Mathematics Program · Real Analysis II · Basic Analysis I & II · Jiří Lebl · sources and attribution

Section 6.1 Pointwise and uniform convergence

Note: 1–1.5 lecture
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Up till now, when we talked about limits of sequences we talked about sequences of numbers. A very useful concept in analysis is a sequence of functions. For example, a solution to some differential equation might be found by finding only approximate solutions. Then the actual solution is some sort of limit of those approximate solutions.
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When talking about sequences of functions, the tricky part is that there are multiple notions of a limit. Let us describe two common notions of a limit of a sequence of functions.
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Subsection 6.1.1 Pointwise convergence

Definition 6.1.1.

For every \(n \in \N\text{,}\) let \(f_n \colon S \to \R\) be a function. The sequence \(\{ f_n \}_{n=1}^\infty\) converges pointwise
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Unless otherwise specified, converges generally means converges pointwise.
to \(f \colon S \to \R\) if for every \(x \in S\text{,}\) we have
\begin{equation} f(x) = \lim_{n\to\infty} f_n(x) . \end{equation}
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Limits of sequences of numbers are unique, and so if a sequence \(\{ f_n \}_{n=1}^\infty\) converges pointwise, the limit function \(f\) is unique. It is common to say that \(f_n \colon S \to \R\) converges pointwise to \(f\) on \(T \subset S\) for some \(f \colon T \to \R\text{.}\) In that case we mean \(f(x) = \lim_{n\to\infty} f_n(x)\) for every \(x \in T\text{.}\) In other words, the restrictions of \(f_n\) to \(T\) converge pointwise to \(f\text{.}\)
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Example 6.1.2.

On \([-1,1]\text{,}\) the sequence of functions defined by \(f_n(x) \coloneqq x^{2n}\) converges pointwise to \(f \colon [-1,1] \to \R\text{,}\) where
\begin{equation} f(x) = \begin{cases} 1 & \text{if } x=-1 \text{ or } x=1, \\ 0 & \text{otherwise.} \end{cases} \end{equation}
See Figure 6.1.
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Graphs of the functions x squared, x to the fourth, x to the sixth, and x to the sixteenth. The graphs are over the region from minus 1 to 1, and all start at the same point on the left-hand side and all end at the same point on the right-hand side (as the functions are all 1 at the endpoints). Inside the interval the higher powers make the graph flatter and closer to the x-axis in the middle, but they have to then rise more steeply back to 1 when we get to the ends of the interval.
Figure 6.1. Graphs of \(f_1\text{,}\) \(f_2\text{,}\) \(f_3\text{,}\) and \(f_8\) for \(f_n(x) \coloneqq x^{2n}\text{.}\)
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To see this is so, first take \(x \in (-1,1)\text{.}\) Then \(0 \leq x^2 < 1\text{.}\) We have seen before that
\begin{equation} \sabs{x^{2n} - 0} = {(x^2)}^n \to 0 \quad \text{as} \quad n \to \infty . \end{equation}
Therefore, \(\lim_{n\to\infty}f_n(x) = 0\text{.}\) When \(x = 1\) or \(x=-1\text{,}\) we have \(x^{2n} = 1\) for all \(n\) and hence \(\lim_{n\to\infty}f_n(x) = 1\text{.}\)
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Note that for \(x\) not in \([-1,1]\text{,}\) the sequence \(\{ x^{2n} \}_{n=1}^\infty\) does not converge. So if we defined \(f_n\) on a set larger than \([-1,1]\text{,}\) then \(\{ f_n \}_{n=1}^\infty\) would converge pointwise only on \([-1,1]\text{.}\)
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Often, functions are given as a series. In this case, we use the notion of pointwise convergence to find the values of the function.
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Example 6.1.3.

We write
\begin{equation} \sum_{k=0}^\infty x^k \end{equation}
to denote the limit of the functions
\begin{equation} f_n(x) \coloneqq \sum_{k=0}^n x^k . \end{equation}
When studying series, we saw that for \((-1,1)\) the \(f_n\) converge pointwise to
\begin{equation} \frac{1}{1-x} . \end{equation}
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The subtle point here is that while \(\frac{1}{1-x}\) is defined for all \(x \neq 1\text{,}\) and \(f_n\) are defined for all \(x\) (even at \(x=1\)), convergence only happens on \((-1,1)\text{.}\) Therefore, when we write
\begin{equation} f(x) \coloneqq \sum_{k=0}^\infty x^k , \end{equation}
we mean that \(f\) is defined on \((-1,1)\) and is the pointwise limit of the partial sums.
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Example 6.1.4.

