(x,y)} < \\frac{\\epsilon}{b-a}\\) for all \\(x \\in [a,b]\\text{.}\\) So suppose \\(\\sabs{z-y} < \\delta\\text{.}\\) Then
\n\n",
"licence": "CC-BY-SA-4.0",
"proof_scope": "Continuity of a compact integral in one parameter."
},
{
"id": "L7.6.2",
"reader": "sec_metpicard.html",
"author_url": "https://www.jirka.org/ra/html/sec_metpicard.html",
"source_page_sha256": "a215f3c653d26eb05da1216407f730a3c437ce2da52e55580149f281cdbdd7e9",
"html": "
\nTheorem7.6.2.Contraction mapping principle or Banach fixed point theorem.\n
\n
\n Let \\((X,d)\\) be a nonempty complete metric space and \\(\\varphi \\colon X \\to X\\) a contraction. Then \\(\\varphi\\) has a unique fixed point.\n
\n\n\n
Proof.
\n
\n
Pick \\(x_0 \\in X\\text{.}\\) Define a sequence \\(\\{ x_n \\}_{n=1}^\\infty\\) by \\(x_{n+1} \\coloneqq \\varphi(x_n)\\text{.}\\) Then
\n
\r\n\\begin{equation}\r\nd(x_{n+1},x_n) = d\\bigl(\\varphi(x_n),\\varphi(x_{n-1})\\bigr)\r\n\\leq k d(x_n,x_{n-1}) .\r\n\\end{equation}\r\n
\n
Repeating \\(n\\) times, we get \\(d(x_{n+1},x_n) \\leq k^n d(x_1,x_0)\\text{.}\\) For \\(m > n\\text{,}\\)\n
In particular, the sequence is Cauchy (why?). Since \\(X\\) is complete, we let \\(x \\coloneqq \\lim_{n\\to\\infty} x_n\\text{,}\\) and we claim that \\(x\\) is our unique fixed point.
\n\n
\n
\n
Fixed point? The function \\(\\varphi\\) is a contraction, so it is Lipschitz continuous:
For \\(n=1\\text{,}\\) the real line, this metric agrees with what we defined above. For \\(n > 1\\text{,}\\) the only tricky part of the definition to check, as before, is the triangle inequality. It is less messy to work with the square of the metric. In the following estimate, note the use of the Cauchy–Schwarz inequality.
Because the square root is an increasing function, the inequality is preserved when we take the square root of both sides, and we obtain the triangle inequality.
\n\n
\n\n",
"licence": "CC-BY-SA-4.0",
"proof_scope": "Euclidean distance satisfies triangle inequality; positivity and symmetry follow from its displayed sum of squares"
},
{
"id": "L7.3.9",
"reader": "sec_metseqs.html",
"author_url": "https://www.jirka.org/ra/html/sec_metseqs.html",
"source_page_sha256": "3e69386e7818abc32cce85ebe4ef2f2bae24a2e5863c034314c0657e1a9abbb3",
"html": "
\nProposition7.3.9.\n
\n
\n
Let \\(\\{ x_m \\}_{m=1}^\\infty\\) be a sequence in \\(\\R^n\\text{,}\\) where \\(x_m = \\bigl(x_{m,1},x_{m,2},\\ldots,x_{m,n}\\bigr) \\in \\R^n\\text{.}\\) Then \\(\\{ x_m \\}_{m=1}^\\infty\\) converges if and only if \\(\\{ x_{m,k} \\}_{m=1}^\\infty\\) converges for every \\(k=1,2,\\ldots,n\\text{,}\\) in which case
Suppose \\(\\{ x_m \\}_{m=1}^\\infty\\) converges to \\(y = (y_1,y_2,\\ldots,y_n) \\in \\R^n\\text{.}\\) Given \\(\\epsilon > 0\\text{,}\\) there exists an \\(M\\) such that for all \\(m \\geq M\\text{,}\\) we have
Hence, the sequence \\(\\{ x_{m,k} \\}_{m=1}^\\infty\\) converges to \\(y_k\\text{.}\\)\n
\n\n
\n
\n
For the other direction, suppose \\(\\{ x_{m,k} \\}_{m=1}^\\infty\\) converges to \\(y_k\\) for every \\(k=1,2,\\ldots,n\\text{.}\\) Given \\(\\epsilon > 0\\text{,}\\) pick an \\(M\\) such that if \\(m \\geq M\\text{,}\\) then \\(\\bigl\\lvert y_k-x_{m,k} \\bigr\\rvert < \\nicefrac{\\epsilon}{\\sqrt{n}}\\) for all \\(k=1,2,\\ldots,n\\text{.}\\) Then
That is, the sequence \\(\\{ x_m \\}_{m=1}^\\infty\\) converges to \\(y = (y_1,y_2,\\ldots,y_n) \\in \\R^n\\text{.}\\)\n
\n\n
\n\n",
"licence": "CC-BY-SA-4.0",
"proof_scope": "Coordinatewise and Euclidean convergence agree for every finite n"
},
{
"id": "L7.3.11",
"reader": "sec_metseqs.html",
"author_url": "https://www.jirka.org/ra/html/sec_metseqs.html",
"source_page_sha256": "3e69386e7818abc32cce85ebe4ef2f2bae24a2e5863c034314c0657e1a9abbb3",
"html": "
\nProposition7.3.11.\n
\n
Let \\((X,d)\\) be a metric space and \\(\\{x_n\\}_{n=1}^\\infty\\) a sequence in \\(X\\text{.}\\) Then \\(\\{ x_n \\}_{n=1}^\\infty\\) converges to \\(p \\in X\\) if and only if for every open neighborhood \\(U\\) of \\(p\\text{,}\\) there exists an \\(M \\in \\N\\) such that for all \\(n \\geq M\\text{,}\\) we have \\(x_n \\in U\\text{.}\\)\n
\n\n\n
Proof.
\n
Suppose \\(\\{ x_n \\}_{n=1}^\\infty\\) converges to \\(p\\text{.}\\) Let \\(U\\) be an open neighborhood of \\(p\\text{.}\\) There exists an \\(\\epsilon > 0\\) such that \\(B(p,\\epsilon) \\subset\r\nU\\text{.}\\) As the sequence converges, find an \\(M \\in \\N\\) such that for all \\(n \\geq\r\nM\\text{,}\\) we have \\(d(p,x_n) < \\epsilon\\text{,}\\) or in other words \\(x_n \\in B(p,\\epsilon)\r\n\\subset U\\text{.}\\)\n
\n
Let us prove the other direction. Given \\(\\epsilon > 0\\text{,}\\) let \\(U \\coloneqq\r\nB(p,\\epsilon)\\) be the neighborhood of \\(p\\text{.}\\) Then there is an \\(M \\in \\N\\) such that for \\(n \\geq M\\text{,}\\) we have \\(x_n \\in U = B(p,\\epsilon)\\text{,}\\) or in other words, \\(d(p,x_n) < \\epsilon\\text{.}\\)\n
Let \\((X,d)\\) be a metric space, \\(E \\subset X\\) a closed set, and \\(\\{ x_n \\}_{n=1}^\\infty\\) a sequence in \\(E\\) that converges to some \\(p \\in X\\text{.}\\) Then \\(p \\in E\\text{.}\\)\n
\n\n\n
Proof.
\n
Let us prove the contrapositive. Suppose \\(\\{ x_n \\}_{n=1}^\\infty\\) is a sequence in \\(X\\) that converges to \\(p \\in E^c\\text{.}\\) As \\(E^c\\) is open, Proposition 7.3.11 says that there is an \\(M\\) such that for all \\(n \\geq M\\text{,}\\)\\(x_n \\in E^c\\text{.}\\) So \\(\\{ x_n \\}_{n=1}^\\infty\\) is not a sequence in \\(E\\text{.}\\)\n
is also open. That is, a union of open sets is open.
\n\n
\n\n
\n\n\n\n
Proof.
\n
The sets \\(\\emptyset\\) and \\(X\\) are obviously open in \\(X\\text{.}\\)\n
\n
Let us prove ii. If \\(x \\in \\bigcap_{j=1}^k V_j\\text{,}\\) then \\(x \\in V_j\\) for all \\(j\\text{.}\\) As \\(V_j\\) are all open, for every \\(j\\) there exists a \\(\\delta_j > 0\\) such that \\(B(x,\\delta_j) \\subset V_j\\text{.}\\) Take \\(\\delta \\coloneqq \\min \\{\r\n\\delta_1,\\delta_2,\\ldots,\\delta_k \\}\\) and notice \\(\\delta > 0\\text{.}\\) We have \\(B(x,\\delta) \\subset B(x,\\delta_j) \\subset V_j\\) for every \\(j\\) and so \\(B(x,\\delta) \\subset \\bigcap_{j=1}^k V_j\\text{.}\\) Consequently the intersection is open.\n
\n
Let us prove iii. If \\(x \\in \\bigcup_{\\lambda \\in I} V_\\lambda\\text{,}\\) then \\(x \\in V_\\lambda\\) for some \\(\\lambda \\in I\\text{.}\\) As \\(V_\\lambda\\) is open, there exists a \\(\\delta > 0\\) such that \\(B(x,\\delta) \\subset V_\\lambda\\text{.}\\) But then \\(B(x,\\delta) \\subset \\bigcup_{\\lambda \\in I} V_\\lambda\\text{,}\\) and so the union is open.\n
Let \\((X,d)\\) be a metric space, \\(x \\in X\\text{,}\\) and \\(\\delta > 0\\text{.}\\) Then \\(B(x,\\delta)\\) is open and \\(C(x,\\delta)\\) is closed.\n
\n\n\n
Proof.
\n
\n
Let \\(y \\in B(x,\\delta)\\text{.}\\) Let \\(\\alpha \\coloneqq \\delta-d(x,y)\\text{.}\\) As \\(\\alpha\r\n> 0\\text{,}\\) consider \\(z \\in B(y,\\alpha)\\text{.}\\) Then
Therefore, \\(z \\in B(x,\\delta)\\) for every \\(z \\in B(y,\\alpha)\\text{.}\\) So \\(B(y,\\alpha) \\subset B(x,\\delta)\\text{,}\\) and so \\(B(x,\\delta)\\) is open. See Figure 7.6.
\n\n
\n\nFigure7.6.Proof that \\(B(x,\\delta)\\) is open: \\(B(y,\\alpha) \\subset\r\nB(x,\\delta)\\) with the triangle inequality illustrated.\n\n
The proof that \\(C(x,\\delta)\\) is closed is left as an exercise.\n
Let \\((X,d)\\) be a metric space and \\(A \\subset X\\text{.}\\) The closure \\(\\widebar{A}\\) is closed, and \\(A \\subset \\widebar{A}\\text{.}\\) Furthermore, if \\(A\\) is closed, then \\(\\widebar{A} = A\\text{.}\\)\n
\n\n\n
Proof.
\n
The closure is an intersection of closed sets, so \\(\\widebar{A}\\) is closed. There is at least one closed set containing \\(A\\text{,}\\) namely \\(X\\) itself, so \\(A \\subset \\widebar{A}\\text{.}\\) If \\(A\\) is closed, then \\(A\\) is a closed set that contains \\(A\\text{.}\\) So \\(\\widebar{A} \\subset A\\text{,}\\) and thus \\(A = \\widebar{A}\\text{.}\\)\n
\n\n",
"licence": "CC-BY-SA-4.0",
"proof_scope": "Closure is closed and contains the set"
},
{
"id": "L7.2.22",
"reader": "sec_mettop.html",
"author_url": "https://www.jirka.org/ra/html/sec_mettop.html",
"source_page_sha256": "ca94b6b1f75aa0b6f627703e67fbf5a4c3b74d3b1a1a3de367831a86f41f3d46",
"html": "
\nProposition7.2.22.\n
\n
Let \\((X,d)\\) be a metric space and \\(A \\subset X\\text{.}\\) Then \\(x \\in \\widebar{A}\\) if and only if for every \\(\\delta > 0\\text{,}\\)\\(B(x,\\delta) \\cap A \\neq \\emptyset\\text{.}\\)\n
\n\n\n
Proof.
\n
Let us prove the two contrapositives. Let us show that \\(x \\notin \\widebar{A}\\) if and only if there exists a \\(\\delta > 0\\) such that \\(B(x,\\delta) \\cap A = \\emptyset\\text{.}\\)\n
\n
First suppose \\(x \\notin \\widebar{A}\\text{.}\\) We know \\(\\widebar{A}\\) is closed. Thus there is a \\(\\delta > 0\\) such that \\(B(x,\\delta) \\subset \\widebar{A}^c\\text{.}\\) As \\(A \\subset \\widebar{A}\\) we see that \\(B(x,\\delta) \\subset \\widebar{A}^c \\subset A^c\\) and hence \\(B(x,\\delta) \\cap A = \\emptyset\\text{.}\\)\n
\n
On the other hand, suppose there is a \\(\\delta > 0\\) such that \\(B(x,\\delta) \\cap A = \\emptyset\\text{.}\\) In other words, \\(A \\subset {B(x,\\delta)}^c\\text{.}\\) As \\({B(x,\\delta)}^c\\) is a closed set, as \\(x \\not \\in {B(x,\\delta)}^c\\text{,}\\) and as \\(\\widebar{A}\\) is the intersection of closed sets containing \\(A\\text{,}\\) we have \\(x \\notin \\widebar{A}\\text{.}\\)\n
\n\n",
"licence": "CC-BY-SA-4.0",
"proof_scope": "Every ball about a point of the closure meets the set"
},
{
"id": "L3.3.7",
"reader": "sec_minmaxint.html",
"author_url": "https://www.jirka.org/ra/html/sec_minmaxint.html",
"source_page_sha256": "9bb6af8b932032834ece8dc7fa4ad5462ab69179603dd49373b3441aada4460b",
"html": "
\nLemma3.3.7.\n
\n
Let \\(f \\colon [a,b] \\to \\R\\) be a continuous function. Suppose \\(f(a) < 0\\) and \\(f(b) > 0\\text{.}\\) Then there exists a number \\(c \\in (a,b)\\) such that \\(f(c) = 0\\text{.}\\)\n
\n\n\n
Proof.
\n
We define two sequences \\(\\{ a_n \\}_{n=1}^\\infty\\) and \\(\\{ b_n \\}_{n=1}^\\infty\\) inductively:\n
\n\n
\n
Let \\(a_1 \\coloneqq a\\) and \\(b_1 \\coloneqq b\\text{.}\\)\n
\n\n
\n
\n
If \\(f\\left(\\frac{a_n+b_n}{2}\\right) \\geq 0\\text{,}\\) let \\(a_{n+1} \\coloneqq a_n\\) and \\(b_{n+1} \\coloneqq \\frac{a_n+b_n}{2}\\text{.}\\)\n
\n\n
\n
\n
If \\(f\\left(\\frac{a_n+b_n}{2}\\right) < 0\\text{,}\\) let \\(a_{n+1} \\coloneqq \\frac{a_n+b_n}{2}\\) and \\(b_{n+1} \\coloneqq b_n\\text{.}\\)\n
See Figure 3.7 for an example of the first five steps. If \\(a_n < b_n\\text{,}\\) then \\(a_n < \\frac{a_n+b_n}{2} < b_n\\text{.}\\) So \\(a_{n+1} < b_{n+1}\\text{.}\\) As \\(a_1 = a < b = b_1\\text{,}\\)induction gives that \\(a_n < b_n\\) for all \\(n\\text{.}\\) Furthermore, \\(a_n \\leq a_{n+1}\\) and \\(b_n \\geq b_{n+1}\\) for all \\(n\\text{,}\\) that is, the sequences are monotone. As \\(a_n < b_n \\leq b_1 = b\\) and \\(b_n > a_n \\geq a_1 = a\\) for all \\(n\\text{,}\\) the sequences are also bounded. Therefore, the sequences converge. Let \\(c \\coloneqq \\lim_{n\\to\\infty} a_n\\) and \\(d \\coloneqq \\lim_{n\\to\\infty} b_n\\text{,}\\) where also \\(a \\leq c \\leq d \\leq b\\text{.}\\) We need to show that \\(c=d\\text{.}\\) Notice
Since \\(\\lim_{n\\to\\infty} a_n = \\lim_{n\\to\\infty} b_n = c\\) and \\(f\\) is continuous at \\(c\\text{,}\\) we may take limits in those inequalities:
As \\(f(c) \\geq 0\\) and \\(f(c) \\leq 0\\text{,}\\) we conclude \\(f(c) = 0\\text{.}\\) Thus also \\(c \\neq a\\) and \\(c \\neq b\\text{,}\\) so \\(a < c < b\\text{.}\\)\n
\n\n
\n\n",
"licence": "CC-BY-SA-4.0",
"proof_scope": "Complete bisection proof of a zero between strictly signed endpoints."
},
{
"id": "L3.3.8",
"reader": "sec_minmaxint.html",
"author_url": "https://www.jirka.org/ra/html/sec_minmaxint.html",
"source_page_sha256": "9bb6af8b932032834ece8dc7fa4ad5462ab69179603dd49373b3441aada4460b",
"html": "
\nTheorem3.3.8.Bolzano’s intermediate value theorem.\n
\n
Let \\(f \\colon [a,b] \\to \\R\\) be a continuous function. Suppose \\(y \\in \\R\\) is such that \\(f(a) < y < f(b)\\) or \\(f(a) > y > f(b)\\text{.}\\) Then there exists a \\(c \\in (a,b)\\) such that \\(f(c) = y\\text{.}\\)\n
\n\n\n
Proof.
