Equality of mixed partial derivatives

Question

Is the following statement $$\frac{\partial^2 f}{\partial x \, \partial y}=\frac{\partial^2 f}{\partial y \, \partial x}$$ always true? If not what are the conditions for this to be true?

Answer 1

Second order partial derivatives commute if $f$ is $\mathcal{C}^2$ (i.e. all the second partial derivatives exist and are continuous). This is sometimes called Schwarz's Theorem or Clairaut's Theorem; see here.

Answer 2

This is true in general if $ f \in \mathcal{C}^2 $. This has a name: symmetry. More formally, it is known as Clariut's Theorem or Schwarz's theorem.

Answer 3

The equality of mixed directional derivatives holds if $f$ is twice differentiable at the given point. There a quick easy (and nice) proof using Fubini's theorem - but it assumes that $f\in C^2$. It is a striking fact (IMHO) that the symmetry already follows from point-wise differentiability. I've learned this from the (excellent) "Advanced calculus" book by Loomis and Sternberg (an electronic copy is freely available at https://people.math.harvard.edu/~shlomo), theorem 16.3 (page 187).

I shall reproduce the proof below, correcting a small error (as I understand it) that confused me a quite a bit.

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We first recall the definition: Let $V,W$ be normed linear spaces (over $\mathbb R$). A function $f:V\longrightarrow W$, defined at some neighborhood of $v_0\in V$, is said to be differentiable at $v_0$ if there exists $df|_{v_0}\in Hom(V,W)$ (necessarily unique) such that $$\lim_{v\to 0_V}\frac{f(v_0+v)-f(v_0) - df|_{v_0}(v)}{\|v\|} = 0_W $$ The uniqueness follows from the observation that the zero map is the only $S\in Hom(V,W)$ such that $\lim_{v\to 0_V}\frac{S(v)}{\|v\|} = 0_W $.

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Recall that for any $v\in V$ the directional derivative is defined as $\frac{\partial f}{\partial v}(v_0)=\lim_{t\to 0}\frac{f(v_0+tv)-f(v_0)}{t}$ assuming the limit exists. It is easy to see that if $df|_{v_0}$ exists than the directional derivatives exist and $$\frac{\partial f}{\partial v}(v_0)=df|_{v_0}(v)$$ Proof: Plug in $tv$ for $v$ in the definition of to get $\frac{f(v_0+tv)-f(v_0) - df|_{v_0}(tv)}{\|tv\|} = \frac{1}{\|v\|}\cdot\Big(\frac{f(v_0+tv)-f(v_0) }{t} - df|_{v_0}(v)\Big)$ and this by assumption tends to $0_W$ as $t\to 0$. $\ $q.e.d.
Assuming that $f$ is differentiable, it follows that the map $\frac{\partial f}{\partial v}:V\longrightarrow W$ is equal to the composition $V\overset{df|_{(\cdot)}}{\longrightarrow} Hom(V,W) \overset{eval_v}{\longrightarrow}W\ $ where $eval_v$ is the (linear) map $T\mapsto T(v)$

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If $f$ is differentiable on some neighborhood of $v_0$, we can ask whether the map $df|_{(\cdot)}:V\longrightarrow Hom(V,W)$ is itself differentiable at $v_0$. If so then $d^2f|_{v_0}:=d\big(df|_{(\cdot)}\big)|_{v_0}$ is an element of $Hom\big(V,Hom(V,W)\big)$. Also : $$ \frac{\partial^2 f}{\partial u \partial v}(v_0)= \frac{\partial\big(\frac{\partial f}{\partial v}\big)}{\partial u}(v_0)= d\Big(\frac{\partial f}{\partial v}\Big)|_{v_0}(u)= d\big(eval_v\circ df|_{(\cdot)}\big)|_{v_0}(u)= \big(eval_v\circ d^2f|_{v_0}\big)(u)= eval_v\big(d^2f|_{v_0}(u)\big)= \big(d^2f|_{v_0}(u)\big)(v) $$