Let \(f_n(x) \coloneqq \sin(nx)\text{.}\) Then \(f_n\) does not converge pointwise to any function on any interval. It may converge at certain points, such as when \(x=0\) or \(x=\pi\text{.}\) It is left as an exercise that in any interval \([a,b]\text{,}\) there exists an \(x\) such that \(\sin(xn)\) does not have a limit as \(n\) goes to infinity. See Figure 6.2.
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Graphs of a number of sine waves, with higher frequencies in lighter gray. It is almost like we are trying to fill out a horizontal strip by drawing wavy lines. All the curves start at the same point on the x-axis on the left-hand side going upwards, and all go through same point on the x-axis on the right hand side going up. All the curves go through the center on the x-axis, but some upwards and some downwards. Similar patterns repeat at other points, but only some of the higher frequency lines go through those points.
Figure 6.2. Graphs of \(\sin(nx)\) for \(n=1,2,\ldots,10\text{,}\) with higher \(n\) in fainter gray.
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Before we move to uniform convergence, let us reformulate pointwise convergence in a different way. We leave the proof to the reader—it is a simple application of the definition of convergence of a sequence of real numbers.
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The key point is that \(N\) can depend on \(x\text{,}\) not just on \(\epsilon\text{.}\) For each \(x\text{,}\) we can pick a different \(N\text{.}\) If we could pick one \(N\) for all \(x\text{,}\) we would have what is called uniform convergence.
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Subsection 6.1.2 Uniform convergence

Definition 6.1.6.

Let \(f_n \colon S \to \R\) and \(f \colon S \to \R\) be functions. The sequence \(\{ f_n \}_{n=1}^\infty\) converges uniformly to \(f\) if for every \(\epsilon > 0\text{,}\) there exists an \(N \in \N\) such that for all \(n \geq N\text{,}\)
\begin{equation} \babs{f_n(x) - f(x)} < \epsilon \qquad \text{for all } x \in S. \end{equation}
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A graph of a function marked f with a strip marked around the function with its boundary marked f minus epsilon and f plus epsilon. A function marked f sub n is shown as a wiggly line that fits entirely within this strip.
Figure 6.3. In uniform convergence, for \(n \geq N\text{,}\) the functions \(f_n\) are within a strip of \(\pm\epsilon\) from \(f\text{.}\)
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In uniform convergence, \(N\) cannot depend on \(x\text{.}\) Given \(\epsilon > 0\text{,}\) we must find an \(N\) that works for all \(x \in S\text{.}\) See Figure 6.3 for an illustration. Uniform convergence implies pointwise convergence, and the proof follows by Proposition 6.1.5:
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The converse does not hold.
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Example 6.1.8.

The functions \(f_n(x) \coloneqq x^{2n}\) do not converge uniformly on \([-1,1]\text{,}\) even though they converge pointwise. To see this, suppose for contradiction that the convergence is uniform. For \(\epsilon \coloneqq \nicefrac{1}{2}\text{,}\) there would have to exist an \(N\) such that \(x^{2N} = \sabs{x^{2N} - 0} < \nicefrac{1}{2}\) for all \(x \in (-1,1)\) (as \(f_n(x)\) converges to 0 on \((-1,1)\)). But that means that for every sequence \(\{ x_k \}_{k=1}^\infty\) in \((-1,1)\) such that \(\lim_{k\to\infty} x_k = 1\text{,}\) we have \(x_k^{2N} < \nicefrac{1}{2}\) for all \(k\text{.}\) On the other hand, \(x^{2N}\) is a continuous function of \(x\) (it is a polynomial). Therefore, we obtain a contradiction
\begin{equation} 1 = 1^{2N} = \lim_{k\to\infty} x_k^{2N} \leq \nicefrac{1}{2} . \end{equation}
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However, if we restrict our domain to \([-a,a]\) where \(0 < a < 1\text{,}\) then \(\{ f_n \}_{n=1}^\infty\) converges uniformly to 0 on \([-a,a]\text{.}\) Note that \(a^{2n} \to 0\) as \(n \to \infty\text{.}\) Given \(\epsilon > 0\text{,}\) pick \(N \in \N\) such that \(a^{2n} < \epsilon\) for all \(n \geq N\text{.}\) If \(x \in [-a,a]\text{,}\) then \(\sabs{x} \leq a\text{.}\) So for all \(n \geq N\) and all \(x \in [-a,a]\text{,}\)
\begin{equation} \sabs{x^{2n}} = \sabs{x}^{2n} \leq a^{2n} < \epsilon . \end{equation}
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Subsection 6.1.3 Convergence in uniform norm

For bounded functions, there is another more abstract way to think of uniform convergence. To every bounded function we assign a certain nonnegative number that measures the “distance” of the function from the constant function \(0\text{.}\) This number allows us to “measure” how far two functions are from each other. We then translate a statement about uniform convergence into a statement about a certain sequence of real numbers converging to zero.
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Definition 6.1.9.