\n
If \\(f(a) < y < f(b)\\text{,}\\) then define \\(g(x) \\coloneqq f(x)-y\\text{.}\\) Then \\(g(a) < 0\\) and \\(g(b) > 0\\text{,}\\) and we apply Lemma 3.3.7 to \\(g\\) to find \\(c\\text{.}\\) If \\(g(c) = 0\\text{,}\\) then \\(f(c) = y\\text{.}\\)\n
\n
Similarly, if \\(f(a) > y > f(b)\\text{,}\\) then define \\(g(x) \\coloneqq y-f(x)\\text{.}\\) Again, \\(g(a) < 0\\) and \\(g(b) > 0\\text{,}\\) and we apply Lemma 3.3.7 to find \\(c\\text{.}\\) As before, if \\(g(c) = 0\\text{,}\\) then \\(f(c) = y\\text{.}\\)\n
\n\n",
"licence": "CC-BY-SA-4.0",
"proof_scope": "Both strict orientations of the intermediate value theorem."
},
{
"id": "L2.6.5",
"reader": "sec_moreonseries.html",
"author_url": "https://www.jirka.org/ra/html/sec_moreonseries.html",
"source_page_sha256": "25a4e33ae0e02ead2786e85912828edaf9a2b208fed4965e840717139b642f4a",
"html": "
\nTheorem2.6.5.Mertens’ theorem.\n
\n
\n
\n Suppose \\(\\sum_{n=0}^\\infty a_n\\) and \\(\\sum_{n=0}^\\infty b_n\\) are two convergent series, converging to \\(A\\) and \\(B\\text{,}\\) respectively. Suppose at least one of the series converges absolutely. Define
Then the series \\(\\sum_{n=0}^\\infty c_n\\) converges to \\(AB\\text{.}\\)\n
\n\n
\n\n\n
Proof.
\n
Suppose \\(\\sum_{n=0}^\\infty a_n\\) converges absolutely, and let \\(\\epsilon > 0\\) be given. In this proof, instead of picking complicated estimates just to make the final estimate come out as less than \\(\\epsilon\\text{,}\\) let us simply obtain an estimate that depends on \\(\\epsilon\\) and can be made arbitrarily small.\n
We can surely make the second term on the right-hand side small. The trick is to handle the first term. Pick \\(K\\) such that for all \\(m \\geq K\\text{,}\\) we have \\(\\sabs{A_m - A} < \\epsilon\\) and also \\(\\sabs{B_m - B} < \\epsilon\\text{.}\\) As \\(\\sum_{n=0}^\\infty a_n\\) converges absolutely, there is a \\(K\\) large enough such that for all \\(m \\geq K\\text{,}\\)\n
As \\(\\sum_{n=0}^\\infty b_n\\) converges, \\(B_{\\text{max}} \\coloneqq \\sup \\bigl\\{ \\sabs{ B_n - B } : n = 0,1,2,\\ldots\r\n\\bigr\\}\\) is finite. Take \\(m \\geq 2K\\text{.}\\) In particular, \\(m-K+1 > K\\text{.}\\) So
\n
\r\n\\begin{equation}\r\n\\begin{split}\r\n\\sum_{n=0}^m\r\n\\sabs{ B_n - B } \\sabs{a_{m-n}}\r\n& =\r\n\\left(\r\n\\sum_{n=0}^{m-K}\r\n\\sabs{ B_n - B } \\sabs{a_{m-n}}\r\n\\right)\r\n+\r\n\\left(\r\n\\sum_{n=m-K+1}^m\r\n\\sabs{ B_n - B } \\sabs{a_{m-n}}\r\n\\right)\r\n\\\\\r\n& \\leq\r\n\\left(\r\n\\sum_{n=K}^m\r\n\\sabs{a_{n}}\r\n\\right)\r\nB_{\\text{max}}\r\n+\r\n\\left(\r\n\\sum_{n=0}^{K-1}\r\n\\epsilon \\sabs{a_{n}}\r\n\\right)\r\n\\\\\r\n& \\leq\r\n\\epsilon\r\nB_{\\text{max}}\r\n+\r\n\\epsilon\r\n\\left(\r\n\\sum_{n=0}^\\infty \\sabs{a_{n}}\r\n\\right) .\r\n\\end{split}\r\n\\end{equation}\r\n
The expression in the parentheses on the right-hand side is a fixed number. Hence, we can make the right-hand side arbitrarily small by picking a small enough \\(\\epsilon> 0\\text{.}\\) So \\(\\sum_{n=0}^\\infty c_n\\) converges to \\(AB\\text{.}\\)\n
\n\n
\n\n",
"licence": "CC-BY-SA-4.0",
"proof_scope": "Full written Mertens proof; real comparisons only on moduli, with the complex extension stated explicitly in P13.2."
},
{
"id": "L10.7.2",
"reader": "sec_mvchangeofvars.html",
"author_url": "https://www.jirka.org/ra/html/sec_mvchangeofvars.html",
"source_page_sha256": "ffdb891e94888fd91da33334e29fbada7376a5f42da566c155c77cb795879f60",
"html": "
\nTheorem10.7.2.\n
\n
\n
Suppose \\(U \\subset \\R^n\\) is open, \\(S \\subset U\\) is a compact Jordan measurable set, and \\(g \\colon U \\to \\R^n\\) is a one-to-one continuously differentiable mapping, such that \\(J_g\\) is never zero on \\(S\\text{.}\\) Suppose \\(f \\colon g(S) \\to \\R\\) is Riemann integrable. Then \\(f \\circ g\\) is Riemann integrable on \\(S\\) and
The set \\(S\\) can be covered by finitely many closed rectangles \\(P_1,P_2,\\ldots,P_k\\text{,}\\) whose interiors do not overlap such that each \\(P_j \\subset U\\) (Exercise 10.7.2). Proving the theorem for \\(P_j \\cap S\\) instead of \\(S\\) is enough as we can simply add up the integrals. Define \\(f(y) \\coloneqq 0\\) for all \\(y \\notin g(S)\\text{.}\\) The new \\(f\\) is Riemann integrable as \\(g(S)\\) is Jordan measurable. We can now replace the integrals over \\(S\\) with integrals over the whole rectangle. We therefore assume without loss of generality that \\(S\\) is a closed rectangle.\n
\n
The matrix \\(g'(x)\\) is invertible for every \\(x \\in S\\text{,}\\) and it is continuous. Therefore, \\({\\bigl(g'(x)\\bigr)}^{-1}\\) and consequently \\(\\bnorm{{\\bigl(g'(x)\\bigr)}^{-1}}\\) is continuous and never zero. As \\(S\\) is compact, then there exists an \\(M > 1\\) so that \\(\\bnorm{{\\bigl(g'(x)\\bigr)}^{-1}} \\leq M\\) for all \\(x \\in S\\text{.}\\)\n
\n
\n
Let \\(\\epsilon > 0\\) be given. For every \\(x \\in S\\text{,}\\) let
\n
\r\n\\begin{equation}\r\nW_x \\coloneqq\r\n\\left\\{ y \\in U : \\bnorm{g'(x)-g'(y)} < \\frac{\\epsilon}{2M} \\right\\} .\r\n\\end{equation}\r\n
\n
By Exercise 10.7.3, \\(W_x\\) is open. As \\(x \\in W_x\\) for every \\(x\\text{,}\\) we have an open cover. By the Lebesgue covering lemma (Lemma 7.4.10), there exists a \\(\\delta > 0\\) such that for every \\(y \\in S\\text{,}\\) there is an \\(x\\) such that \\(B(y,\\delta) \\subset W_x\\text{.}\\) In other words, if \\(Q\\) is a rectangle of maximum side length less than \\(\\frac{\\delta}{\\sqrt{n}}\\) and \\(y \\in Q\\text{,}\\) then \\(Q \\subset\r\nB(y,\\delta) \\subset W_x\\text{.}\\) By the triangle inequality, \\(\\bnorm{g'(\\xi)-g'(\\eta)} < \\nicefrac{\\epsilon}{M}\\) for all \\(\\xi, \\eta \\in Q\\text{.}\\)\n
\n\n
\n
Let \\(\\varphi(x) \\coloneqq f\\bigl(g(x)\\bigr) \\babs{J_g(x)}\\text{.}\\) There exists a partition \\(P\\) of \\(S\\) such that \\(\\epsilon + \\int_S \\varphi \\geq U(P,\\varphi)\\text{.}\\) We can assume \\(\\delta\\) is sufficiently small relative to the side of each subrectangle of the partition \\(P\\) so that we can cut each such subrectangle into further subrectangles each of whose sides \\(s\\) satisfies \\(\\frac{\\delta}{2\\sqrt{n}} \\leq s \\leq \\frac{\\delta}{\\sqrt{n}}\\) (Exercise 10.7.4). Denote these subrectangles by \\(R_1,R_2,\\ldots,R_N\\text{.}\\) For each \\(j=1,2,\\dots,N\\text{,}\\) find \\(x_j \\in R_j\\) so that \\(\\sabs{J_g(x_j)} \\leq \\sabs{J_g(x)}\\) for all \\(x \\in R_j\\text{,}\\) which is possible as \\(\\sabs{J_g(x)}\\) is continuous and \\(R_j\\) is compact.\n
\n
\n
Consider some \\(R_j\\text{.}\\) First suppose \\(x_j=0\\text{,}\\)\\(g(0) = 0\\text{,}\\) and \\(g'(0) = I\\text{.}\\) We claim that \\(g(R_j)\\) is contained in a rectangle of volume at most \\(V(R_j) \\, {\\bigl(1+4\\sqrt{n} \\, \\epsilon\\bigr)}^n\\text{.}\\) Let us prove this claim. For any given \\(y \\in R_j\\text{,}\\) apply the fundamental theorem of calculus to the function \\(t \\mapsto g(ty)\\) to find \\(g(y) = \\int_0^1 g'(ty)y \\,dt\\text{.}\\) As the side of \\(R_j\\) is at most \\(\\frac{\\delta}{\\sqrt{n}}\\text{,}\\) we have \\(\\snorm{y} \\leq \\delta\\text{.}\\) We note that \\(\\bnorm{g'(x)-I} < \\epsilon\\) as \\(M > 1\\) and so
\n
\r\n\\begin{equation}\r\n\\begin{aligned}\r\n\\bnorm{g(y)-y}\r\n=\r\n\\norm{\\int_0^1 \\bigl(g'(ty) y - y\\bigr) \\,dt}\r\n&\r\n\\leq\r\n\\int_0^1 \\bnorm{g'(ty) y - y} \\,dt\r\n\\\\\r\n&\r\n\\leq\r\n\\snorm{y} \\int_0^1 \\bnorm{g'(ty) - I} \\,dt\r\n\\leq\r\n\\delta \\epsilon .\r\n\\end{aligned}\r\n\\end{equation}\r\n
\n
Therefore, \\(g(R_j) \\subset \\widetilde{R}_j\\text{,}\\) where \\(\\widetilde{R}_j\\) is a rectangle obtained from \\(R_j\\) by extending by \\(\\delta \\epsilon\\) on all sides. See Figure 10.17.
\n\n
\n\nFigure10.17.Image of \\(R_j\\) under \\(g\\) lies inside \\(\\widetilde{R}_j\\text{.}\\) A sample point \\(y \\in R_j\\) (on the boundary of \\(R_j\\) in fact) is marked and \\(g(y)\\) must lie within with a radius of \\(\\delta\\epsilon\\) (also marked).\n\n
\n
If the sides of \\(R_j\\) are \\(s_1,s_2,\\ldots,s_n\\text{,}\\) then \\(V(R_j) = s_1 s_2 \\cdots s_n\\text{.}\\) Recall \\(\\delta \\leq 2\\sqrt{n} \\, s_j\\text{.}\\) Thus,
Next, suppose \\(A \\coloneqq g'(0)\\) is not necessarily the identity. Write \\(g = A \\circ \\widetilde{g}\\) where \\(\\widetilde{g}'(0) = I\\text{.}\\) We have that \\(\\snorm{A^{-1}} \\leq M\\) and so
\n
\r\n\\begin{equation}\r\n\\bnorm{\\widetilde{g}'(x)-I} =\r\n\\bnorm{A^{-1}g'(x)-A^{-1}A} \\leq\r\n\\bnorm{A^{-1}}\\,\\bnorm{g'(x)-A} < M \\frac{\\epsilon}{M} = \\epsilon .\r\n\\end{equation}\r\n
\n
In other words, the claim above applies to \\(\\widetilde{g}\\) as what we needed in its proof was precisely that \\(\\bnorm{\\widetilde{g}'(x)-I} < \\epsilon\\text{.}\\) Therefore, we have that \\(\\widetilde{g}(R_j)\\) is contained in a rectangle \\(\\widetilde{R}_j\\) with \\(V(\\widetilde{R}_j) \\leq V(R_j) \\, {\\bigl(1+4\\sqrt{n} \\, \\epsilon\\bigr)}^n\\text{.}\\)\n
\n\n
\n
\n
By Proposition 10.7.1, \\(V\\bigl(A(\\widetilde{R}_j)\\bigr) = \\babs{\\det(A)} \\, V(\\widetilde{R}_j)\\text{,}\\) and hence
Translation does not change volume, and therefore for every \\(R_j\\text{,}\\) and \\(x_j \\in R_j\\text{,}\\) including when \\(x_j \\neq 0\\) and \\(g(x_j)\r\n\\neq 0\\text{,}\\) we find
So, if we prove the theorem for a nonnegative \\(f\\text{,}\\) we obtain the theorem for arbitrary \\(f\\text{.}\\) Therefore, suppose without loss of generality that \\(f(u) \\geq 0\\) for all \\(u \\in g(S)\\text{.}\\)\n
\n\n
\n
\n
As the rectangles \\(R_1,R_2,\\ldots,R_N\\) give a refinement of the partition \\(P\\text{,}\\)\n
The last equality follows because the overlaps of the rectangles are their boundaries, which are of measure zero, and hence the image of their boundaries is also measure zero. Let \\(\\epsilon\\) go to zero to find
\n\n",
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\nProposition8.6.2.\n
\n
\n
Suppose \\(U \\subset \\R^n\\) is open and \\(f \\colon U \\to \\R\\) is a \\(C^2\\) function, and \\(\\ell\\) and \\(m\\) are two integers from \\(1\\) to \\(n\\text{.}\\) Then
Fix a \\(p \\in U\\text{,}\\) and let \\(e_{\\ell}\\) and \\(e_m\\) be the standard basis vectors. Pick two positive numbers \\(s\\) and \\(t\\) small enough so that \\(p+s_0e_{\\ell} +t_0e_m \\in U\\) whenever \\(0 < s_0 \\leq s\\) and \\(0 < t_0 \\leq t\\text{.}\\) Any small enough \\(s\\) and \\(t\\) work as \\(U\\) is open and so contains a small open ball (or a box if you wish) around \\(p\\text{.}\\)\n
\n\nFigure8.14.Using the mean value theorem to estimate a second order partial derivative by a certain difference quotient.\n\n
\n
See Figure 8.14. The \\(s_0\\) and \\(t_0\\) depend on \\(s\\) and \\(t\\text{,}\\) but \\(0 < s_0 < s\\) and \\(0 < t_0 < t\\text{.}\\) Let the domain of the function \\(g\\) be the set \\((0,\\epsilon) \\times\r\n(0,\\epsilon)\\) for some small \\(\\epsilon > 0\\text{.}\\) As \\((s,t) \\in (0,\\epsilon) \\times (0,\\epsilon)\\) goes to \\((0,0)\\text{,}\\) the point \\((s_0,t_0)\\) also goes to \\((0,0)\\text{.}\\) By continuity of the second partial derivatives,
Now reverse the roles of \\(s\\) and \\(t\\) (and \\(\\ell\\) and \\(m\\)). Use the mean value theorem on the function \\(\\sigma \\mapsto f(p+\\sigma e_{\\ell} + te_m)-f(p + \\sigma\r\ne_{\\ell})\\) to find an \\(s_1 \\in (0,s)\\) such that
Suppose \\(f \\colon (a,b) \\to \\R\\) is differentiable at \\(c \\in (a,b)\\text{,}\\) and \\(f\\) has a relative minimum or a relative maximum at \\(c\\text{.}\\) Then \\(f'(c) = 0\\text{.}\\)\n
\n\n\n
Proof.
\n
\n
Suppose \\(c\\) is a relative maximum of \\(f\\text{.}\\) That is, there is a \\(\\delta > 0\\) such that for every \\(x \\in (a,b)\\) where \\(\\sabs{x-c} < \\delta\\text{,}\\) we have \\(f(x)-f(c) \\leq 0\\text{.}\\) Consider the difference quotient. If \\(c < x < c+\\delta\\text{,}\\) then
\n\nFigure4.3.Slopes of secants at a relative maximum.\n\n
\n
As \\(a < c < b\\text{,}\\) there exist sequences \\(\\{ x_n \\}_{n=1}^\\infty\\) and \\(\\{ y_n \\}_{n=1}^\\infty\\) in \\((a,b)\\) such that \\(c < x_n < c+\\delta\\) and \\(c-\\delta < y_n < c\\) for all \\(n \\in \\N\\text{,}\\) and such that \\(\\lim_{n\\to\\infty} x_n = \\lim_{n\\to\\infty} y_n = c\\text{.}\\) Since \\(f\\) is differentiable at \\(c\\text{,}\\)\n
Let \\(f \\colon [a,b] \\to \\R\\) be a continuous function differentiable on \\((a,b)\\) such that \\(f(a) = f(b)\\text{.}\\) Then there exists a \\(c \\in (a,b)\\) such that \\(f'(c) = 0\\text{.}\\)\n
\n\n\n
Proof.
