Mixed derivatives and the compact-interval integral

Prerequisite companion. This is an attributed adaptation and extension of Jiří Lebl, Basic Analysis, version 6.3, freely accessible author edition, under CC BY-SA 4.0. The source sections actually read for these arguments are §§5.1–5.3, 7.5, 8.6 and 9.1. Complete earlier programme proofs are retained and bound in integration-proof-chain.json; this companion supplies their used omissions and the integral Taylor formula needed by the Morse reduction.

The earlier topology and differential chains, including the ordered complete real-field axioms, finite-dimensional norms, derivative rules, mean-value theorem and smooth inverse theorem, remain in force. This chapter concerns bounded-interval Riemann integration. It does not claim the general improper or Lebesgue integration, multidimensional substitution or exponential contracts needed elsewhere in the full stationary-phase lesson.

P11. Higher derivatives and Hessian symmetry

P11.1. Equivalence of the two definitions of $C^r$

The earlier companion defines $C^r$ recursively by requiring the total derivative to be $C^{r-1}$. Lebl §8.6 instead requires every ordered partial derivative through order $r$ to exist and be continuous. These definitions agree for maps between finite-dimensional real spaces.

First, a map $f=(f_1,\ldots,f_m)$ is differentiable exactly when all its components are differentiable. If $Df(p)=A$, applying a coordinate projection to its remainder gives the component derivative, since a coordinate's absolute value is at most the vector norm. Conversely assemble the component derivatives into a matrix $A$. If $r_j(h)$ is the $j$th component remainder, then

The finite maximum tends to zero by taking the minimum of the component thresholds. Empty coordinate blocks give the unique zero map. The same coordinate inequalities characterize continuity. By induction on $r$, they characterize recursive $C^r$ regularity as well, because the derivative matrices consist of finitely many component derivatives.

For $r=1$, the equivalence with continuous partials is the complete Proposition 8.4.6, including P10.3's omitted base case. Suppose the equivalence holds at order $r-1$. If $f$ is recursively $C^r$, every entry of $Df$ is $C^{r-1}$. Those entries are the first partials by Proposition 8.3.9, so the induction hypothesis supplies all their ordered partials through order $r-1$, continuously. This gives all ordered partials of $f$ through order $r$.

Conversely, if those ordered partials exist continuously, Proposition 8.4.6 makes $f$ a $C^1$ map. Each first partial has continuous ordered partials through order $r-1$, so is recursively $C^{r-1}$ by induction. Assembling the entries shows $Df\in C^{r-1}$, hence $f\in C^r$. This proves the equivalence for every finite order; being smooth means satisfying it for every order. No commutation of derivatives was assumed in this argument. $\square$

P11.2. The exact scope of the mixed-partial proof

The programme contains the full double-mean-value proof of Proposition 8.6.2 at sec_mvhighordders.html#sec_mvhighordders-10. It compares the same rectangle difference quotient in both orders. For distinct indices, choose $B(p,r)\subset U$ and $0<s,t<r/3$; all points of the closed rectangle, including its edges, lie in this ball, since their distance from $p$ is at most $s+t<2r/3$.

The source supplies points $p_0,p_1$ in that rectangle with

Each point tends to $p$ because its distance is bounded by $s+t$. Given $\varepsilon>0$, continuity therefore makes both respective errors from their values at $p$ less than $\varepsilon$, once $s,t$ are sufficiently small. Subtracting the two equal expressions for $G$ gives

If the left side were positive, taking $\varepsilon$ smaller than half of it would contradict this bound. Thus it is zero. Equal indices already give the same expression. This records the boundary and limit steps in the existing proof without replacing its mean-value argument. P11.1 verifies that its $C^2$ hypothesis agrees with the one used here.