Let \(f \colon S \to \R\) be a bounded function. Define
\begin{equation} \snorm{f}_S \coloneqq \sup \Bigl\{ \babs{f(x)} : x \in S \Bigr\} . \end{equation}
We call \(\snorm{\cdot}_S\) the uniform norm. Sometimes other notation
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The notation and terminology are not completely standardized. The norm is also called the sup norm or infinity norm, and in addition to \(\snorm{f}_u\) and \(\snorm{f}_S\) it is sometimes written as \(\snorm{f}_{\infty}\) or \(\snorm{f}_{\infty,S}\text{.}\)
is used, such as \(\snorm{f}_u\text{.}\)
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The subscript is the set over which the supremum is taken. So if \(K \subset S\text{,}\) then
\begin{equation} \snorm{f}_K = \sup \Bigl\{ \babs{f(x)} : x \in K \Bigr\} . \end{equation}
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Proof.

First suppose \(\lim_{n\to\infty} \snorm{f_n - f}_S = 0\text{.}\) Let \(\epsilon > 0\) be given. There exists an \(N\) such that for \(n \geq N\text{,}\) we have \(\snorm{f_n - f}_S < \epsilon\text{.}\) As \(\snorm{f_n-f}_S\) is the supremum of \(\babs{f_n(x)-f(x)}\text{,}\) we see that for all \(x \in S\text{,}\) we have \(\babs{f_n(x)-f(x)} \leq \snorm{f_n - f}_S < \epsilon\text{.}\)
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On the other hand, suppose \(\{ f_n \}_{n=1}^\infty\) converges uniformly to \(f\text{.}\) Let \(\epsilon > 0\) be given. Then find \(N\) such that for all \(n \geq N\text{,}\) we have \(\babs{f_n(x)-f(x)} < \epsilon\) for all \(x \in S\text{.}\) Taking the supremum over \(x \in S\text{,}\) we see that \(\snorm{f_n - f}_S \leq \epsilon\text{.}\) Hence, \(\lim_{n\to\infty} \snorm{f_n-f}_S = 0\text{.}\)
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Sometimes it is said that \(\{ f_n \}_{n=1}^\infty\) converges to \(f\) in uniform norm instead of converges uniformly if \(\snorm{f_n-f}_S \to 0\text{.}\) The proposition says that the two notions are the same thing for bounded functions.
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Example 6.1.11.

Let \(f_n \colon [0,1] \to \R\) be defined by \(f_n(x) \coloneqq \frac{nx+ \sin(nx^2)}{n}\text{.}\) We claim \(\{ f_n \}_{n=1}^\infty\) converges uniformly to \(f(x) \coloneqq x\text{.}\) Let us compute:
\begin{equation} \begin{split} \snorm{f_n-f}_{[0,1]} & = \sup \left\{ \abs{\frac{nx+ \sin(nx^2)}{n} - x} : x \in [0,1] \right\} \\ & = \sup \left\{ \frac{\babs{\sin(nx^2)}}{n} : x \in [0,1] \right\} \\ & \leq \sup \bigl\{ \nicefrac{1}{n} : x \in [0,1] \bigr\} \\ & = \nicefrac{1}{n}. \end{split} \end{equation}
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Using the uniform norm, we define Cauchy sequences similarly to how we define Cauchy sequences of real numbers.
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Definition 6.1.12.

Let \(f_n \colon S \to \R\) be bounded functions. The sequence is Cauchy in the uniform norm or uniformly Cauchy if for every \(\epsilon > 0\text{,}\) there exists an \(N \in \N\) such that for all \(m,k \geq N\text{,}\)
\begin{equation} \snorm{f_m-f_k}_S < \epsilon . \end{equation}
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Proof.