\n
As \\(f\\) is continuous on \\([a,b]\\text{,}\\) it attains an absolute minimum and an absolute maximum in \\([a,b]\\text{.}\\) We wish to apply Lemma 4.2.2, and so we need to find some \\(c \\in (a,b)\\) where \\(f\\) attains a minimum or a maximum. Write \\(K \\coloneqq f(a) = f(b)\\text{.}\\) If there exists an \\(x\\) such that \\(f(x) > K\\text{,}\\) then the absolute maximum is larger than \\(K\\) and hence occurs at some \\(c \\in (a,b)\\text{,}\\) and therefore \\(f'(c) = 0\\text{.}\\) On the other hand, if there exists an \\(x\\) such that \\(f(x) < K\\text{,}\\) then the absolute minimum occurs at some \\(c \\in (a,b)\\text{,}\\) and so \\(f'(c) = 0\\text{.}\\) If there is no \\(x\\) such that \\(f(x) > K\\) or \\(f(x) < K\\text{,}\\) then \\(f(x) = K\\) for all \\(x\\) and then \\(f'(x) = 0\\) for all \\(x \\in [a,b]\\text{,}\\) so any \\(c \\in (a,b)\\) works.\n
Let \\(f \\colon [a,b] \\to \\R\\) be a continuous function differentiable on \\((a,b)\\text{.}\\) Then there exists a point \\(c \\in (a,b)\\) such that
The function \\(g\\) is differentiable on \\((a,b)\\text{,}\\) continuous on \\([a,b]\\text{,}\\) such that \\(g(a) = 0\\) and \\(g(b) = 0\\text{.}\\) Thus there exists a \\(c \\in (a,b)\\) such that \\(g'(c) = 0\\text{,}\\) that is,
In other words, \\(f(b)-f(a) = f'(c)(b-a)\\text{.}\\)\n
\n\n
\n\n",
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\nProposition4.2.6.\n
\n
Let \\(I\\) be an interval and let \\(f \\colon I \\to \\R\\) be a differentiable function such that \\(f'(x) = 0\\) for all \\(x \\in I\\text{.}\\) Then \\(f\\) is constant.\n
\n\n\n
Proof.
\n
\n
Take arbitrary \\(x,y \\in I\\) with \\(x < y\\text{.}\\) As \\(I\\) is an interval, \\([x,y] \\subset I\\text{.}\\) Then \\(f\\) restricted to \\([x,y]\\) satisfies the hypotheses of the mean value theorem. Therefore, there is a \\(c \\in (x,y)\\) such that
Let \\(X\\) and \\(Y\\) be normed vector spaces with \\(X\\) finite-dimensional, and let \\(A \\in L(X,Y)\\text{.}\\) Then \\(\\snorm{A} < \\infty\\text{,}\\) and \\(A\\) is uniformly continuous (Lipschitz with constant \\(\\snorm{A}\\)).\n
\n\n\n
Proof.
\n
As we said we only prove the proposition for euclidean spaces, so suppose that \\(X = \\R^n\\) and the norm is the standard euclidean norm. The general case is left as an exercise.\n
\n
\n
Let \\(\\{ e_1,e_2,\\ldots,e_n \\}\\) be the standard basis of \\(\\R^n\\text{.}\\) Write \\(x \\in \\R^n\\text{,}\\) with \\(\\snorm{x} = 1\\text{,}\\) as
Since \\(e_k \\cdot e_\\ell = 0\\) whenever \\(k \\neq \\ell\\) and \\(e_k \\cdot e_k = 1\\text{,}\\) we have \\(c_k = x \\cdot e_k\\text{.}\\) By Cauchy–Schwarz,
The right-hand side does not depend on \\(x\\text{.}\\) We found a finite upper bound for \\(\\snorm{Ax}\\) independent of \\(x\\text{,}\\) so \\(\\snorm{A} < \\infty\\text{.}\\)\n
\n\n
\n
\n
Take normed vector spaces \\(X\\) and \\(Y\\text{,}\\) and \\(A \\in L(X,Y)\\) with \\(\\snorm{A} < \\infty\\text{.}\\) For \\(v,w \\in X\\text{,}\\)\n
Thus \\(\\snorm{Bx} \\neq 0\\) for all \\(x \\neq 0\\text{,}\\) and consequently \\(Bx \\neq 0\\) for all \\(x \\neq 0\\text{.}\\) So \\(B\\) is one-to-one; if \\(Bx = By\\text{,}\\) then \\(B(x-y) = 0\\text{,}\\) so \\(x=y\\text{.}\\) As \\(B\\) is a one-to-one linear mapping from \\(X\\) to \\(X\\text{,}\\) which is finite-dimensional, it is also onto by Proposition 8.1.18. Therefore, \\(B\\) is invertible. It follows that, in particular, \\(GL(X)\\) is open.
\n\n
\n
\n
Let us prove ii. We must show that the inverse is continuous. Fix an \\(A \\in GL(X)\\text{.}\\) Let \\(B\\) be near \\(A\\text{,}\\) specifically \\(\\snorm{A-B} < \\frac{1}{2 \\snorm{A^{-1}}}\\text{.}\\) Then (8.2) is satisfied and \\(B\\) is invertible. A similar computation as above (using \\(B^{-1}y\\) instead of \\(x\\)) gives
Therefore, as \\(B\\) tends to \\(A\\text{,}\\)\\(\\snorm{B^{-1}-A^{-1}}\\) tends to 0, and so the inverse operation is a continuous function at \\(A\\text{.}\\)\n
\n\n
\n\n",
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\nSubsection8.2.2Matrices\n
\n
\n
Once we fix a basis in a finite-dimensional vector space \\(X\\text{,}\\) we can represent a vector of \\(X\\) as an \\(n\\)-tuple of numbers—a vector in \\(\\R^n\\text{.}\\) The same can be done with \\(L(X,Y)\\text{,}\\) bringing us to matrices, which are a convenient way to represent finite-dimensional linear transformations. Suppose \\(\\{ x_1, x_2, \\ldots, x_n \\}\\) and \\(\\{ y_1, y_2, \\ldots, y_m \\}\\) are bases for vector spaces \\(X\\) and \\(Y\\) respectively. A linear operator is determined by its values on the basis. Given \\(A \\in L(X,Y)\\text{,}\\)\\(A x_j\\) is an element of \\(Y\\text{.}\\) Define the numbers \\(a_{i,j}\\) via
We sometimes write \\(A\\) as \\([a_{i,j}]\\text{.}\\) We say \\(A\\) is an \\(m\\)-by-\\(n\\) matrix. The \\(j\\)th column of the matrix contains precisely the coefficients that represent \\(A x_j\\) in terms of the basis \\(\\{ y_1,y_2,\\ldots,y_m \\}\\text{.}\\) Given the numbers \\(a_{i,j}\\text{,}\\) then via the formula (8.3), we find the corresponding linear operator, as it is determined by the action on a basis. Hence, once we fix bases on \\(X\\) and \\(Y\\text{,}\\) we have a one-to-one correspondence between \\(L(X,Y)\\) and the \\(m\\)-by-\\(n\\) matrices. When
which gives rise to the familiar rule for matrix multiplication, thinking of \\(z\\) as a column vector, that is, an \\(n\\)-by-1 matrix. More generally, if \\(B\\) is an \\(n\\)-by-\\(r\\) matrix with entries \\(b_{j,k}\\text{,}\\) then the matrix for \\(C = AB\\) is an \\(m\\)-by-\\(r\\) matrix whose \\((i,k)\\)th entry \\(c_{i,k}\\) is
A way to remember it is if you order the indices as we do—row, column—and put the elements in the same order as the matrices, then the “middle index” is “summed-out.”
\n\n
\n
There is a one-to-one correspondence between matrices and linear operators in \\(L(X,Y)\\text{,}\\) once we fix bases in \\(X\\) and \\(Y\\text{.}\\) If we choose different bases, we get different matrices. This is an important distinction. The operator \\(A\\) acts on elements of \\(X\\text{,}\\) while the matrix is something that works with \\(n\\)-tuples of numbers, that is, vectors of \\(\\R^n\\text{.}\\) By convention, we use standard bases in \\(\\R^n\\) unless otherwise specified, and we identify \\(L(\\R^n,\\R^m)\\) with the set of \\(m\\)-by-\\(n\\) matrices.\n
\n
A linear mapping changing one basis to another is represented by a square matrix in which the columns represent vectors of the second basis in terms of the first basis. We call such a linear mapping a change of basis. So for two choices of a basis in an \\(n\\)-dimensional vector space, there is a linear mapping (a change of basis) taking one basis to the other, and this corresponds to an \\(n\\)-by-\\(n\\) matrix which does the corresponding operation on \\(\\R^n\\text{.}\\)\n
\n
\n
Suppose \\(X=\\R^n\\text{,}\\)\\(Y=\\R^m\\text{,}\\) and all the bases are just the standard bases. Using the Cauchy–Schwarz inequality, with \\(c=(c_1,c_2,\\ldots,c_n) \\in \\R^n\\text{,}\\) compute
The right-hand side is the euclidean norm on \\(\\R^{nm}\\text{,}\\) the space of all the entries of the matrix. If the entries go to zero, then \\(\\snorm{A}\\) goes to zero. Conversely,
\n
\r\n\\begin{equation}\r\n\\sum_{i=1}^m \\sum_{j=1}^n {(a_{i,j})}^2\r\n=\r\n\\sum_{j=1}^n \\snorm{A e_j}^2\r\n\\leq\r\n\\sum_{j=1}^n \\snorm{A}^2\r\n= n \\snorm{A}^2 .\r\n\\end{equation}\r\n
\n
So if the operator norm of \\(A\\) goes to zero, so do the entries. In particular, if \\(A\\) is fixed and \\(B\\) is changing, then the entries of \\(B\\) go to the entries of \\(A\\) if and only if \\(B\\) goes to \\(A\\) in operator norm (\\(\\snorm{A-B}\\) goes to zero). We have proved:
\n\n
\n
\nProposition8.2.7.\n
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The topology (the set of open sets) on \\(L(\\R^n,\\R^m)\\) is the same whether we consider \\(L(\\R^n,\\R^m)\\) as a metric space using the operator norm, or the euclidean metric of \\(\\R^{nm}\\text{.}\\)\n
\n
In particular, let \\(S\\) be a metric space and let \\(\\pi \\colon L(\\R^n,\\R^m) \\to \\R^{nm}\\) identify an operator with the \\(nm\\)-tuple of entries of the corresponding matrix. Then \\(f \\colon S \\to L(\\R^n,\\R^m)\\) is continuous if and only if \\(\\pi \\circ f \\colon S \\to \\R^{nm}\\) is continuous. Similarly for \\(g \\colon L(\\R^n,\\R^m) \\to S\\) and \\(g \\circ \\pi^{-1} \\colon \\R^{nm} \\to S\\text{.}\\)\n
For every \\(j=1,2,\\ldots,n\\text{,}\\) the function \\(x_j\r\n\\mapsto \\det\\bigl([x_1 ~~~ x_2 ~~~ \\cdots ~~~ x_n ]\\bigr)\\) is linear.\n
\n\n
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If two columns of a matrix are interchanged, then the determinant changes sign.\n
\n\n
\n
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If two columns of \\(A\\) are equal, then \\(\\det(A) = 0\\text{.}\\)\n
\n\n
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If a column is zero, then \\(\\det(A) = 0\\text{.}\\)\n
\n\n
\n
\n
\n\\(A \\mapsto \\det(A)\\) is a continuous function on \\(L(\\R^n)\\text{.}\\)\n
\n\n
\n
\n
\n\\(\\det\\left( \\left[\\begin{smallmatrix} a & b \\\\ c\r\n&d \\end{smallmatrix}\\right] \\right)\r\n= ad-bc\\text{,}\\) and \\(\\det \\bigl( [a] \\bigr) = a\\text{.}\\)\n
\n\n
\n\n\n\n
Proof.
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We go through the proof quickly, as you have likely seen it before. Item i is trivial. For ii, note that each term in the definition of the determinant contains exactly one factor from each column. Item iii follows as switching two columns is switching the two corresponding numbers in every element in \\(S_n\\text{.}\\) Hence, all the signs are changed. Item iv follows because if two columns are equal, and we switch them, we get the same matrix back. So item iii says the determinant must be 0. Item v follows because the product in each term in the definition includes one element from the zero column. Item vi follows as \\(\\det\\) is a polynomial in the entries of the matrix and hence continuous (as a function of the entries of the matrix). A function defined on matrices is continuous in the operator norm if and only if it is continuous as a function of the entries (Proposition 8.2.7). Finally, item vii is a direct computation.\n
If \\(A\\) and \\(B\\) are \\(n\\)-by-\\(n\\) matrices, then \\(\\det(AB) = \\det(A)\\det(B)\\text{.}\\) Furthermore, \\(A\\) is invertible if and only if \\(\\det(A) \\neq 0\\) and in this case, \\(\\det(A^{-1}) = \\frac{1}{\\det(A)}\\text{.}\\)\n
\n\n\n
Proof.
\n
\n
Let \\(b_1,b_2,\\ldots,b_n\\) be the columns of \\(B\\text{.}\\) Then
In the last equality, we sum over the elements of \\(S_n\\) instead of all \\(n\\)-tuples for integers between 1 and \\(n\\text{,}\\) because when two columns in the determinant are the same, then the determinant is zero. Reordering the columns to the original ordering obtains the sgn.
\n\n
\n
The conclusion that \\(\\det(AB) = \\det(A)\\det(B)\\) follows by recognizing that the expression in parentheses above is the determinant of \\(B\\text{.}\\) We obtain this by plugging in \\(A=I\\text{.}\\) The expression we get for the determinant of \\(B\\) has rows and columns swapped, so as a bonus, we have also just proved that the determinant of a matrix and its transpose are equal.\n
\n
\n
Let us prove the “Furthermore.” If \\(A\\) is invertible, then \\(A^{-1}A = I\\text{.}\\) Consequently \\(\\det(A^{-1})\\det(A) = \\det(A^{-1}A) = \\det(I) = 1\\text{.}\\) If \\(A\\) is not invertible, then it is not one-to-one, and so \\(A\\) takes some nonzero vector to zero. In other words, the columns of \\(A\\) are linearly dependent. Suppose
Using the definition of the determinant (there is only a single permutation \\(\\sigma\\) for which \\(\\prod_{i=1}^n b_{i,\\sigma_i}\\) is nonzero) we find \\(\\det(B) = \\gamma_1 \\neq 0\\text{.}\\) Then \\(\\det(AB) = \\det(A)\\det(B) = \\gamma_1\\det(A)\\text{.}\\) The first column of \\(AB\\) is zero, and hence \\(\\det(AB) = 0\\text{.}\\) We conclude \\(\\det(A) = 0\\text{.}\\)\n
Let \\(\\delta > 0\\) be given. A set \\(S \\subset \\R^n\\) is of measure zero if and only if for every \\(\\epsilon >\r\n0\\text{,}\\) there exists a sequence of open balls \\(\\{ B_k \\}_{k=1}^\\infty\\text{,}\\) where the radius of \\(B_k\\) is \\(r_k < \\delta\\text{,}\\) and such that
If \\(C\\) is a closed cube (rectangle with all sides equal) of side \\(s\\text{,}\\) then \\(C\\) is contained in a closed ball of radius \\(\\sqrt{n}\\, s\\) by Proposition 10.1.14 and so in an open ball of radius \\(2 \\sqrt{n}\\, s\\text{.}\\) 1 \n
A closed ball of radius \\(\\frac{\\sqrt{n}\\,s}{2}\\) could be used but that requires an argument and \\(\\sqrt{n}\\, s\\) is good enough.\r\n
\n
\n
\n
Suppose \\(R\\) is a rectangle of positive volume. Let \\(s > 0\\) be a number less than the smallest side of \\(R\\) and such that \\(2\\sqrt{n} \\, s < \\delta\\text{.}\\) If each side of \\(R\\) is an integer multiple of \\(s\\text{,}\\) then \\(R\\) is contained in a union of closed cubes \\(C_1, C_2, \\ldots, C_m\\) of side \\(s\\) such that \\(\\sum_{k=1}^m V(C_k) = V(R)\\text{.}\\) So suppose the sides of \\(R\\) are not integer multiples of \\(s\\text{.}\\) Consider a side of length \\((\\ell+\\alpha) s\\text{,}\\) for an integer \\(\\ell\\) and \\(0 \\leq \\alpha < 1\\text{.}\\) As \\(s\\) is less than the smallest side, \\(\\ell \\geq 1\\text{,}\\) and so \\((\\ell+\\alpha)s \\leq 2\\ell s\\text{.}\\) Increasing this side to \\(2\\ell s\\text{,}\\) and similarly increasing every side of \\(R\\text{,}\\) we obtain a new larger rectangle of volume at most \\(2^n\\) times larger, whose sides are multiples of \\(s\\text{.}\\) See Figure 10.7. Thus \\(R\\) is contained in a union of closed cubes \\(C_1, C_2, \\ldots, C_m\\) of side \\(s\\) such that
\n\nFigure10.7.Covering a rectangle by cubes of total size at most \\(2^n V(R)\\text{.}\\)\n\n
\n
So suppose that \\(S\\) is a null set and there exist open rectangles \\(\\{ R_j \\}_{j=1}^\\infty\\) whose union contains \\(S\\) and such that (10.2) is true. Choose closed cubes \\(\\{ C_k \\}_{k=1}^\\infty\\) with \\(C_k\\) of side \\(s_k\\) as above that cover all the rectangles \\(\\{ R_j \\}_{j=1}^\\infty\\) and so that
As \\(S \\subset\\bigcup_{j} R_j \\subset \\bigcup_{k} C_k \\subset \\bigcup_{k}\r\nB_k\\) and \\({\\bigl(4\\sqrt{n}\\bigr)}^n \\epsilon\\) can be arbitrarily small, the forward direction follows.