For a $C^k$ function, any two orders of a fixed list of at most $k$ differentiations give the same result. To swap two adjacent operators, first apply the inner operators to obtain a function with at least two continuous derivatives. Apply the just-proved second-order identity on its open domain. Any remaining outer derivatives preserve this equality, because equal functions have equal difference quotients. Any permutation is a finite sequence of adjacent swaps: successively move the desired first, second and later entries into position. This proves the assertion for every order, including repeated indices. In particular the smooth Hessians in P4 and M1 are symmetric. $\square$

P12. The omitted integral inputs

P12.1. Suprema and infima used by Darboux sums

All sets in this paragraph are nonempty and bounded when both extrema are used. If every $a\in A$ is at most every $b\in B$, each $b$ is an upper bound for $A$, so $\sup A\leq b$. Thus $\sup A$ is a lower bound for $B$, and $\sup A\leq\inf B$.

For two sets, write $A+B=\{a+b:a\in A,b\in B\}$. The number $\sup A+\sup B$ is an upper bound for $A+B$. By P6.2, for each $\varepsilon>0$ choose $a>\sup A-\varepsilon/2$ and $b>\sup B-\varepsilon/2$. Their sum exceeds $\sup A+\sup B-\varepsilon$. No smaller upper bound is therefore possible, proving $\sup(A+B)=\sup A+\sup B$. Negating the sets proves $\inf(A+B)=\inf A+\inf B$, since $\inf(-A)=-\sup A$ and $\sup(-A)=-\inf A$, directly by the order definitions.

For $c>0$, $c\sup A$ bounds $cA$ above. Choosing $a>\sup A-\varepsilon/c$ proves it is the least such bound. Thus $\sup(cA)=c\sup A$; negation proves the corresponding infimum formula. For $c=0$, the image set is $\{0\}$; for $c<0$, factor out the minus sign and use the preceding negation identities. These statements justify the extrema operations used in Lebl's integral-linearity proof.

If $A\subset B$ and every $b\in B$ is at most some $a\in A$, then $\sup A=\sup B$: inclusion gives one inequality, and the second condition makes every upper bound for $A$ an upper bound for $B$. Reversing order proves the analogous infimum assertion. This supplies the cofinal-partition step left to an exercise in Lemma 5.2.1.

Finally, for bounded functions $f\leq g$ on the same nonempty set, $\inf f\leq\inf g$ because $\inf f$ is a lower bound for all values of $g$; likewise $\sup f\leq\sup g$. For functions on the same set, every value of $f+g$ lies between $\inf f+\inf g$ and $\sup f+\sup g$. These give inequalities, without asserting equality when the two extrema require different points. $\square$

P12.2. Upper refinement and the necessary integrability criterion

Use Lebl's definitions of a partition $P$, its lower and upper sums $L(P,f),U(P,f)$, and the lower and upper Darboux integrals. The complete lower-refinement proof is Proposition 5.1.7. For its omitted upper half, each refined interval is contained in its original interval, so its supremum $\widetilde M_q$ is at most the original $M_i$. Multiplying by positive refined lengths and summing over that original interval gives

Summing over $i$ proves $U(\widetilde P,f)\leq U(P,f)$. The index $q$ in the source definition of $\Delta\widetilde x_q$ must start at 1, not 0; no undefined $\widetilde x_{-1}$ is used.

Proposition 5.1.13 proves that arbitrarily small gaps $U(P,f)-L(P,f)$ imply Riemann integrability. Conversely suppose the common integral is $I$. By the defining supremum and infimum choose partitions $P_1,P_2$ with $L(P_1,f)>I-\varepsilon/2$ and $U(P_2,f)<I+\varepsilon/2$. Their union is a finite partition and refines both, so

This proves the necessary half that will be used below. For any finite list of integrable functions, take the union of the chosen partitions to obtain all their small-gap bounds on one common partition. $\square$

P12.3. Negative scalars, addition, restrictions and orientation

The positive-scalar part of Proposition 5.2.4 already has a written proof. For its negative-scalar exercise, P12.1 gives

Taking the defining extrema gives $\underline\int(-f)=-\overline\int f$ and $\overline\int(-f)=-\underline\int f$. Thus $-f$ is integrable when $f$ is, with integral $-\int f$. Combining this with the proved nonnegative-scalar case gives every real scalar.