First suppose \(\{ f_n \}_{n=1}^\infty\) is Cauchy in the uniform norm. Let us define \(f\text{.}\) Fix \(x\text{.}\) The sequence \(\bigl\{ f_n(x) \bigr\}_{n=1}^\infty\) is Cauchy because
\begin{equation} \babs{f_m(x)-f_k(x)} \leq \snorm{f_m-f_k}_S . \end{equation}
Thus \(\bigl\{ f_n(x) \bigr\}_{n=1}^\infty\) converges to some real number. Define \(f \colon S \to \R\) by
\begin{equation} f(x) \coloneqq \lim_{n \to \infty} f_n(x) . \end{equation}
The sequence \(\{ f_n \}_{n=1}^\infty\) converges pointwise to \(f\text{.}\) To show that the convergence is uniform, let \(\epsilon > 0\) be given. Find an \(N\) such that for all \(m, k \geq N\text{,}\) we have \(\snorm{f_m-f_k}_S < \nicefrac{\epsilon}{2}\text{.}\) In other words, for all \(x\text{,}\) we have \(\babs{f_m(x)-f_k(x)} < \nicefrac{\epsilon}{2}\text{.}\) For any fixed \(x\text{,}\) take the limit as \(k\) goes to infinity. Then \(\babs{f_m(x)-f_k(x)}\) goes to \(\babs{f_m(x)-f(x)}\text{.}\) Consequently for all \(x\text{,}\)
\begin{equation} \babs{f_m(x)-f(x)} \leq \nicefrac{\epsilon}{2} < \epsilon . \end{equation}
Hence, \(\{ f_n \}_{n=1}^\infty\) converges uniformly.
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Next, we prove the other direction. Suppose \(\{ f_n \}_{n=1}^\infty\) converges uniformly to \(f\text{.}\) Given \(\epsilon > 0\text{,}\) find \(N\) such that for all \(n \geq N\text{,}\) we have \(\babs{f_n(x)-f(x)} < \nicefrac{\epsilon}{4}\) for all \(x \in S\text{.}\) Therefore, for all \(m, k \geq N\) and all \(x\text{,}\)
\begin{multline*} \babs{f_m(x)-f_k(x)} = \babs{f_m(x)-f(x)+f(x)-f_k(x)} \\ \leq \babs{f_m(x)-f(x)}+\babs{f(x)-f_k(x)} < \nicefrac{\epsilon}{4} + \nicefrac{\epsilon}{4} = \nicefrac{\epsilon}{2} . \end{multline*}
Take the supremum over all \(x\) to obtain
\begin{equation} \snorm{f_m-f_k}_S \leq \nicefrac{\epsilon}{2} < \epsilon . \qedhere \end{equation}
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Exercises 6.1.4 Exercises

6.1.1.

Let \(f\) and \(g\) be bounded functions on \([a,b]\text{.}\) Prove
\begin{equation} \snorm{f+g}_{[a,b]} \leq \snorm{f}_{[a,b]} + \snorm{g}_{[a,b]} . \end{equation}
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6.1.2.

  1. Find the pointwise limit of \(\left\{ \dfrac{e^{x/n}}{n} \right\}_{n=1}^\infty\) for \(x \in \R\text{.}\)
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  2. Is the limit uniform on \(\R\text{?}\)
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  3. Is the limit uniform on \([0,1]\text{?}\)
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6.1.3.

Suppose \(f_n \colon S \to \R\) are functions that converge uniformly to \(f \colon S \to \R\text{.}\) Suppose \(A \subset S\text{.}\) Show that the sequence of restrictions \(\{ f_n|_A \}_{n=1}^\infty\) converges uniformly to \(f|_A\text{.}\)
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6.1.4.

Suppose \(\{ f_n \}_{n=1}^\infty\) and \(\{ g_n \}_{n=1}^\infty\) defined on some set \(A\) converge to \(f\) and \(g\) respectively pointwise. Show that \(\{ f_n+g_n \}_{n=1}^\infty\) converges pointwise to \(f+g\text{.}\)
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6.1.5.

Suppose \(\{ f_n \}_{n=1}^\infty\) and \(\{ g_n \}_{n=1}^\infty\) defined on some set \(A\) converge to \(f\) and \(g\) respectively uniformly on \(A\text{.}\) Show that \(\{ f_n+g_n \}_{n=1}^\infty\) converges uniformly to \(f+g\) on \(A\text{.}\)
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6.1.6.

Find an example of a pair of sequences of functions \(\{ f_n \}_{n=1}^\infty\) and \(\{ g_n \}_{n=1}^\infty\) that converge uniformly to some \(f\) and \(g\) on some set \(A\text{,}\) but such that \(\{ f_ng_n \}_{n=1}^\infty\) (the multiple) does not converge uniformly to \(fg\) on \(A\text{.}\) Hint: Let \(A \coloneqq \R\text{,}\) let \(f(x)\coloneqq g(x) \coloneqq x\text{.}\) You can even pick \(f_n = g_n\text{.}\)
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6.1.7.