\n\n
\n
\n
For the other direction, suppose \\(S\\) is covered by balls \\(B_j\\) of radii \\(r_j\\text{,}\\) such that \\(\\sum_{j=1}^\\infty r_j^n < \\epsilon\\text{,}\\) as in the statement of the proposition. Each \\(B_j\\) is contained in an open cube \\(R_j\\) of side \\(2r_j\\text{.}\\) So \\(V(R_j) = {(2 r_j)}^n = 2^n r_j^n\\text{.}\\) Therefore,
where \\(S_j\\) are all measure zero sets. Let \\(\\epsilon > 0\\) be given. For each \\(j\\text{,}\\) there exists a sequence of open rectangles \\(\\{ R_{j,k} \\}_{k=1}^\\infty\\) such that
All \\(V(R_{j,k})\\) are nonnegative, so the sum over all \\(j\\) and \\(k\\) can be done by summing first over the \\(k\\) and then over the \\(j\\text{,}\\) see Exercise 2.6.15. In particular,
\n\n",
"licence": "CC-BY-SA-4.0",
"proof_scope": "Countable union theorem with the nonnegative double-series exercise completed in P19.1."
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\nProposition10.3.7.\n
\n
\n
Suppose \\(E \\subset \\R^n\\) is a compact set of measure zero. Then for every \\(\\epsilon > 0\\text{,}\\) there exist finitely many open rectangles \\(R_1,R_2,\\ldots,R_k\\) such that
Moreover, for every \\(\\epsilon > 0\\) and every \\(\\delta > 0\\text{,}\\) there exist finitely many open balls \\(B_1,B_2,\\ldots,B_{\\ell}\\) of radii \\(r_1,r_2,\\ldots,r_{\\ell} < \\delta\\) such that
By compactness, there are finitely many of these rectangles that still contain \\(E\\text{.}\\) That is, there is some \\(k\\) such that \\(E \\subset R_1 \\cup R_2 \\cup \\cdots \\cup R_k\\text{.}\\) Hence
The proof that we can choose balls instead of rectangles is left as an exercise.\n
\n\n",
"licence": "CC-BY-SA-4.0",
"proof_scope": "Finite open rectangle and small-ball covers of compact null sets; omitted ball case supplied."
},
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"id": "L10.3.10",
"reader": "sec_outermeasure.html",
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\nProposition10.3.10.\n
\n
Suppose \\(U \\subset \\R^n\\) is open and \\(f \\colon U \\to \\R^n\\) is continuously differentiable. If \\(E \\subset U\\) is a measure zero set, then \\(f(E)\\) is measure zero.\n
\n\n\n
Proof.
\n
\n
We prove the proposition for a compact \\(E\\) and leave the general case as an exercise. Suppose \\(E\\) is compact and of measure zero. First, we will replace \\(U\\) by a smaller open set to make \\(\\bnorm{f'(x)}\\) bounded. At each point \\(x \\in\r\nE\\) pick an open ball \\(B(x,r_x)\\) such that the closed ball \\(C(x,r_x) \\subset\r\nU\\text{.}\\) By compactness, we only need to take finitely many points \\(x_1,x_2,\\ldots,x_q\\) to cover \\(E\\) with the balls \\(B(x_j,r_{x_j})\\text{.}\\) Define
We have \\(E \\subset U' \\subset K \\subset U\\text{.}\\) The set \\(K\\text{,}\\) being a finite union of compact sets, is compact. The function that takes \\(x\\) to \\(\\bnorm{f'(x)}\\) is continuous, and therefore there exists an \\(M > 0\\) such that \\(\\bnorm{f'(x)} \\leq M\\) for all \\(x \\in K\\text{.}\\) So without loss of generality, we may replace \\(U\\) by \\(U'\\) and from now on suppose that \\(\\bnorm{f'(x)} \\leq M\\) for all \\(x \\in U\\text{.}\\)\n
\n\n
\n
At each \\(x \\in E\\text{,}\\) take the maximum radius \\(\\delta_x\\) such that \\(B(x,\\delta_x) \\subset U\\) (we may assume \\(U \\neq \\R^n\\)). Let \\(\\delta \\coloneqq \\inf_{x\\in E} \\delta_x\\text{.}\\) We want to show that \\(\\delta > 0\\text{.}\\) Take a sequence \\(\\{ x_j \\}_{j=1}^\\infty\\) in \\(E\\) so that \\(\\delta_{x_j} \\to \\delta\\text{.}\\) As \\(E\\) is compact, we can pick the sequence to be convergent to some \\(y \\in\r\nE\\text{.}\\) Once \\(\\snorm{x_j-y} < \\frac{\\delta_y}{2}\\text{,}\\) then \\(\\delta_{x_j} > \\frac{\\delta_y}{2}\\) by the triangle inequality. Thus, \\(\\delta > 0\\text{.}\\)\n
\n
\n
Given \\(\\epsilon > 0\\text{,}\\) there exist balls \\(B_1,B_2,\\ldots,B_k\\) of radii \\(r_1,r_2,\\ldots,r_k < \\nicefrac{\\delta}{2}\\) such that
We can assume that each ball contains a point of \\(E\\) and so the balls are contained in \\(U\\text{.}\\) Suppose \\(B_1', B_2', \\ldots, B_k'\\) are the balls of radius \\(Mr_1, Mr_2, \\ldots, Mr_k\\) from Lemma 10.3.9, such that \\(f(B_j) \\subset B_j'\\) for all \\(j\\text{.}\\) Then,
\n\n",
"licence": "CC-BY-SA-4.0",
"proof_scope": "Full null-image theorem; the omitted noncompact case and positive ball-bound convention are P19.4."
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"id": "L10.4.1",
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\nProposition10.4.1.\n
\n
A function \\(f \\colon D \\to \\R\\) is continuous at \\(x \\in D\\) if and only if \\(o(f,x) = 0\\text{.}\\)\n
\n\n\n
Proof.
\n
\n
First suppose that \\(f\\) is continuous at \\(x \\in D\\text{.}\\) Given \\(\\epsilon > 0\\text{,}\\) there exists a \\(\\delta > 0\\) such that for \\(y \\in B_D(x,\\delta)\\text{,}\\) we have \\(\\babs{f(x)-f(y)} < \\epsilon\\text{.}\\) Therefore, if \\(y_1,y_2 \\in\r\nB_D(x,\\delta)\\text{,}\\) then
As \\(o(f,x) \\leq o(f,x,\\delta) \\leq 2\\epsilon\\text{,}\\) and \\(\\epsilon > 0\\) was arbitrary, \\(o(f,x) = 0\\text{.}\\)\n
\n\n
\n
\n
On the other hand, suppose \\(o(f,x) = 0\\text{.}\\) Given \\(\\epsilon > 0\\text{,}\\) find a \\(\\delta > 0\\) such that \\(o(f,x,\\delta) < \\epsilon\\text{.}\\) If \\(y \\in B_D(x,\\delta)\\text{,}\\) then
Let \\(D \\subset \\R^n\\) be closed, \\(f \\colon D \\to \\R\\text{,}\\) and \\(\\epsilon > 0\\text{.}\\) The set \\(\\bigl\\{ x \\in D : o(f,x) \\geq \\epsilon \\bigr\\}\\) is closed.\n
\n\n\n
Proof.
\n
\n
Equivalently, we want to show that \\(G \\coloneqq \\bigl\\{ x \\in D : o(f,x) < \\epsilon \\bigr\\}\\) is open in the subspace topology. Consider \\(x \\in G\\text{.}\\) As \\(\\inf_{\\delta > 0} o(f,x,\\delta) < \\epsilon\\text{,}\\) find a \\(\\delta > 0\\) such that
Take any \\(\\xi \\in B_D(x,\\nicefrac{\\delta}{2})\\text{.}\\) Notice that \\(B_D(\\xi,\\nicefrac{\\delta}{2}) \\subset B_D(x,\\delta)\\text{.}\\) Therefore,
So \\(o(f,\\xi) < \\epsilon\\) as well. As this is true for all \\(\\xi \\in\r\nB_D(x,\\nicefrac{\\delta}{2})\\text{,}\\) we get that \\(G\\) is open in the subspace topology, and \\(D \\setminus G\\) is closed as claimed.
\n\n
\n\n",
"licence": "CC-BY-SA-4.0",
"proof_scope": "Closed positive oscillation level sets on a closed domain; full proof."
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"id": "L10.4.3",
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\nTheorem10.4.3.Riemann–Lebesgue or Lebesgue–Vitali.\n
\n
\n Let \\(R \\subset \\R^n\\) be a closed rectangle and \\(f \\colon R \\to \\R\\) bounded. Then \\(f\\) is Riemann integrable if and only if the set of discontinuities of \\(f\\) is of measure zero.\n
\n\n\n
Proof.
\n
\n
Let \\(S \\subset R\\) be the set of discontinuities of \\(f\\text{,}\\) that is, \\(S = \\bigl\\{ x \\in R : o(f,x) > 0 \\bigr\\}\\text{.}\\) Suppose \\(S\\) is a measure zero set: \\(m^*(S) = 0\\text{.}\\) The trick to proving that \\(f\\) is integrable is to isolate the bad set into a small set of subrectangles of a partition. A partition has finitely many subrectangles, so we need compactness. If \\(S\\) were closed, then it would be compact and we could cover it by finitely many small rectangles. Unfortunately, \\(S\\) itself is not closed in general, but the following set is. Given \\(\\epsilon > 0\\text{,}\\) define
\n
\r\n\\begin{equation}\r\nS_\\epsilon \\coloneqq \\bigl\\{ x \\in R : o(f,x) \\geq \\epsilon \\bigr\\} .\r\n\\end{equation}\r\n
\n
By Proposition 10.4.2, \\(S_\\epsilon\\) is closed, and as it is also a subset of the bounded \\(R\\text{,}\\)\\(S_\\epsilon\\) is compact. Moreover, \\(S_\\epsilon \\subset S\\) and \\(S\\) is of measure zero, so \\(S_\\epsilon\\) is of measure zero. Via Proposition 10.3.7, there exist finitely many open rectangles \\(O_1,O_2,\\ldots,O_k\\) that cover \\(S_\\epsilon\\) and \\(\\sum_{j=1}^\\infty V(O_j) < \\epsilon\\text{.}\\)\n
\n\n
\n
\n
The set \\(T \\coloneqq R \\setminus ( O_1 \\cup \\cdots \\cup O_k )\\) is closed, bounded, and thus compact. As \\(o(f,x) < \\epsilon\\) for all \\(x \\in T\\text{,}\\) for each \\(x \\in T\\text{,}\\) there is a \\(\\delta > 0\\) such that \\(o(f,x,\\delta) < \\epsilon\\text{,}\\) so there exists a small closed rectangle \\(T_x \\subset B(x,\\delta)\\) with \\(x\\) in the interior of \\(T_x\\text{,}\\) such that
The interiors of the rectangles \\(T_x\\) cover \\(T\\text{.}\\) As \\(T\\) is compact, finitely many such rectangles \\(T_1, T_2, \\ldots, T_m\\) cover \\(T\\text{.}\\) Construct a partition \\(P\\) out of the endpoints of the rectangles \\(T_1,T_2,\\ldots,T_m\\) and \\(O_1,O_2,\\ldots,O_k\\) (ignoring those that are outside the endpoints of \\(R\\)). The subrectangles \\(R_1,R_2,\\ldots,R_p\\) of \\(P\\) are such that every \\(R_j\\) is contained in \\(T_\\ell\\) for some \\(\\ell\\) or the closure of \\(O_\\ell\\) for some \\(\\ell\\text{.}\\) Order the rectangles so that \\(R_1,R_2,\\ldots,R_q\\) are those that are contained in some \\(T_\\ell\\text{,}\\) and \\(R_{q+1},R_{q+2},\\ldots,R_{p}\\) are the rest. See Figure 10.12. So
The second estimate holds because the \\(R_j\\) that are subsets of \\(\\widebar{O}_\\ell\\) give a partition of \\(\\widebar{O}_\\ell\\) and hence their volumes sum to \\(V(O_\\ell)\\text{.}\\) Let \\(m_j\\) and \\(M_j\\) be the inf and sup of \\(f\\) over \\(R_j\\) as usual. If \\(R_j \\subset T_\\ell\\) for some \\(\\ell\\text{,}\\) then \\(M_j-m_j < \\epsilon\\text{.}\\) Let \\(B \\in \\R\\) be such that \\(\\babs{f(x)} \\leq B\\) for all \\(x \\in R\\text{,}\\) so \\(M_j-m_j \\leq 2B\\) over all rectangles. Then
We can make the right-hand side as small as we want, and hence \\(f\\) is integrable.
\n\n
\n\nFigure10.12.A rectangle \\(R\\) with \\(S_\\epsilon\\) marked as thick black line, and the \\(O_\\ell\\) as shaded rectangles. The partition is given by the dotted lines. Note how the \\(R_j\\) partition the \\(O_\\ell\\text{.}\\)\n\n
\n
For the other direction, suppose \\(f\\) is Riemann integrable on \\(R\\text{.}\\) Let \\(S\\) be the set of discontinuities of \\(f\\) again. Consider the sequence of sets
\n
\r\n\\begin{equation}\r\nS_{1/k} = \\bigl\\{ x \\in R : o(f,x) \\geq \\nicefrac{1}{k} \\bigr\\}.\r\n\\end{equation}\r\n
\n
Fix a \\(k \\in \\N\\text{.}\\) Given an \\(\\epsilon > 0\\text{,}\\) find a partition \\(P\\) with subrectangles \\(R_1,R_2,\\ldots,R_p\\) such that
Suppose \\(R_1,R_2,\\ldots,R_p\\) are ordered so that the interiors of \\(R_1,R_2,\\ldots,R_{q}\\) intersect \\(S_{1/k}\\text{,}\\) while the interiors of \\(R_{q+1},R_{q+2},\\ldots,R_p\\) are disjoint from \\(S_{1/k}\\text{.}\\) Let \\(R_j^\\circ\\) denote the interior of \\(R_j\\text{.}\\) Suppose \\(j \\leq q\\) and consider \\(x \\in R_j^\\circ \\cap S_{1/k}\\text{.}\\) Let \\(\\delta > 0\\) be small enough so that \\(B(x,\\delta) \\subset R_j\\text{.}\\) As \\(x \\in S_{1/k}\\text{,}\\) we get \\(o(f,x,\\delta) \\geq o(f,x) \\geq \\nicefrac{1}{k}\\text{,}\\) which, along with \\(B(x,\\delta) \\subset R_j\\text{,}\\) implies \\(M_j-m_j \\geq \\nicefrac{1}{k}\\text{.}\\) Then
In other words, \\(\\sum_{j=1}^q V(R_j) < k \\epsilon\\text{.}\\) Let \\(G\\) be the set of all boundaries of all the subrectangles of \\(P\\text{.}\\) The set \\(G\\) is of measure zero (it can be covered by finitely many sets from Example 10.3.5). We find
Let \\(f \\colon [a,b] \\to \\R\\) be a bounded function. Let \\(m, M \\in \\R\\) be such that for all \\(x \\in [a,b]\\text{,}\\) we have \\(m \\leq f(x) \\leq M\\text{.}\\) Then for every partition \\(P\\) of \\([a,b]\\text{,}\\)\n
Let \\(P\\) be a partition of \\([a,b]\\text{.}\\) Note that for all \\(i\\text{,}\\) we have \\(m \\leq m_i \\leq M_i \\leq M\\text{.}\\) We also have \\(\\sum_{i=1}^n \\Delta x_i = (b-a)\\text{.}\\) Therefore,
Hence, we get (5.1). In particular, the sets of lower and upper sums are bounded sets.
\n\n
\n\n",
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"proof_scope": "Darboux sum bounds and existence of their defining finite extrema."