For the addition exercise, choose a common partition with the sum of the two Darboux gaps smaller than $\varepsilon$, using P12.2. The pointwise extrema inequalities of P12.1 imply

Consequently the Darboux gap of $f+g$ is less than $\varepsilon$, so Proposition 5.1.13 proves its integrability. Both $\int(f+g)$ and $\int f+\int g$ lie in the displayed outside interval, whose length is less than $\varepsilon$. Their difference is therefore zero, proving full linearity. Repeated application proves linearity for any finite sum. The unused general Darboux subadditivity assertion is not needed in this proof.

For Corollary 5.2.3, let $[c,d]\subset[a,b]$ with $c<d$. If $c>a$, split at $c$ by Proposition 5.2.2 to obtain integrability on $[c,b]$; if $c=a$, it is already known. If $d<b$, split that interval at $d$; if $d=b$, no second split is needed. This proves the restriction exercise in all endpoint cases. Define the integral on a singleton to be 0 and define $\int_y^x f=-\int_x^y f$ for $x<y$. These are orientation conventions, not limits of undefined partitions.

Within a fixed interval $[a,b]$, put $J(x)=\int_a^x f$. Additivity gives $\int_x^y f=J(y)-J(x)$ when $x<y$; the definitions give it when $x=y$ and $x>y$. Subtracting these identities proves $\int_x^z f=\int_x^y f+\int_y^z f$ in every order of the three points. In particular, changing the base point of an antiderivative integral changes it by a constant, completing Remark 5.3.4. $\square$

P12.4. The norm bound and passing a uniform error through an integral

For $f:[a,b]\to\mathbb R^m$ with Riemann-integrable components, define its integral componentwise. Its norm is also integrable. To see this, on any partition interval the reverse triangle inequality and the coordinate bound give

Taking a supremum in $x$ and an infimum in $y$, by P12.1, bounds the oscillation of $|f|$ by that sum. Therefore its Darboux gap is at most the sum of the component gaps. A common partition from P12.2 makes this arbitrarily small, proving integrability of $|f|$.

Write $I=\int_a^b f$. If $I\ne0$, let $v=I/|I|$. Finite linearity, the Euclidean Cauchy–Schwarz bound and integral monotonicity (Proposition 5.2.6) give

For $I=0$ the same inequality follows from nonnegativity of the last integral. In particular $|\int_a^b f|\leq(b-a)\sup|f|$. Complex-valued integrals are the case $m=2$, using their real and imaginary components.

If $f_\lambda$ and $f$ are integrable and $\sup_{[a,b]}|f_\lambda-f|\to0$, the same bound and linearity give

This is the precise uniform-error result used in Theorem 9.1.1. It does not assume that an arbitrary pointwise limit can pass through an integral. $\square$

P12.5. Endpoint and domain details in the fundamental theorem

The programme supplies the full proofs of both forms of the fundamental theorem, Theorems 5.3.1 and 5.3.3. In the latter, a derivative at an endpoint of $[a,b]$ means the appropriate one-sided derivative. At interior points it is the ordinary two-sided derivative. P12.3 justifies its signed integral differences when the increment is negative. Its final non-strict error estimate proves the strict limit definition by starting with half the requested error. A different integration base point changes the primitive by a constant, by P12.3, and hence has the same derivative.

In the substitution proof of Theorem 5.3.5, the primitive of a continuous $f:[c,d]\to\mathbb R$ may be evaluated at an endpoint of that interval. To apply the earlier open-domain chain rule without a boundary assumption, extend $f$ to the real line by the constant value $f(c)$ to the left and $f(d)$ to the right. This extension is continuous: away from the junctions the assertion is inherited or constant, and at a junction the original one-sided continuity and the identical constant value give the same bound on both sides. If $c=d$, use the constant function $f(c)$ everywhere.