Suppose there exists a sequence of functions \(\{ g_n \}_{n=1}^\infty\) uniformly converging to \(0\) on \(A\text{.}\) Now suppose we have a sequence of functions \(\{ f_n \}_{n=1}^\infty\) and a function \(f\) on \(A\) such that
\begin{equation} \babs{f_n(x) - f(x)} \leq g_n(x) \end{equation}
for all \(x \in A\text{.}\) Show that \(\{ f_n \}_{n=1}^\infty\) converges uniformly to \(f\) on \(A\text{.}\)
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6.1.8.

Let \(\{ f_n \}_{n=1}^\infty\text{,}\) \(\{ g_n \}_{n=1}^\infty\) and \(\{ h_n \}_{n=1}^\infty\) be sequences of functions on \([a,b]\text{.}\) Suppose \(\{ f_n \}_{n=1}^\infty\) and \(\{ h_n \}_{n=1}^\infty\) converge uniformly to some function \(f \colon [a,b] \to \R\) and suppose \(f_n(x) \leq g_n(x) \leq h_n(x)\) for all \(x \in [a,b]\text{.}\) Show that \(\{ g_n \}_{n=1}^\infty\) converges uniformly to \(f\text{.}\)
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6.1.9.

Let \(f_n \colon [0,1] \to \R\) be a sequence of increasing functions (that is, \(f_n(x) \geq f_n(y)\) whenever \(x \geq y\)). Suppose \(f_n(0) = 0\) and \(\lim\limits_{n \to \infty} f_n(1) = 0\text{.}\) Show that \(\{ f_n \}_{n=1}^\infty\) converges uniformly to \(0\text{.}\)
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6.1.10.

Consider a sequence of functions \(f_n \colon [0,1] \to \R\) so that there is a sequence of distinct numbers \(x_n \in [0,1]\) such that for all \(n\text{,}\)
\begin{equation} f_n(x_n) = 1. \end{equation}
Prove or disprove the following statements:
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  1. True or false: There exists \(\{ f_n \}_{n=1}^\infty\) as above that converges pointwise to \(0\text{.}\)
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  2. True or false: There exists \(\{ f_n \}_{n=1}^\infty\) as above that converges uniformly to \(0\text{.}\)
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6.1.11.

Fix a continuous \(h \colon [a,b] \to \R\text{.}\) Let \(f(x) \coloneqq h(x)\) for \(x \in [a,b]\text{,}\) \(f(x) \coloneqq h(a)\) for \(x < a\) and \(f(x) \coloneqq h(b)\) for all \(x > b\text{.}\) First show that \(f \colon \R \to \R\) is continuous. Now let \(f_n\) be the function \(g\) from Exercise 5.3.7 with \(\epsilon = \nicefrac{1}{n}\text{,}\) defined on the interval \([a,b]\text{.}\) That is,
\begin{equation} f_n(x) \coloneqq \frac{n}{2} \int_{x-1/n}^{x+1/n} f . \end{equation}
Show that \(\{ f_n \}_{n=1}^\infty\) converges uniformly to \(h\) on \([a,b]\text{.}\)
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6.1.12.

Prove that if a sequence of functions \(f_n \colon S \to \R\) converge uniformly to a bounded function \(f \colon S \to \R\text{,}\) then there exists an \(N\) such that for all \(n \geq N\text{,}\) the \(f_n\) are bounded.
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6.1.13.

Suppose there is a single constant \(B\) and a sequence of functions \(f_n \colon S \to \R\) that are bounded by \(B\text{,}\) that is \(\babs{f_n(x)} \leq B\) for all \(x \in S\text{.}\) Suppose that \(\{ f_n \}_{n=1}^\infty\) converges pointwise to \(f \colon S \to \R\text{.}\) Prove that \(f\) is bounded.
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6.1.14.

(requires Section 2.6)   In Example 6.1.3 we saw \(\sum_{k=0}^\infty x^k\) converges pointwise to \(\frac{1}{1-x}\) on \((-1,1)\text{.}\)
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  1. Show that whenever \(0 \leq c < 1\text{,}\) the series \(\sum_{k=0}^\infty x^k\) converges uniformly on \([-c,c]\text{.}\)
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  2. Show that the series \(\sum_{k=0}^\infty x^k\) does not converge uniformly on \((-1,1)\text{.}\)
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