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\nProposition5.1.7.\n
\n
\n
Let \\(f \\colon [a,b] \\to \\R\\) be a bounded function, and let \\(P\\) be a partition of \\([a,b]\\text{.}\\) Let \\(\\widetilde{P}\\) be a refinement of \\(P\\text{.}\\) Then
The tricky part of this proof is to get the notation correct. Let \\(\\widetilde{P} = \\{ \\widetilde{x}_0, \\widetilde{x}_1, \\ldots,\r\n\\widetilde{x}_{\\ell} \\}\\) be a refinement of \\(P = \\{ x_0, x_1, \\ldots, x_n \\}\\text{.}\\) Then \\(x_0 = \\widetilde{x}_0\\) and \\(x_n = \\widetilde{x}_{\\ell}\\text{.}\\) In fact, there are integers \\(k_0 < k_1 < \\cdots < k_n\\) such that \\(x_i = \\widetilde{x}_{k_i}\\) for \\(i=0,1,2,\\ldots,n\\text{.}\\)\n
\n
\n
Let \\(\\Delta \\widetilde{x}_q \\coloneqq \\widetilde{x}_q - \\widetilde{x}_{q-1}\\) for \\(q=0,1,2,\\ldots,\\ell\\text{.}\\) See Figure 5.2. We get
\n\nFigure5.2.Refinement of a subinterval. Notice \\(\\Delta x_i =\r\n\\Delta \\widetilde{x}_{q-2} +\r\n\\Delta \\widetilde{x}_{q-1} +\r\n\\Delta \\widetilde{x}_{q}\\text{,}\\) and also \\(k_{i-1}+1 = q-2\\) and \\(k_{i} = q\\text{.}\\)\n\n
\n
Let \\(m_i\\) be as before and correspond to the partition \\(P\\text{.}\\) Let \\(\\widetilde{m}_q \\coloneqq \\inf \\bigl\\{ f(x) : \\widetilde{x}_{q-1} \\leq x \\leq\r\n\\widetilde{x}_q \\bigr\\}\\text{.}\\) Now, \\(m_i \\leq \\widetilde{m}_q\\) for \\(k_{i-1} < q \\leq k_i\\text{.}\\) Therefore,
Let \\(f \\colon [a,b] \\to \\R\\) be a bounded function. Let \\(m, M \\in \\R\\) be such that for all \\(x \\in [a,b]\\text{,}\\) we have \\(m \\leq f(x) \\leq M\\text{.}\\) Then
\n
\r\n\\begin{equation}\r\nm(b-a) \\leq\r\n\\underline{\\int_a^b} f \\leq \\overline{\\int_a^b} f\r\n\\leq M(b-a) .\\tag{5.2}\r\n\\end{equation}\r\n
The inequality \\(m(b-a) \\leq L(P,f)\\) implies \\(m(b-a) \\leq \\underline{\\int_a^b} f\\text{.}\\) The inequality \\(U(P,f) \\leq M(b-a)\\) implies \\(\\overline{\\int_a^b} f \\leq M(b-a)\\text{.}\\)\n
\n\n
\n
\n
The middle inequality in (5.2) is the main point of the proposition. Let \\(P_1, P_2\\) be partitions of \\([a,b]\\text{.}\\) Define \\(\\widetilde{P} \\coloneqq P_1 \\cup P_2\\text{.}\\) The set \\(\\widetilde{P}\\) is a partition of \\([a,b]\\text{,}\\) which is a refinement of \\(P_1\\) and a refinement of \\(P_2\\text{.}\\) By Proposition 5.1.7, \\(L(P_1,f) \\leq L(\\widetilde{P},f)\\) and \\(U(\\widetilde{P},f) \\leq U(P_2,f)\\text{.}\\) So
In other words, for two arbitrary partitions \\(P_1\\) and \\(P_2\\text{,}\\) we have \\(L(P_1,f) \\leq U(P_2,f)\\text{.}\\) Recall Proposition 1.2.7, and take the supremum and infimum over all partitions:
\n
\r\n\\begin{multline*}\r\n\\underline{\\int_a^b} f = \r\n\\sup \\, \\bigl\\{ L(P,f) : P \\text{ a partition of } [a,b] \\bigr\\}\r\n\\\\\r\n\\leq\r\n\\inf \\, \\bigl\\{ U(P,f) : P \\text{ a partition of } [a,b] \\bigr\\}\r\n=\r\n\\overline{\\int_a^b} f . \\qedhere\r\n\\end{multline*}\r\n
Let \\(f \\colon [a,b] \\to \\R\\) be a Riemann integrable function. Let \\(m, M \\in \\R\\) be such that \\(m \\leq f(x) \\leq M\\) for all \\(x \\in [a,b]\\text{.}\\) Then
Proof. For an integrable function the lower and upper Darboux integrals are equal to its integral. Substitute this equality in the preceding bound (5.2), proved above. This gives the two asserted inequalities.
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"proof_scope": "Integral bound is the just-proved Darboux bound with equal lower/upper integrals, as Definition 5.1.9 specifies."
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\nExample5.1.11.\n
\n
We integrate constant functions using Proposition 5.1.8. If \\(f(x) \\coloneqq c\\) for some constant \\(c\\text{,}\\) then we take \\(m = M = c\\text{.}\\) In inequality (5.2) all the inequalities must be equalities. Thus \\(f\\) is integrable on \\([a,b]\\) and \\(\\int_a^b f = c(b-a)\\text{.}\\)\n
Let \\(f \\colon [a,b] \\to \\R\\) be a bounded function. Then \\(f\\) is Riemann integrable if for every \\(\\epsilon > 0\\text{,}\\) there exists a partition \\(P\\) of \\([a,b]\\) such that
Suppose \\(a < b < c\\) and \\(f \\colon [a,c] \\to \\R\\) is a bounded function. Then
\n
\r\n\\begin{equation}\r\n\\underline{\\int_a^c} f\r\n=\r\n\\underline{\\int_a^b} f\r\n+\r\n\\underline{\\int_b^c} f\r\n\\quad \\text{and} \\quad\r\n\\overline{\\int_a^c} f\r\n=\r\n\\overline{\\int_a^b} f\r\n+\r\n\\overline{\\int_b^c} f .\r\n\\end{equation}\r\n
\n\n
\n\n\n
Proof.
\n
\n
If we have partitions \\(P_1 = \\{ x_0,x_1,\\ldots,x_k \\}\\) of \\([a,b]\\) and \\(P_2 = \\{ x_k, x_{k+1}, \\ldots, x_n \\}\\) of \\([b,c]\\text{,}\\) then the set \\(P \\coloneqq P_1 \\cup P_2 = \\{ x_0, x_1, \\ldots, x_n \\}\\) is a partition of \\([a,c]\\text{.}\\) We find
When we take the supremum of the right-hand side over all \\(P_1\\) and \\(P_2\\text{,}\\) we are taking a supremum of the left-hand side over all partitions \\(P\\) of \\([a,c]\\) that contain \\(b\\text{.}\\) If \\(Q\\) is a partition of \\([a,c]\\) and \\(P = Q \\cup \\{ b \\}\\text{,}\\) then \\(P\\) is a refinement of \\(Q\\) and so \\(L(Q,f) \\leq L(P,f)\\text{.}\\) Therefore, taking a supremum only over the \\(P\\) that contain \\(b\\) is sufficient to find the supremum of \\(L(P,f)\\) over all partitions \\(P\\text{,}\\) see Exercise 1.1.9. Finally, recall Exercise 1.2.9 to compute
\n
\r\n\\begin{equation}\r\n\\begin{split}\r\n\\underline{\\int_a^c} f\r\n& =\r\n\\sup \\, \\bigl\\{ L(P,f) : P \\text{ a partition of } [a,c] \\bigr\\}\r\n\\\\\r\n& =\r\n\\sup \\, \\bigl\\{ L(P,f) : P \\text{ a partition of } [a,c], b \\in P \\bigr\\}\r\n\\\\\r\n& =\r\n\\sup \\, \\bigl\\{ L(P_1,f) + L(P_2,f) :\r\nP_1 \\text{ a partition of } [a,b], P_2 \\text{ a partition of } [b,c] \\bigr\\}\r\n\\\\\r\n& =\r\n\\sup \\, \\bigl\\{ L(P_1,f) : P_1 \\text{ a partition of } [a,b] \\bigr\\}\r\n+\r\n\\sup \\, \\bigl\\{ L(P_2,f) : P_2 \\text{ a partition of } [b,c] \\bigr\\}\r\n\\\\\r\n&=\r\n\\underline{\\int_a^b} f + \\underline{\\int_b^c} f .\r\n\\end{split}\r\n\\end{equation}\r\n
\n\n
\n
\n
Similarly, for \\(P\\text{,}\\)\\(P_1\\text{,}\\) and \\(P_2\\) as above, we obtain
We wish to take the infimum on the right over all \\(P_1\\) and \\(P_2\\text{,}\\) and so we are taking the infimum over all partitions \\(P\\) of \\([a,c]\\) that contain \\(b\\text{.}\\) If \\(Q\\) is a partition of \\([a,c]\\) and \\(P = Q \\cup \\{ b \\}\\text{,}\\) then \\(P\\) is a refinement of \\(Q\\) and so \\(U(Q,f) \\geq U(P,f)\\text{.}\\) Therefore, taking an infimum only over the \\(P\\) that contain \\(b\\) is sufficient to find the infimum of \\(U(P,f)\\) for all \\(P\\text{.}\\) We obtain
\n
\r\n\\begin{equation}\r\n\\overline{\\int_a^c} f\r\n=\r\n\\overline{\\int_a^b} f + \\overline{\\int_b^c} f . \\qedhere\r\n\\end{equation}\r\n
\n\n
\n\n",
"licence": "CC-BY-SA-4.0",
"proof_scope": "Additivity of upper and lower integrals on adjacent intervals."
},
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"id": "L5.2.2",
"reader": "sec_rintprop.html",
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\nProposition5.2.2.\n
\n
\n
Let \\(a < b < c\\text{.}\\) A function \\(f \\colon [a,c] \\to \\R\\) is Riemann integrable if and only if \\(f\\) is Riemann integrable on \\([a,b]\\) and \\([b,c]\\text{.}\\) If \\(f\\) is Riemann integrable, then
\n
\r\n\\begin{equation}\r\n\\int_a^c f\r\n=\r\n\\int_a^b f\r\n+\r\n\\int_b^c f .\r\n\\end{equation}\r\n
\n\n
\n\n\n
Proof.
\n
\n
Suppose \\(f \\in \\sR\\bigl([a,c]\\bigr)\\text{.}\\) Then it is bounded and \\(\\overline{\\int_a^c} f = \r\n\\underline{\\int_a^c} f = \r\n\\int_a^c f\\text{.}\\) The lemma gives
\n
\r\n\\begin{equation}\r\n\\int_a^c f\r\n=\r\n\\underline{\\int_a^c} f\r\n=\r\n\\underline{\\int_a^b} f + \\underline{\\int_b^c} f\r\n\\leq\r\n\\overline{\\int_a^b} f + \\overline{\\int_b^c} f\r\n=\r\n\\overline{\\int_a^c} f\r\n=\r\n\\int_a^c f .\r\n\\end{equation}\r\n
\n
Thus the inequality is an equality:
\n
\r\n\\begin{equation}\r\n\\underline{\\int_a^b} f + \\underline{\\int_b^c} f\r\n=\r\n\\overline{\\int_a^b} f + \\overline{\\int_b^c} f .\r\n\\end{equation}\r\n
\n
As we also know \\(\\underline{\\int_a^b} f \\leq \\overline{\\int_a^b} f\\) and \\(\\underline{\\int_b^c} f \\leq \\overline{\\int_b^c} f\\text{,}\\) we conclude
\n
\r\n\\begin{equation}\r\n\\underline{\\int_a^b} f\r\n=\r\n\\overline{\\int_a^b} f\r\n\\qquad \\text{and} \\qquad\r\n\\underline{\\int_b^c} f\r\n=\r\n\\overline{\\int_b^c} f .\r\n\\end{equation}\r\n
\n
Thus \\(f\\) is Riemann integrable on \\([a,b]\\) and \\([b,c]\\) and the desired formula holds.
\n\n
\n
\n
Now assume \\(f\\) is Riemann integrable on \\([a,b]\\) and on \\([b,c]\\text{.}\\) Again it is bounded, and the lemma gives
\n
\r\n\\begin{equation}\r\n\\underline{\\int_a^c} f\r\n=\r\n\\underline{\\int_a^b} f + \\underline{\\int_b^c} f\r\n=\r\n\\int_a^b f + \\int_b^c f\r\n=\r\n\\overline{\\int_a^b} f + \\overline{\\int_b^c} f\r\n=\r\n\\overline{\\int_a^c} f .\r\n\\end{equation}\r\n
\n
Therefore, \\(f\\) is Riemann integrable on \\([a,c]\\text{,}\\) and the integral is computed as indicated.
Let us prove the first item for \\(\\alpha \\geq 0\\text{.}\\) Let \\(P\\) be a partition of \\([a,b]\\text{,}\\) and \\(m_i \\coloneqq \\inf \\bigl\\{ f(x) : x \\in [x_{i-1},x_i] \\bigr\\}\\) as usual. As \\(\\alpha \\geq 0\\text{,}\\) the multiplication by \\(\\alpha\\) moves past the infimum,
The conclusion now follows for \\(\\alpha \\geq 0\\text{.}\\)\n
\n\n
\n
To finish the proof of the first item (for \\(\\alpha < 0\\)), we need to show that \\(-f\\) is Riemann integrable and \\(\\int_a^b - f(x)\\,dx =\r\n-\r\n\\int_a^b f(x)\\,dx\\text{.}\\) The proof of this fact is left as Exercise 5.2.1.\n
\n
The proof of the second item is left as Exercise 5.2.2. It is not difficult, but it is not as trivial as it may appear at first glance.\n
\n\n",
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"proof_scope": "The actually written nonnegative-scalar proof only; omitted negative and sum cases are supplied locally."
},
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"id": "L5.2.6",
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\nProposition5.2.6.Monotonicity.\n
\n
\n
Let \\(f \\colon [a,b] \\to \\R\\) and \\(g \\colon [a,b] \\to \\R\\) be bounded, and \\(f(x) \\leq g(x)\\) for all \\(x \\in [a,b]\\text{.}\\) Then
\n
\r\n\\begin{equation}\r\n\\underline{\\int_a^b} f \r\n\\leq\r\n\\underline{\\int_a^b} g \r\n\\qquad \\text{and} \\qquad\r\n\\overline{\\int_a^b} f \r\n\\leq\r\n\\overline{\\int_a^b} g .\r\n\\end{equation}\r\n
\n
Moreover, if \\(f\\) and \\(g\\) are in \\(\\sR\\bigl([a,b]\\bigr)\\text{,}\\) then
\n
\r\n\\begin{equation}\r\n\\int_a^b f \r\n\\leq\r\n\\int_a^b g .\r\n\\end{equation}\r\n
\n\n
\n\n\n
Proof.
\n
\n
Let \\(P = \\{ x_0, x_1, \\ldots, x_n \\}\\) be a partition of \\([a,b]\\text{.}\\) Then let
We take the supremum over all \\(P\\) (see Proposition 1.3.7) to obtain
\n
\r\n\\begin{equation}\r\n\\underline{\\int_a^b} f \r\n\\leq\r\n\\underline{\\int_a^b} g .\r\n\\end{equation}\r\n
\n
Similarly, we obtain the same conclusion for the upper integrals. Finally, if \\(f\\) and \\(g\\) are Riemann integrable all the integrals are equal, and the conclusion follows.
If \\(f \\colon [a,b] \\to \\R\\) is a continuous function, then \\(f \\in \\sR\\bigl([a,b]\\bigr)\\text{.}\\)\n
\n\n\n
Proof.