The primitive $\widetilde F(y)=\int_{g(a)}^y\widetilde f(u)\,du$ is consequently $C^1$ on an open interval containing the whole range of $g$, by the proved fundamental theorem on each bounded subinterval. Its derivative is $\widetilde f$. The open-domain chain rule now applies to $\widetilde F\circ g$ at every interior point of $[a,b]$, including points where $g$ reaches $c$ or $d$. The source's first-fundamental-theorem argument gives exactly its stated oriented substitution formula. No monotonicity or injectivity of $g$ is required. This is only the one-variable formula; multidimensional change of variables remains a separate proof obligation. $\square$

P12.6. Integration by parts and the required Taylor formula

For $u,v\in C^1([a,b])$, the product derivative is continuous. The product rule and the first fundamental theorem therefore give

This follows by integrating $(uv)'=u'v+uv'$ and using linearity; all integrands are continuous and hence integrable by Lemma 5.2.7. Vector-valued $v$ is handled componentwise.

Let $g\in C^N([0,1];\mathbb R^m)$, $N\geq1$. Define

The first fundamental theorem gives $R_1=g(1)-g(0)$. For $N\geq2$, apply (P12.3) with $u(t)=(1-t)^{N-1}$ and $v=g^{(N-1)}$. The endpoint term at 1 vanishes, and the one at 0 is $-g^{(N-1)}(0)$. Since $u'=-(N-1)(1-t)^{N-2}$, this gives

Induction now proves the full integral-remainder identity

The polynomial derivative used here follows by the product rule applied to its $N-1$ equal factors. Applying the fundamental theorem to $-(1-t)^N/N$ also gives $\int_0^1(1-t)^{N-1}\,dt=1/N$. Thus (P12.1) bounds $|R_N|$ by $\sup|g^{(N)}|/N!$.

For M1, take $g(t)=f(q+tv,u',s)$ and $N=2$. Repeated chain rules give $g'(0)=v\partial_1f(q,u',s)=0$ and $g''(t)=v^2\partial_1^2f(q+tv,u',s)$. Formula (P12.4) is precisely the identity (M1.3), including its factor 2 in the definition of $A$. It holds for positive, zero and negative $v$ on any segment contained in the chart. No exchange of two integrals is used. $\square$

P12.7. Smooth dependence on all parameters of a compact integral

Let $U\subset\mathbb R^p$ be open, and suppose that $f(t,z)$ and all its ordered $z$-partial derivatives through order $r$ are continuous on $[a,b]\times U$. Then

is $C^r$ and every such parameter derivative is the integral of the corresponding derivative of $f$. The assertion holds for all finite $r$, for $r=\infty$, and for finite-dimensional vector values.

Fix $z_0$, and choose a closed ball $K$ centred at $z_0$ with positive radius, contained in $U$. The product $[a,b]\times K$ is closed and bounded in a Euclidean space, hence compact. To verify closedness directly, a point outside either factor has a positive-distance neighbourhood still outside that factor; coordinate projections do not increase distance. Boundedness follows from the sum of the squared coordinate bounds. F0-COMP therefore makes each continuous integrand on this product uniformly continuous.

For $z\to z_0$, this gives $\sup_t|f(t,z)-f(t,z_0)|\to0$. P12.4 proves continuity of $H$. For a coordinate increment $he_j$ sufficiently small to stay in the interior of $K$, the one-variable mean-value argument of the existing Theorem 9.1.1 gives, for each scalar component, a point $\theta$ between 0 and $h$ with

Uniform continuity of $\partial_j f$ makes the difference between this quotient and $\partial_j f(t,z)$ tend uniformly in $t$ to zero. The intermediate point may depend on $t$; the uniform bound does not. By linearity and (P12.2),

The earlier continuity argument applied to $\partial_j f$ shows that this partial derivative is continuous jointly in $z$. Repeat with each ordered derivative of $f$. Finite induction gives every ordered partial through order $r$, continuously; P11.1 identifies this with $C^r$ regularity. Applying the argument at every finite order proves the smooth case. Vector values follow componentwise. When $p=0$ there are no parameter derivatives; when $a=b$ every integral is the zero map. These cases satisfy the same conclusion directly. $\square$

Use in the Morse reduction

P11 supplies the mixed-partial input in the Schur-complement calculation P4. P12 supplies the precise Taylor and compact-parameter integration inputs in M1. The full earlier inverse/implicit and signature proofs remain unchanged. The global Fourier integrals, exponential identities and multidimensional Jacobian substitutions needed by the analytic module are still separate obligations.