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Now we can finally state the theorem: If $f$ is twice differentiable at $v_0$ then for all $u,v\in V\ $ $$\frac{\partial^2 f}{\partial u \partial v}(v_0) =\frac{\partial^2 f}{\partial v \partial u}(v_0)$$ Proof: We need to show that the map $\big(d^2f|_{v_0}(u)\big)(v):V\times V\longrightarrow W$ is symmetric.
Fix some $\epsilon > 0$.
By the definition of $d^2f|_{v_0}$, $\ \exists\delta>0$ such that $\forall x\in V$ $$ \|x\|<\delta\ \Rightarrow\ \big\| df|_{v_0+x}-df|_{v_0} - d^2f|_{v_0}(x) \big\|_{Hom(V,W)} <\epsilon\cdot\|x\| $$ Apply this also to some (small) $x+y$ and subtract to conclude that $\forall x,y\in V$ $$ \Big(\|x\|<\delta\ \& \ \|x+y\|<\delta \Big)\ \Rightarrow\ \big\| df|_{v_0+x}-df|_{v_0+x+y} - d^2f|_{v_0}(-y) \big\| <\epsilon\cdot\big(\|x\| + \|x+y\|\big) $$ Now fix some (non-zero) $y\in V$ such that $0<\|y\|<\delta/3$.
Clearly $\|x\|\leq 2\|y\| \Rightarrow\ \|x\|<\delta\ \& \ \|x+y\|<\delta \ \& \ \big(\|x\| + \|x+y\|\big) \leq 5\|y\| $
So setting $$g(x):=f(v_0+x)-f(v_0+x+y)-d^2f|_{v_0}(-y)(x)$$ we have that $\forall x\in B_{2\|y\|} \ \| dg|_x \| < 5 \epsilon\cdot \|y\| $ (I'm using $B_a := \{x\in V:\|x\|\leq a\})$).
Therefore (from the mean value theorem) $$\forall a,b\in B_{2\|y\|}\quad \|g(a)-g(b)\|_W < 5 \epsilon\cdot \|y\|_V \cdot \|a-b\|_V $$ i.e. $$ \big\| f(v_0+a) - f(v_0+a+y) - f(v_0+b) + f(v_0+b+y) - d^2f|_{v_0}(-y)(a-b) \big\| < 5 \epsilon\cdot \|y\| \cdot \|a-b\| $$ For any $x\in B_{\|y\|}$ we can apply the above with $a=-y,b=-y-x$ (note that indeed $a,b\in B_{2\|y\|}$) and we get that $$ \big\| f(v_0-y) - f(v_0) - f(v_0-y-x) + f(v_0-x) - d^2f|_{v_0}(-y)(x) \big\| < 5 \epsilon\cdot \|y\| \cdot \|x\| $$ Let's recap: Denoting $$ S(x,y) = f(v_0) + f(v_0-y-x) - f(v_0-y) - f(v_0-x) $$ we have that for any $0<\|y\|<\delta/3$ and $\|x\|\leq \|y\|$ it holds that $$\big\| d^2f|_{v_0}(y)(x) - S(x,y) \big\| < 5 \epsilon\cdot \|y\| \cdot \|x\|$$ Clearly $S(x,y)=S(y,x)$ and therefore combined with the same inequality with reversed order we conclude that $$ 0< \|x\| = \|y\|<\delta/3 \Rightarrow \big\| d^2f|_{v_0}(y)(x) - d^2f|_{v_0}(x)(y) \big\| < 10 \epsilon\cdot \|y\| \cdot \|x\| $$ Using the homogeneity of $d^2f|_{v_0}(\cdot)(\cdot)$, it follows that $$ \|x\| = \|y\| = 1 \Rightarrow \big\| d^2f|_{v_0}(y)(x) - d^2f|_{v_0}(x)(y) \big\| \leq 10 \epsilon\cdot \delta^2/9 $$ (apply the previous inequality to $tx,ty$ with $t<\delta/3$)
Since $\epsilon$ was arbitrary (and we may assume $\delta<1$), it follows that $d^2f|_{v_0}(y)(x)=d^2f|_{v_0}(x)(y)$ for any pair of unit vectors. Applying homogeneity once more we see that it holds without restriction.
q.e.d.