\n
As \\(f\\) is continuous on a closed bounded interval, it is bounded and uniformly continuous. Given \\(\\epsilon > 0\\text{,}\\) find a \\(\\delta > 0\\) such that \\(\\sabs{x-y} < \\delta\\) implies \\(\\babs{f(x)-f(y)} < \\frac{\\epsilon}{b-a}\\text{.}\\)\n
\n
\n
Let \\(P = \\{ x_0, x_1, \\ldots, x_n \\}\\) be a partition of \\([a,b]\\) such that \\(\\Delta x_i < \\delta\\) for all \\(i = 1,2,\r\n\\ldots, n\\text{.}\\) For example, take \\(n\\) such that \\(\\frac{b-a}{n} < \\delta\\text{,}\\) and let \\(x_i \\coloneqq \\frac{i}{n}(b-a) + a\\text{.}\\) Then for all \\(x, y \\in [x_{i-1},x_i]\\text{,}\\) we have \\(\\sabs{x-y} \\leq \\Delta x_i < \\delta\\text{,}\\) and so
As \\(f\\) is continuous on \\([x_{i-1},x_i]\\text{,}\\) it attains a maximum and a minimum on this interval. Let \\(x\\) be a point where \\(f\\) attains the maximum and \\(y\\) be a point where \\(f\\) attains the minimum. Then \\(f(x) = M_i\\) and \\(f(y) = m_i\\) in the notation from the definition of the integral. Therefore,
Suppose \\(R \\subset \\R^n\\) is a closed rectangle and \\(f \\colon R \\to \\R\\) is a bounded function. Let \\(m, M \\in \\R\\) be such that for all \\(x \\in R\\text{,}\\) we have \\(m \\leq f(x) \\leq M\\text{.}\\) Then for every partition \\(P\\) of \\(R\\text{,}\\)\n
Let \\(P\\) be a partition of \\(R\\text{.}\\) For all \\(i\\text{,}\\) we have \\(m \\leq m_i \\leq M_i \\leq M\\text{.}\\) Also \\(\\sum_{i=1}^N V(R_i) = V(R)\\text{.}\\) Therefore,
\n\n",
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\nProposition10.1.5.\n
\n
\n
Suppose \\(R \\subset \\R^n\\) is a closed rectangle, \\(P\\) is a partition of \\(R\\text{,}\\) and \\(\\widetilde{P}\\) is a refinement of \\(P\\text{.}\\) If \\(f \\colon R \\to \\R\\) is bounded, then
We prove the first inequality, and the second follows similarly. Let \\(R_1,R_2,\\ldots,R_N\\) be the subrectangles of \\(P\\) and \\(\\widetilde{R}_1,\\widetilde{R}_2,\\ldots,\\widetilde{R}_{\\widetilde{N}}\\) be the subrectangles of \\(\\widetilde{P}\\text{.}\\) Let \\(I_k\\) be the set of all indices \\(j\\) such that \\(\\widetilde{R}_j \\subset R_k\\text{.}\\) For example, in figures 10.1 and 10.2, \\(I_4 = \\{ 6, 7, 8, 9 \\}\\) as \\(R_4 =\r\n\\widetilde{R}_6 \\cup \\widetilde{R}_7 \\cup\r\n\\widetilde{R}_8 \\cup \\widetilde{R}_9\\text{.}\\) Then,
Let \\(m_k \\coloneqq \\inf \\bigl\\{ f(x) : x \\in R_k \\bigr\\}\\text{,}\\) and \\(\\widetilde{m}_j \\coloneqq \\inf \\bigl\\{ f(x) : x \\in \\widetilde{R}_j \\bigr\\}\\) as usual. If \\(j \\in I_k\\text{,}\\) then \\(m_k \\leq \\widetilde{m}_j\\text{.}\\) Then
Let \\(R \\subset \\R^n\\) be a closed rectangle and \\(f \\colon R \\to \\R\\) a bounded function. Let \\(m, M \\in \\R\\) be such that for all \\(x \\in R\\text{,}\\) we have \\(m \\leq f(x) \\leq M\\text{.}\\) Then
\n
\r\n\\begin{equation}\r\nm \\, V(R) \\leq\r\n\\underline{\\int_R} f \\leq \\overline{\\int_R} f\r\n\\leq M \\, V(R).\\tag{10.1}\r\n\\end{equation}\r\n
Taking the supremum of \\(L(P,f)\\) and the infimum of \\(U(P,f)\\) over all partitions \\(P\\text{,}\\) we obtain the first and the last inequality in (10.1).
\n\n
\n
\n
The key inequality in (10.1) is the middle one. Let \\(P=(P_1,P_2,\\ldots,P_n)\\) and \\(Q=(Q_1,Q_2,\\ldots,Q_n)\\) be partitions of \\(R\\text{.}\\) Define \\(\\widetilde{P} = ( \\widetilde{P}_1,\\widetilde{P}_2,\\ldots,\\widetilde{P}_n )\\) by letting \\(\\widetilde{P}_k \\coloneqq P_k \\cup Q_k\\) for every \\(k\\text{.}\\) Then \\(\\widetilde{P}\\) is a partition of \\(R\\text{,}\\) and \\(\\widetilde{P}\\) is a refinement of \\(P\\) and also a refinement of \\(Q\\text{.}\\) By Proposition 10.1.5, \\(L(P,f) \\leq L(\\widetilde{P},f)\\) and \\(U(\\widetilde{P},f) \\leq U(Q,f)\\text{.}\\) Therefore,
In other words, for two arbitrary partitions \\(P\\) and \\(Q\\text{,}\\) we have \\(L(P,f) \\leq U(Q,f)\\text{.}\\) Via Proposition 1.2.7, we obtain
\n
\r\n\\begin{equation}\r\n\\sup \\, \\bigl\\{ L(P,f) : P \\text{ a partition of } R \\bigr\\}\r\n\\leq\r\n\\inf \\, \\bigl\\{ U(P,f) : P \\text{ a partition of } R \\bigr\\} .\r\n\\end{equation}\r\n
\n
In other words, \\(\\underline{\\int_R} f \\leq \\overline{\\int_R} f\\text{.}\\)\n
\n\n
\n\n",
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\nProposition10.1.12.\n
\n
\n
Let \\(R \\subset \\R^n\\) be a closed rectangle and \\(f \\colon R \\to \\R\\) a bounded function. Then \\(f \\in \\sR(R)\\) if and only if for every \\(\\epsilon > 0\\text{,}\\) there exists a partition \\(P\\) of \\(R\\) such that
First, if \\(f\\) is integrable, then the supremum of \\(L(P,f)\\) and infimum of \\(U(Q,f)\\) over all partitions \\(P\\) and \\(Q\\) are equal and hence the infimum of \\(U(P,f)-L(Q,f)\\) is zero. Taking a common refinement \\(\\widetilde{P}\\) of \\(P\\) and \\(Q\\) we find \\(U(\\widetilde{P},f)-L(\\widetilde{P},f) \\leq U(P,f)-L(Q,f)\\text{.}\\) Hence the infimum of \\(U(P,f)-L(P,f)\\) over all partitions \\(P\\) is zero, and so for every \\(\\epsilon > 0\\text{,}\\) there must be some partition \\(P\\) such that \\(U(P,f) - L(P,f) < \\epsilon\\text{.}\\)\n
\n
\n
For the other direction, given an \\(\\epsilon > 0\\) find \\(P\\) such that \\(U(P,f) - L(P,f) < \\epsilon\\text{.}\\)\n
\n
\r\n\\begin{equation}\r\n\\overline{\\int_R} f - \r\n\\underline{\\int_R} f \r\n\\leq\r\nU(P,f) - L(P,f)\r\n< \\epsilon .\r\n\\end{equation}\r\n
\n
As \\(\\overline{\\int_R} f \\geq \\underline{\\int_R} f\\) and the above holds for every \\(\\epsilon > 0\\text{,}\\) we conclude \\(\\overline{\\int_R} f = \\underline{\\int_R} f\\) and \\(f \\in \\sR(R)\\text{.}\\)\n
\n\n
\n\n",
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"proof_scope": "Both directions of the small-gap criterion with supremum/infimum operations explicitly available."
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\nProposition10.1.13.\n
\n
Let \\(S \\subset \\R^n\\) be a closed rectangle. If \\(f \\colon S \\to \\R\\) is integrable and \\(R \\subset S\\) is a closed rectangle, then \\(f\\) is integrable on \\(R\\text{.}\\)\n
\n\n\n
Proof.
\n
\n
Given \\(\\epsilon > 0\\text{,}\\) find a partition \\(P=(P_1,\\ldots,P_n)\\) of \\(S\\) such that \\(U(P,f)-L(P,f) < \\epsilon\\text{.}\\) By making a refinement of \\(P\\) if necessary, assume that the endpoints of \\(R\\) are in \\(P\\text{.}\\) That is, if \\(R = [a_1,b_1] \\times [a_2,b_2] \\times \\cdots \\times [a_n,b_n]\\text{,}\\) then \\(a_i,b_i \\in P_i\\text{.}\\) Let \\(\\widetilde{P} = (\\widetilde{P}_1,\\ldots,\\widetilde{P}_n)\\) be the partition of \\(R\\) given by \\(\\widetilde{P}_i = P_i \\cap [a_i,b_i]\\text{.}\\) Subrectangles of \\(\\widetilde{P}\\) are subrectangles of \\(P\\text{,}\\) that is, \\(R\\) is a union of subrectangles of \\(P\\text{.}\\) Divide the subrectangles of \\(P\\) into two collections: Let \\(R_1,R_2\\ldots,R_K\\) be the subrectangles of \\(P\\) that are also subrectangles of \\(\\widetilde{P}\\) and let \\(R_{K+1},\\ldots, R_N\\) be the rest. See Figure 10.3. Let \\(m_k\\) and \\(M_k\\) be the infimum and supremum of \\(f\\) on \\(R_k\\) as usual. Then,
Let \\(R \\subset \\R^n\\) be a closed rectangle. If \\(f \\colon R \\to \\R\\) is continuous, then \\(f \\in \\sR(R)\\text{.}\\)\n
\n\n\n
Proof.
\n
The proof is analogous to the one-variable proof with some complications. The set \\(R\\) is a closed and bounded subset of \\(\\R^n\\text{,}\\) and hence compact. So \\(f\\) is uniformly continuous by Theorem 7.5.11. Let \\(\\epsilon > 0\\) be given. Find a \\(\\delta > 0\\) such that \\(\\snorm{x-y} < \\delta\\) implies \\(\\babs{f(x)-f(y)} < \\frac{\\epsilon}{V(R)}\\text{.}\\)\n
\n
\n
Let \\(P\\) be a partition of \\(R\\text{,}\\) where the longest side of every subrectangle is strictly less than \\(\\frac{\\delta}{\\sqrt{n}}\\text{.}\\) If \\(x, y \\in R_k\\) for a subrectangle \\(R_k\\) of \\(P\\text{,}\\) then, by the proposition, \\(\\snorm{x-y} < \\sqrt{n} \\frac{\\delta}{\\sqrt{n}} = \\delta\\text{.}\\) Therefore,
As \\(f\\) is continuous on \\(R_k\\text{,}\\) which is compact, \\(f\\) attains a maximum and a minimum on this subrectangle. Let \\(x\\) be a point where \\(f\\) attains the maximum and \\(y\\) be a point where \\(f\\) attains the minimum. Then \\(f(x) = M_k\\) and \\(f(y) = m_k\\) in the notation from the definition of the integral. Thus,
\n\n",
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\nProposition10.1.19.\n
\n
\n
Suppose \\(f \\colon \\R^n \\to \\R\\) is a continuous function with compact support. If \\(R\\) and \\(S\\) are closed rectangles such that \\(\\operatorname{supp}(f) \\subset R\\) and \\(\\operatorname{supp}(f) \\subset S\\text{,}\\) then
\n
\r\n\\begin{equation}\r\n\\int_S f = \\int_R f .\r\n\\end{equation}\r\n
\n\n
\n\n\n
Proof.
\n
As \\(f\\) is continuous, it is automatically integrable on the rectangles \\(R\\text{,}\\)\\(S\\text{,}\\) and \\(R\r\n\\cap S\\text{.}\\) Applying Exercise 10.1.7 twice, \\(\\int_S f = \\int_{S \\cap R} f = \\int_R f\\text{.}\\)\n
\n\n",
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"proof_scope": "Independence of support-containing rectangle, with the exact zero-extension exercise and empty-support case supplied in P17.3."
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"html": "Figure10.1.Example partition of a rectangle in \\(\\R^2\\text{.}\\) The order of the subrectangles is not important.\n",
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"html": "Figure10.2.Example refinement of the partition from Figure 10.1. New “cuts” are marked in dashed lines. The exact order of the new subrectangles does not matter.\n",
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"html": "Figure10.3.A partition of a large rectangle \\(S\\text{,}\\) that also gives a partition of a smaller rectangle (shaded and outlined) \\(R \\subset S\\text{.}\\) The subrectangles \\(R_1,R_2,R_3,R_4\\) are the subrectangles of \\(\\widetilde{P} = \\bigl( \\{ x_{1,1}, x_{1,2} , x_{1,3} \\} ,\r\n\\{ x_{2,1}, x_{2,2} , x_{2,3} \\} \\bigr)\\text{.}\\)\n",
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\nTheorem2.1.10.Monotone convergence theorem.\n
\n
A monotone sequence \\(\\{ x_n \\}_{n=1}^\\infty\\) is bounded if and only if it is convergent.\n
\n
\n
Furthermore, if \\(\\{ x_n \\}_{n=1}^\\infty\\) is monotone increasing and bounded, then
Consider a monotone increasing sequence \\(\\{ x_n \\}_{n=1}^\\infty\\text{.}\\) Suppose first the sequence is bounded, that is, the set \\(\\{ x_n : n \\in \\N \\}\\) is bounded. Let
Let \\(\\epsilon > 0\\) be arbitrary. As \\(x\\) is the supremum, there must be at least one \\(M \\in \\N\\) such that \\(x_{M} > x-\\epsilon\\text{.}\\) As \\(\\{ x_n \\}_{n=1}^\\infty\\) is monotone increasing, it is easy to see (by induction) that \\(x_n \\geq x_{M}\\) for all \\(n \\geq M\\text{.}\\) Hence, for all \\(n \\geq M\\text{,}\\)\n
So \\(\\{ x_n \\}_{n=1}^\\infty\\) converges to \\(x\\text{.}\\) Therefore, a bounded monotone increasing sequence converges. For the other direction, we already proved that a convergent sequence is bounded.
\n\n
\n
The proof for monotone decreasing sequences is left as an exercise.\n
\n\n",
"licence": "CC-BY-SA-4.0",
"proof_scope": "The fully written increasing half only; decreasing half is P6.2"
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"id": "L2.1.17",
"reader": "sec_seqsandlims.html",
"author_url": "https://www.jirka.org/ra/html/sec_seqsandlims.html",
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\nProposition2.1.17.\n
\n
\n
If \\(\\{ x_n \\}_{n=1}^\\infty\\) is a convergent sequence, then every subsequence \\(\\{ x_{n_i} \\}_{i=1}^\\infty\\) is also convergent, and
Suppose \\(\\lim_{n\\to \\infty} x_n = x\\text{.}\\) So for every \\(\\epsilon > 0\\text{,}\\) there is an \\(M \\in \\N\\) such that for all \\(n \\geq M\\text{,}\\)\n
It is not hard to prove (do it!) by induction that \\(n_i \\geq i\\) for all \\(i \\in \\N\\text{.}\\) Hence, \\(i \\geq M\\) implies \\(n_i \\geq M\\text{.}\\) Thus, for all \\(i \\geq M\\text{,}\\)\n
\n\n",
"licence": "CC-BY-SA-4.0",
"proof_scope": "Subsequences preserve real limits; index induction supplied in P6.1"
},
{
"id": "L8.5.1",
"reader": "sec_svinvfuncthm.html",
"author_url": "https://www.jirka.org/ra/html/sec_svinvfuncthm.html",
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\nTheorem8.5.1.Inverse function theorem.\n
\n
\n
Let \\(U \\subset \\R^n\\) be an open set and let \\(f \\colon U \\to \\R^n\\) be a continuously differentiable function. Suppose \\(p \\in U\\) and \\(f'(p)\\) is invertible (that is, \\(J_f(p) \\neq 0\\)). Then there exist open sets \\(V, W \\subset \\R^n\\) such that \\(p \\in V \\subset U\\text{,}\\)\\(f(V) = W\\text{,}\\) and \\(f|_V\\) is one-to-one. Hence a function \\(g \\colon W \\to V\\) exists such that \\(g(y) \\coloneqq (f|_V)^{-1}(y)\\text{.}\\) Furthermore, \\(g\\) is continuously differentiable and
\n
\r\n\\begin{equation}\r\ng'(y) = {\\bigl(f'(x)\\bigr)}^{-1}, \\qquad \\text{for all } x \\in V, y = f(x).\r\n\\end{equation}\r\n
\n\n
\n\n\n
Proof.
\n
\n
Write \\(A = f'(p)\\text{.}\\) As \\(f'\\) is continuous, there is an open ball \\(V\\) centered at \\(p\\) such that
\n
\r\n\\begin{equation}\r\n\\bnorm{A-f'(x)} < \\frac{1}{2\\snorm{A^{-1}}}\r\n\\qquad \\text{for all } x \\in V.\r\n\\end{equation}\r\n
\n
Consequently, the derivative \\(f'(x)\\) is invertible for all \\(x \\in V\\) by Proposition 8.2.6.
\n\n
\n
\n
Given \\(y \\in \\R^n\\text{,}\\) define \\(\\varphi_y \\colon V \\to \\R^n\\) by
\n
\r\n\\begin{equation}\r\n\\varphi_y (x) \\coloneqq x + A^{-1}\\bigl(y-f(x)\\bigr) .\r\n\\end{equation}\r\n
\n
As \\(A^{-1}\\) is one-to-one, \\(\\varphi_y(x) = x\\) (\\(x\\) is a fixed point) if and only if \\(y-f(x) = 0\\text{,}\\) or in other words \\(f(x)=y\\text{.}\\) Using the chain rule, we obtain
\r\n\\begin{equation}\r\n\\bnorm{\\varphi_y(x_1)-\\varphi_y(x_2)} \\leq \\frac{1}{2} \\snorm{x_1-x_2} \r\n\\qquad\r\n\\text{for all } x_1,x_2 \\in V.\r\n\\end{equation}\r\n
\n
In other words, \\(\\varphi_y\\) is a contraction defined on \\(V\\text{,}\\) though we so far do not know what is the range of \\(\\varphi_y\\text{.}\\) We cannot yet apply the fixed point theorem, but we can say that \\(\\varphi_y\\) has at most one fixed point in \\(V\\text{:}\\) If \\(\\varphi_y(x_1) = x_1\\) and \\(\\varphi_y(x_2) = x_2\\text{,}\\) then \\(\\snorm{x_1-x_2} = \\bnorm{\\varphi_y(x_1)-\\varphi_y(x_2)} \\leq\r\n\\frac{1}{2} \\snorm{x_1-x_2}\\text{,}\\) so \\(x_1 = x_2\\text{.}\\) That is, there exists at most one \\(x \\in V\\) such that \\(f(x) = y\\text{,}\\) and so \\(f|_V\\) is one-to-one.
\n\n
\n
Let \\(W \\coloneqq f(V)\\) and let \\(g \\colon W \\to V\\) be the inverse of \\(f|_V\\text{.}\\) We need to show that \\(W\\) is open. Take a \\(y_0 \\in W\\text{.}\\) There is a unique \\(x_0 \\in V\\) such that \\(f(x_0) = y_0\\text{.}\\) Let \\(r > 0\\) be small enough such that the closed ball \\(C(x_0,r) \\subset V\\) (such \\(r > 0\\) exists as \\(V\\) is open).\n
So \\(\\varphi_y\\) takes \\(C(x_0,r)\\) into \\(B(x_0,r) \\subset C(x_0,r)\\text{.}\\) It is a contraction on \\(C(x_0,r)\\) and \\(C(x_0,r)\\) is complete (closed subset of \\(\\R^n\\) is complete). Apply the contraction mapping principle to obtain a fixed point \\(x\\text{,}\\) i.e. \\(\\varphi_y(x) = x\\text{.}\\) That is, \\(f(x) = y\\text{,}\\) and \\(y \\in\r\nf\\bigl(C(x_0,r)\\bigr) \\subset f(V) = W\\text{.}\\) Therefore, \\(W\\) is open.
\n\n
\n
Next we need to show that \\(g\\) is continuously differentiable and compute its derivative. First, let us show that it is differentiable. Let \\(y \\in W\\) and \\(k \\in \\R^n\\text{,}\\)\\(k \\neq 0\\text{,}\\) such that \\(y+k \\in W\\text{.}\\) Because \\(f|_V\\) is a one-to-one and onto mapping of \\(V\\) onto \\(W\\text{,}\\) there are unique \\(x \\in V\\) and \\(h \\in \\R^n\\text{,}\\)\\(h \\neq 0\\) and \\(x+h \\in V\\text{,}\\) such that \\(f(x) = y\\) and \\(f(x+h) = y+k\\text{.}\\) In other words, \\(g(y) = x\\) and \\(g(y+k) = x+h\\text{.}\\) See Figure 8.12.\n
\n\nFigure8.12.Proving that \\(g\\) is differentiable.\n\n
\n
We can still squeeze some information from the fact that \\(\\varphi_y\\) is a contraction.
\n
\r\n\\begin{equation}\r\n\\varphi_y(x+h)-\\varphi_y(x) = h + A^{-1} \\bigl( f(x)-f(x+h) \\bigr) = h - A^{-1} k .\r\n\\end{equation}\r\n
In particular, as \\(k\\) goes to 0, so does \\(h\\text{.}\\)\n
\n\n
\n
\n
As \\(x \\in V\\text{,}\\) we find that \\(f'(x)\\) is invertible. Let \\(B \\coloneqq \\bigl(f'(x)\\bigr)^{-1}\\text{,}\\) which is what we think the derivative of \\(g\\) at \\(y\\) is. Then
As \\(k\\) goes to 0, so does \\(h\\text{.}\\) So the right-hand side goes to 0 as \\(f\\) is differentiable, and hence the left-hand side also goes to 0. And \\(B\\) is precisely what we wanted \\(g'(y)\\) to be.
\n\n
\n
We have \\(g\\) is differentiable, let us show it is \\(C^1(W)\\text{.}\\) The function \\(g \\colon W \\to V\\) is continuous (it is differentiable), \\(f'\\) is a continuous function from \\(V\\) to \\(L(\\R^n)\\text{,}\\) and \\(X \\mapsto X^{-1}\\) is a continuous function on the set of invertible operators. As \\(g'(y) = {\\bigl( f'\\bigl(g(y)\\bigr)\\bigr)}^{-1}\\) is the composition of these three continuous functions, it is continuous.\n
\n\n",
"licence": "CC-BY-SA-4.0",
"proof_scope": "C1 inverse theorem, open image, derivative formula and continuity."
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"id": "L8.5.6",
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"author_url": "https://www.jirka.org/ra/html/sec_svinvfuncthm.html",
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\nTheorem8.5.6.Implicit function theorem.\n
\n
\n
Let \\(U \\subset \\R^{n+m}\\) be an open set and let \\(f \\colon U \\to \\R^m\\) be a \\(C^1(U)\\) mapping. Let \\((p,q) \\in U\\) be a point such that \\(f(p,q) = 0\\) and such that
Then there exists an open set \\(W \\subset \\R^n\\) with \\(p \\in W\\text{,}\\) an open set \\(W' \\subset \\R^m\\) with \\(q \\in W'\\text{,}\\) where \\(W \\times W' \\subset U\\text{,}\\) and a \\(C^1(W)\\) map \\(g \\colon W \\to W'\\text{,}\\) with \\(g(p) = q\\text{,}\\) and for all \\(x \\in W\\text{,}\\) the point \\(g(x)\\) is the unique point in \\(W'\\) such that
Define \\(F \\colon U \\to \\R^{n+m}\\) by \\(F(x,y) \\coloneqq \\bigl(x,f(x,y)\\bigr)\\text{.}\\) It is clear that \\(F\\) is \\(C^1\\text{,}\\) and we want to show that its derivative at \\((p,q)\\) is invertible. Let us compute the derivative. The quotient
\n
\r\n\\begin{equation}\r\n\\frac{\\bnorm{f(p+h,q+k) - f(p,q) - A_x h - A_y k}}{\\bnorm{(h,k)}}\r\n\\end{equation}\r\n
\n
goes to zero as \\(\\bnorm{(h,k)} = \\sqrt{\\snorm{h}^2+\\snorm{k}^2}\\) goes to zero. But then so does
So the derivative of \\(F\\) at \\((p,q)\\) takes \\((h,k)\\) to \\((h,A_x h+A_y k)\\text{.}\\) In block matrix form, it is \\(\\left[\\begin{smallmatrix}I & 0\\\\A_x & A_y\\end{smallmatrix}\\right]\\text{.}\\) If \\((h,A_x h+A_y k) = (0,0)\\text{,}\\) then \\(h=0\\text{,}\\) and so \\(A_y k = 0\\text{.}\\) As \\(A_y\\) is one-to-one, \\(k=0\\text{.}\\) Thus \\(F'(p,q)\\) is one-to-one, and hence invertible. We apply the inverse function theorem.
\n\n
\n
\n
That is, there exists an open set \\(V \\subset \\R^{n+m}\\) with \\(F(p,q) = (p,0) \\in V\\text{,}\\) and a \\(C^1\\) mapping \\(G \\colon V \\to \\R^{n+m}\\text{,}\\) such that \\(F\\bigl(G(x,s)\\bigr) = (x,s)\\) for all \\((x,s) \\in V\\text{,}\\)\\(G\\) is one-to-one, and \\(G(V)\\) is open. Write \\(G = (G_1,G_2)\\) (the first \\(n\\) and the next \\(m\\) components of \\(G\\)). Then
As the set \\(G(V)\\) is open and \\((p,q) \\in G(V)\\text{,}\\) there exist some open sets \\(\\widetilde{W}\\) and \\(W'\\) such that \\(\\widetilde{W} \\times W' \\subset G(V)\\) with \\(p\r\n\\in \\widetilde{W}\\) and \\(q \\in W'\\text{.}\\) Take \\(W \\coloneqq \\bigl\\{ x \\in \\widetilde{W} : G_2(x,0) \\in W' \\bigr\\}\\text{.}\\) The function that takes \\(x\\) to \\(G_2(x,0)\\) is continuous and therefore \\(W\\) is open. Define \\(g \\colon W \\to \\R^m\\) by \\(g(x) \\coloneqq G_2(x,0)\\text{,}\\) which is the \\(g\\) in the theorem. The fact that \\(g(x)\\) is the unique point in \\(W'\\) follows because \\(W \\times\r\nW' \\subset G(V)\\) and \\(G\\) is one-to-one.
for all \\(h \\in \\R^{n}\\text{,}\\) and we obtain the desired derivative for \\(g\\text{.}\\)\n
\n\n
\n\n",
"licence": "CC-BY-SA-4.0",
"proof_scope": "C1 implicit theorem on a genuine product neighbourhood, with P10.6 domain correction."
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"id": "L8.3.2",
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\nProposition8.3.2.\n
\n
\n
Let \\(U \\subset \\R^n\\) be an open subset and \\(f \\colon U \\to \\R^m\\) a function. Suppose \\(x \\in U\\) and there exist \\(A,B \\in L(\\R^n,\\R^m)\\) such that
So \\(\\frac{\\snorm{(A-B)h}}{\\snorm{h}} \\to 0\\) as \\(h \\to 0\\text{.}\\) Given \\(\\epsilon > 0\\text{,}\\) for all nonzero \\(h\\) in some \\(\\delta\\)-ball around the origin we have
For any given \\(v \\in \\R^n\\) with \\(\\snorm{v}=1\\text{,}\\) if \\(h = (\\nicefrac{\\delta}{2}) \\, v\\text{,}\\) then \\(\\snorm{h} < \\delta\\) and \\(\\frac{h}{\\snorm{h}} = v\\text{.}\\) So \\(\\bnorm{(A-B)v} < \\epsilon\\text{.}\\) Taking the supremum over all \\(v\\) with \\(\\snorm{v} = 1\\text{,}\\) we get the operator norm \\(\\snorm{A-B} \\leq \\epsilon\\text{.}\\) As \\(\\epsilon > 0\\) was arbitrary, \\(\\snorm{A-B} = 0\\text{,}\\) or in other words \\(A = B\\text{.}\\)\n
\n\n
\n\n",
"licence": "CC-BY-SA-4.0",
"proof_scope": "Uniqueness of the total derivative."
},
{
"id": "L8.3.3",
"reader": "sec_svtheder.html",
"author_url": "https://www.jirka.org/ra/html/sec_svtheder.html",
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\nExample8.3.3.\n
\n
\n
If \\(f(x) = Ax\\) for a linear mapping \\(A\\text{,}\\) then \\(f'(x) = A\\text{:}\\)\n
\n\n",
"licence": "CC-BY-SA-4.0",
"proof_scope": "Full zero-remainder computation for a linear map."
},
{
"id": "L8.3.5",
"reader": "sec_svtheder.html",
"author_url": "https://www.jirka.org/ra/html/sec_svtheder.html",
"source_page_sha256": "d60a9ddf0c5a87c998ccdd108af8b325aa209f82dcf431d249ec34bc1d873062",
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\nProposition8.3.5.\n
\n
Let \\(U \\subset \\R^n\\) be open and \\(f \\colon U \\to \\R^m\\) be differentiable at \\(p \\in U\\text{.}\\) Then \\(f\\) is continuous at \\(p\\text{.}\\)\n
\n\n\n
Proof.
\n
\n
Another way to write the differentiability of \\(f\\) at \\(p\\) is to consider
\n
\r\n\\begin{equation}\r\nr(h) \\coloneqq f(p+h)-f(p) - f'(p) h .\r\n\\end{equation}\r\n
\n
The function \\(f\\) is differentiable at \\(p\\) if \\(\\frac{\\snorm{r(h)}}{\\snorm{h}}\\) goes to zero as \\(h \\to 0\\text{,}\\) so \\(r(h)\\) itself goes to zero. The mapping \\(h \\mapsto f'(p) h\\) is a linear mapping between finite-dimensional spaces, hence continuous and \\(f'(p) h \\to 0\\) as \\(h \\to 0\\text{.}\\) Thus, \\(f(p+h)\\) must go to \\(f(p)\\) as \\(h \\to 0\\text{.}\\) That is, \\(f\\) is continuous at \\(p\\text{.}\\)\n
Suppose \\(U \\subset \\R^n\\) is open, \\(f \\colon U \\to \\R^m\\) and \\(g \\colon U \\to \\R^m\\) are differentiable at \\(p \\in U\\text{,}\\) and \\(\\alpha \\in \\R\\text{.}\\) Then the functions \\(f+g\\) and \\(\\alpha f\\) are differentiable at \\(p\\text{,}\\)\n
The limits as \\(h\\) goes to zero of the right-hand sides are zero by hypothesis. The result follows.
\n\n
\n\n",
"licence": "CC-BY-SA-4.0",
"proof_scope": "Derivative sum and scalar rules; extra-parenthesis typo explicitly corrected in P10.2."
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{
"id": "L8.3.7",
"reader": "sec_svtheder.html",
"author_url": "https://www.jirka.org/ra/html/sec_svtheder.html",
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\nTheorem8.3.7.Chain rule.\n
\n
\n
Let \\(U \\subset \\R^n\\) and \\(V \\subset \\R^m\\) be open sets, \\(f \\colon U \\to\r\n\\R^m\\) be differentiable at \\(p \\in U\\text{,}\\)\\(f(U) \\subset V\\text{,}\\) and let \\(g \\colon V \\to \\R^\\ell\\) be differentiable at \\(f(p)\\text{.}\\) Then \\(F \\colon U \\to \\R^{\\ell}\\) defined by
Let \\(A \\coloneqq f'(p)\\) and \\(B \\coloneqq g'\\bigl(f(p)\\bigr)\\text{.}\\) Take a nonzero \\(h \\in \\R^n\\) and write \\(q \\coloneqq f(p)\\text{,}\\)\\(k \\coloneqq f(p+h)-f(p)\\text{.}\\) Let
\n
\r\n\\begin{equation}\r\nr(h) \\coloneqq f(p+h)-f(p) - A h . \r\n\\end{equation}\r\n
\n
Then \\(r(h) = k-Ah\\) or \\(Ah = k-r(h)\\text{,}\\) and \\(f(p+h) = q+k\\text{.}\\) We look at the quantity we need to go to zero:
We need both terms on the right to go to 0 as \\(h\\) goes to 0. First, \\(\\snorm{B}\\) is a constant and \\(f\\) is differentiable at \\(p\\text{,}\\) so the term \\(\\snorm{B}\\frac{\\snorm{r(h)}}{\\snorm{h}}\\) goes to 0. Next, if \\(k=0\\text{,}\\) then \\(\\frac\r\n{\\snorm{g(q+k)-g(q) - Bk}}\r\n{\\snorm{h}} = 0\\text{.}\\) So suppose that \\(k \\neq 0\\text{.}\\) Then
Because \\(f\\) is continuous at \\(p\\text{,}\\)\\(k\\) goes to 0 as \\(h\\) goes to 0. Thus \\(\\frac\r\n{\\snorm{g(q+k)-g(q) - Bk}}\r\n{\\snorm{k}}\\) goes to 0, because \\(g\\) is differentiable at \\(q\\text{.}\\) We have,
As \\(f\\) is differentiable at \\(p\\text{,}\\) for small enough \\(h\\text{,}\\) the quantity \\(\\frac{\\snorm{f(p+h)-f(p)-Ah}}{\\snorm{h}}\\) is bounded. Hence, the term \\(\\frac\r\n{\\snorm{f(p+h)-f(p)}}\r\n{\\snorm{h}}\\) stays bounded as \\(h\\) goes to 0. In other words, the term \\(\\frac\r\n{\\snorm{g(q+k)-g(q) - Bk}}\r\n{\\snorm{h}}\\) goes to zero as \\(h\\) goes to 0. Therefore, \\(\\frac{\\snorm{F(p+h)-F(p) - BAh}}{\\snorm{h}}\\) goes to zero, and \\(F'(p) = BA\\text{,}\\) which is what was claimed.
Let \\(U \\subset \\R^n\\) be open and let \\(f \\colon U \\to \\R^m\\) be differentiable at \\(p \\in U\\text{.}\\) Then all the partial derivatives at \\(p\\) exist and, in terms of the standard bases of \\(\\R^n\\) and \\(\\R^m\\text{,}\\)\\(f'(p)\\) is represented by the matrix
The limit is in \\(\\R^m\\text{.}\\) Represent \\(f\\) in components \\(f = (f_1,f_2,\\ldots,f_m)\\text{.}\\) Taking a limit in \\(\\R^m\\) is the same as taking the limit in each component separately. So for every \\(k\\text{,}\\) the partial derivative
exists and is equal to the \\(k\\)th component of \\(f'(p)\\, e_j\\text{,}\\) which is the \\(j\\)th column of \\(f'(p)\\text{,}\\) and we are done.
\n\n
\n\n",
"licence": "CC-BY-SA-4.0",
"proof_scope": "Partial derivatives as columns of the total derivative."
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"id": "L8.4.1",
"reader": "sec_svthedercont.html",
"author_url": "https://www.jirka.org/ra/html/sec_svthedercont.html",
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\nLemma8.4.1.\n
\n
\n
If \\(\\varphi \\colon [a,b] \\to \\R^n\\) is differentiable on \\((a,b)\\) and continuous on \\([a,b]\\text{,}\\) then there exists a \\(t_0 \\in (a,b)\\) such that
By the mean value theorem on the scalar-valued function \\(t \\mapsto \\bigl(\\varphi(b)-\\varphi(a) \\bigr) \\cdot \\varphi(t)\\text{,}\\) where the dot is the dot product, we obtain a \\(t_0 \\in (a,b)\\) such that
where we treat \\(\\varphi'\\) as a vector in \\(\\R^n\\) by the abuse of notation we mentioned in the previous section. If we think of \\(\\varphi'(t)\\) as a vector, then by Exercise 8.2.6, \\(\\bnorm{\\varphi'(t)}_{L(\\R,\\R^n)} = \\bnorm{\\varphi'(t)}_{\\R^n}\\text{.}\\) That is, the euclidean norm of the vector is the same as the operator norm of \\(\\varphi'(t)\\text{.}\\)\n
\n\n",
"licence": "CC-BY-SA-4.0",
"proof_scope": "Derivative bound implies Lipschitz bound on a convex open domain."
},
{
"id": "L8.4.6",
"reader": "sec_svthedercont.html",
"author_url": "https://www.jirka.org/ra/html/sec_svthedercont.html",
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\nProposition8.4.6.\n
\n
Let \\(U \\subset \\R^n\\) be open and \\(f \\colon U \\to \\R^m\\text{.}\\) The function \\(f\\) is continuously differentiable if and only if the partial derivatives \\(\\frac{\\partial f_k}{\\partial x_j}\\) exist for all \\(k\\) and \\(j\\) and are continuous.\n
\n\n\n
Proof.
\n
We proved that if \\(f\\) is differentiable, then the partial derivatives exist. The partial derivatives are the entries of the matrix representing \\(f'(x)\\text{.}\\) If \\(f' \\colon U \\to L(\\R^n,\\R^m)\\) is continuous, then the entries are continuous, and hence the partial derivatives are continuous.\n
\n
To prove the opposite direction, suppose the partial derivatives exist and are continuous. Fix \\(x \\in U\\text{.}\\) If we show that \\(f'(x)\\) exists, we are done, because the entries of the matrix representing \\(f'(x)\\) are the partial derivatives and if the entries are continuous functions, the matrix-valued function \\(f'\\) is continuous.\n
\n
We do induction on dimension. First, the conclusion is true when \\(n=1\\) (exercise, note that \\(f\\) is vector-valued). In this case, \\(f'(x)\\) is essentially the derivative of Chapter 4. Suppose the conclusion is true for \\(\\R^{n-1}\\text{.}\\) That is, if we restrict to the first \\(n-1\\) variables, the function is differentiable. When taking the partial derivatives in \\(x_1\\) through \\(x_{n-1}\\text{,}\\) it does not matter if we consider \\(f\\) or \\(f\\) restricted to the set where \\(x_n\\) is fixed. In the following, by a slight abuse of notation, we think of \\(\\R^{n-1}\\) as a subset of \\(\\R^n\\text{,}\\) that is, the set in \\(\\R^n\\) where \\(x_n = 0\\text{.}\\) In other words, we identify the vectors \\((x_1,x_2,\\ldots,x_{n-1})\\) and \\((x_1,x_2,\\ldots,x_{n-1},0)\\text{.}\\)\n
Let \\(\\epsilon > 0\\) be given. By the induction hypothesis, there is a \\(\\delta > 0\\) such that for every \\(h' \\in \\R^{n-1}\\) with \\(\\snorm{h'} < \\delta\\text{,}\\) we have
\n
\r\n\\begin{equation}\r\n\\frac{\\bnorm{f(p+h') - f(p) - A' h'}}{\\snorm{h'}} < \\epsilon .\r\n\\end{equation}\r\n
\n
By continuity of the partial derivatives, suppose \\(\\delta\\) is small enough so that
for all \\(k\\) and all \\(h \\in \\R^n\\) with \\(\\snorm{h} < \\delta\\text{.}\\)\n
\n\n
\n
\n
Suppose \\(h = h' + t e_n\\) is a vector in \\(\\R^n\\text{,}\\) where \\(h' \\in \\R^{n-1}\\text{,}\\)\\(t \\in \\R\\text{,}\\) such that \\(\\snorm{h} < \\delta\\text{.}\\) Then \\(\\snorm{h'} \\leq \\snorm{h} < \\delta\\text{.}\\) Note that \\(Ah = A' h' + tv\\text{.}\\)\n
\n
\r\n\\begin{equation}\r\n\\begin{split}\r\n&\r\n\\bnorm{f(p+h) - f(p) - Ah}\r\n\\\\\r\n& \\qquad \\qquad\r\n= \\bnorm{f(p+h' + t e_n) - f(p+h') - tv + f(p+h') - f(p) - A' h'}\r\n\\\\\r\n&\r\n\\qquad \\qquad\r\n\\leq \\bnorm{f(p+h' + t e_n) - f(p+h') -tv} + \\bnorm{f(p+h') - f(p) -\r\nA' h'}\r\n\\\\\r\n&\r\n\\qquad \\qquad\r\n\\leq \\bnorm{f(p+h' + t e_n) - f(p+h') -tv} + \\epsilon \\snorm{h'} .\r\n\\end{split}\r\n\\end{equation}\r\n
\n
As all the partial derivatives exist, by the mean value theorem, for each \\(k\\) there is some \\(\\theta_k \\in [0,t]\\) (or \\([t,0]\\) if \\(t < 0\\)), such that
\n\n",
"licence": "CC-BY-SA-4.0",
"proof_scope": "C1 iff all partial derivatives exist and are continuous."
},
{
"id": "L8.1.13",
"reader": "sec_vectorspaces.html",
"author_url": "https://www.jirka.org/ra/html/sec_vectorspaces.html",
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\nProposition8.1.13.\n
\n
\n
Suppose a vector space \\(X\\) has a basis \\(B = \\{ x_1, x_2, \\ldots, x_n \\}\\text{.}\\) Then every \\(y \\in X\\) has a unique representation of the form
Let \\(X\\) be a vector space and \\(d\\) a nonnegative integer.\n
\n\n
\n
If \\(X\\) is spanned by \\(d\\) vectors, then \\(\\dim \\, X \\leq d\\text{.}\\)\n
\n\n
\n
\n
If \\(T\\) is a linearly independent set and \\(v \\in X \\setminus \\spn (T)\\text{,}\\) then \\(T \\cup \\{ v \\}\\) is linearly independent.\n
\n\n
\n
\n
\n\\(\\dim \\, X = d\\) if and only if \\(X\\) has a basis of \\(d\\) vectors. In particular, \\(\\dim \\, \\R^n = n\\text{.}\\)\n
\n\n
\n
\n
If \\(Y \\subset X\\) is a vector subspace and \\(\\dim \\, X = d\\text{,}\\) then \\(\\dim \\, Y \\leq d\\text{.}\\)\n
\n\n
\n
\n
If \\(\\dim \\, X = d\\) and a set \\(T\\) of \\(d\\) vectors spans \\(X\\text{,}\\) then \\(T\\) is linearly independent.\n
\n\n
\n
\n
If \\(\\dim \\, X = d\\) and a set \\(T\\) of \\(m\\) vectors is linearly independent, then there is a set \\(S\\) of \\(d-m\\) vectors such that \\(T \\cup S\\) is a basis of \\(X\\text{.}\\)\n
\n\n
\n\n\n\n
Proof.
\n
All statements hold trivially when \\(d=0\\text{,}\\) so assume \\(d \\geq 1\\text{.}\\)\n
\n
\n
We start with i. Suppose \\(S \\coloneqq \\{ x_1 , x_2, \\ldots, x_d \\}\\) spans \\(X\\text{,}\\) and \\(T \\coloneqq \\{ y_1, y_2, \\ldots, y_m \\}\\) is a linearly independent subset of \\(X\\text{.}\\) We wish to show that \\(m \\leq d\\text{.}\\) As \\(S\\) spans \\(X\\text{,}\\) write
for some numbers \\(a_{1,1},a_{2,1},\\ldots,a_{d,1}\\text{.}\\) One of the \\(a_{k,1}\\) is nonzero, otherwise \\(y_1\\) would be zero. Without loss of generality, suppose \\(a_{1,1} \\neq 0\\text{.}\\) Solve
In particular, \\(\\{ y_1 , x_2, \\ldots, x_d \\}\\) spans \\(X\\text{,}\\) since \\(x_1\\) can be obtained from \\(\\{ y_1 , x_2, \\ldots, x_d \\}\\text{.}\\) Therefore, there are some numbers \\(a_{1,2},a_{2,2},\\ldots,a_{d,2}\\text{,}\\) such that
As \\(T\\) is linearly independent—and so \\(\\{ y_1, y_2 \\}\\) is linearly independent—one of the \\(a_{k,2}\\) for \\(k \\geq 2\\) must be nonzero. Without loss of generality suppose \\(a_{2,2} \\neq 0\\text{.}\\) Solve
We continue this procedure. If \\(m < d\\text{,}\\) we are done. Suppose \\(m \\geq d\\text{.}\\) After \\(d\\) steps, we obtain that \\(\\{ y_1 , y_2, \\ldots, y_d \\}\\) spans \\(X\\text{.}\\) Any other vector \\(v\\) in \\(X\\) is a linear combination of \\(\\{ y_1 , y_2, \\ldots, y_d \\}\\) and hence cannot be in \\(T\\) as \\(T\\) is linearly independent. So \\(m = d\\text{.}\\)\n
\n
We continue with ii. Suppose \\(T = \\{x_1,x_2,\\ldots,x_m\\}\\) is linearly independent, does not span \\(X\\text{,}\\) and \\(v \\in X \\setminus \\spn (T)\\text{.}\\) Suppose \\(a_1 x_1 + a_2 x_2 + \\cdots + a_m x_m + a_{m+1} v = 0\\) for some scalars \\(a_1,a_2,\\ldots,a_{m+1}\\text{.}\\) If \\(a_{m+1} \\neq 0\\text{,}\\) then \\(v\\) would be a linear combination of \\(T\\text{,}\\) so \\(a_{m+1} = 0\\text{.}\\) Then, as \\(T\\) is linearly independent, \\(a_1=a_2=\\cdots=a_m = 0\\text{.}\\) So \\(T \\cup \\{ v \\}\\) is linearly independent.\n
\n
We move to iii. If \\(\\dim \\, X = d\\text{,}\\) then there must exist some linearly independent set \\(T\\) of \\(d\\) vectors, and \\(T\\) must span \\(X\\text{,}\\) otherwise we could choose a larger set of linearly independent vectors via ii. So we have a basis of \\(d\\) vectors. On the other hand, if we have a basis of \\(d\\) vectors, the dimension is at least \\(d\\) as a basis is linearly independent. A basis also spans \\(X\\text{,}\\) and so by i we know that dimension is at most \\(d\\text{.}\\) Hence the dimension of \\(X\\) must equal \\(d\\text{.}\\) The “in particular” follows by noting that \\(\\{ e_1, e_2, \\ldots, e_n \\}\\) is a basis of \\(\\R^n\\text{.}\\)\n
\n
To see iv, suppose \\(Y \\subset X\\) is a vector subspace, where \\(\\dim \\, X = d\\text{.}\\) As \\(X\\) cannot contain \\(d+1\\) linearly independent vectors, neither can \\(Y\\text{.}\\)\n
\n
For v, suppose \\(T\\) is a set of \\(m\\) vectors that is linearly dependent and spans \\(X\\text{.}\\) We will show that \\(m > d\\text{.}\\) One of the vectors is a linear combination of the others. If we remove it from \\(T\\text{,}\\) we obtain a set of \\(m-1\\) vectors that still span \\(X\\text{.}\\) Hence \\(d = \\dim \\, X \\leq m-1\\) by i.\n
\n
For vi suppose \\(T = \\{ x_1, x_2, \\ldots, x_m \\}\\) is a linearly independent set. First, \\(m \\leq d\\) by definition of dimension. If \\(m=d\\text{,}\\) the set \\(T\\) must span \\(X\\) as in the proof of iii, otherwise we could add another vector to \\(T\\text{.}\\) If \\(m < d\\text{,}\\)\\(T\\) cannot span \\(X\\) by iii. So find \\(v\\) not in the span of \\(T\\text{.}\\) Via ii, the set \\(T \\cup \\{ v \\}\\) is a linearly independent set of \\(m+1\\) elements. Therefore, we repeat this procedure \\(d-m\\) times to find a set of \\(d\\) linearly independent vectors. Again, they must span \\(X\\text{,}\\) otherwise we could add yet another vector.\n
\n\n",
"licence": "CC-BY-SA-4.0",
"proof_scope": "All six finite-dimensional exchange, basis and subspace assertions."
},
{
"id": "L8.1.16",
"reader": "sec_vectorspaces.html",
"author_url": "https://www.jirka.org/ra/html/sec_vectorspaces.html",
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\nProposition8.1.16.\n
\n
Let \\(X\\text{,}\\)\\(Y\\text{,}\\) and \\(Z\\) be vector spaces.\n
\n\n
\n
If \\(A \\in L(X,Y)\\text{,}\\) then \\(A0 = 0\\text{.}\\)\n
\n\n
\n
\n
If \\(A,B \\in L(X,Y)\\text{,}\\) then \\(A+B \\in L(X,Y)\\text{.}\\)\n
\n\n
\n
\n
If \\(A \\in L(X,Y)\\) and \\(a \\in \\R\\text{,}\\) then \\(aA \\in L(X,Y)\\text{.}\\)\n
\n\n
\n
\n
If \\(A \\in L(Y,Z)\\) and \\(B \\in L(X,Y)\\text{,}\\) then \\(AB \\in L(X,Z)\\text{.}\\)\n
\n\n
\n
\n
If \\(A \\in L(X,Y)\\) is invertible, then \\(A^{-1} \\in L(Y,X)\\text{.}\\)\n
\n\n
\n\n\n\n
Proof.
\n
\n
We leave the first four items as a quick exercise, Exercise 8.1.20. Let us prove the last item. Let \\(a \\in \\R\\) and \\(y \\in Y\\text{.}\\) As \\(A\\) is onto, there is an \\(x \\in X\\) such that \\(y = Ax\\text{.}\\) As it is also one-to-one, \\(A^{-1}(Az) = z\\) for all \\(z \\in X\\text{.}\\) So
\n\n",
"licence": "CC-BY-SA-4.0",
"proof_scope": "Inverse linearity, with the other four parts completed locally."
},
{
"id": "L8.1.17",
"reader": "sec_vectorspaces.html",
"author_url": "https://www.jirka.org/ra/html/sec_vectorspaces.html",
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\nProposition8.1.17.\n
\n
If \\(A \\in L(X,Y)\\) is linear, then it is completely determined by its values on a basis of \\(X\\text{.}\\) Furthermore, if \\(B\\) is a basis of \\(X\\text{,}\\) then every function \\(\\widetilde{A} \\colon B \\to Y\\) extends to a linear function \\(A\\) on \\(X\\text{.}\\)\n
\n\n\n
Proof.
\n
\n
Let \\(\\{ x_1, x_2, \\ldots, x_n \\}\\) be a basis of \\(X\\text{,}\\) and let \\(y_k \\coloneqq A x_k\\text{.}\\) Every \\(x \\in X\\) has a unique representation
The “furthermore” follows by setting \\(y_k \\coloneqq \\widetilde{A}(x_k)\\text{,}\\) and then for \\(x = \\sum_{k=1}^n b_k \\, x_k\\text{,}\\) defining the extension as \\(A(x) \\coloneqq \\sum_{k=1}^n b_k \\, y_k\\text{.}\\) The function is well-defined by uniqueness of the representation of \\(x\\text{.}\\) We leave it to the reader to check that \\(A\\) is linear.
\n\n
\n\n",
"licence": "CC-BY-SA-4.0",
"proof_scope": "Determination and extension from basis values, including the omitted linearity check."
},
{
"id": "L8.1.18",
"reader": "sec_vectorspaces.html",
"author_url": "https://www.jirka.org/ra/html/sec_vectorspaces.html",
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\nProposition8.1.18.\n
\n
If \\(X\\) is a finite-dimensional vector space and \\(A \\in L(X)\\text{,}\\) then \\(A\\) is one-to-one if and only if it is onto.\n
\n\n\n
Proof.
\n
\n
Let \\(\\{ x_1,x_2,\\ldots,x_n \\}\\) be a basis for \\(X\\text{.}\\) First suppose \\(A\\) is one-to-one. Let \\(c_1,c_2,\\ldots,c_n\\) be scalars such that
and so \\(c_k = 0\\) for all \\(k\\) as \\(\\{ x_1,x_2,\\ldots,x_n \\}\\) is a basis. So \\(\\{ Ax_1, Ax_2, \\ldots, Ax_n \\}\\) is linearly independent. By Proposition 8.1.14 and the fact that the dimension is \\(n\\text{,}\\) we conclude \\(\\{ Ax_1, Ax_2, \\ldots, Ax_n \\}\\) spans \\(X\\text{.}\\) Consequently, \\(A\\) is onto, as any \\(y \\in X\\) can be written as
As \\(A\\) is determined by the action on the basis, \\(\\{ Ax_1, Ax_2, \\ldots, Ax_n \\}\\) spans \\(X\\text{.}\\) So by Proposition 8.1.14, the set is linearly independent, and \\(c_k = 0\\) for all \\(k\\text{.}\\) In other words, if \\(Ax = 0\\text{,}\\) then \\(x=0\\text{.}\\) Thus, \\(A\\) is one-to-one.
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