Twice differentiable implies equality of mixed partials?

Question

I have seen the following theorem claimed in several places:

Claim: If a function $f$ of two variables has partial derivatives $f_x$ and $f_y$ which are defined in a neighborhood of $(a,b)$ and are also differentiable at $(a,b)$, then the mixed partials are equal: $f_{xy}= f_{yx}$.

This is a stronger claim than what I find in most textbooks and also on Wikipedia, which also assumes that the second partial derivatives are continuous.

If the above claim is true, where can I find a proof of it? And if not, what is a counterexample? (The latter is basically this question.)

It is stated on page 201 or 220 (maybe depending on edition) of Advanced Calculus by Angus Taylor, but no proof is given.

I have also found it in a 1908 paper called On Differentials by W. H. Young in Proc. LMS, but I don't believe the proof given. He uses differentiability of $f_x$ to show that

$$ \frac{f_x(x+h,y+k) - f_x(x+h,y)}{k} = \frac{h}{k}e + f_{yx} + e' $$

where $e,e'\to 0$ as $(h,k)\to 0$, and therefore that the LHS goes to $f_{yx}$ as $(h,k)\to 0$ "provided $h/k$ does not become infinite". Similarly, $\frac{f_y(x+h,y+k) - f_y(x,y+k)}{h} \to f_{xy}$ as $(h,k)\to 0$ "provided $k/h$ does not become infinite". Then he roughly follows the usual proof of equality of mixed partials, using the mean value theorem to show that

$$ \frac{f(x+h,y+k)-f(x+h,y) - f(x,y+k) + f(x,y)}{hk} = \frac{f_y(x+h,y+\theta k) - f_y(x,y+\theta k)}{h} $$

for some $0<\theta<1$, and dually. Then he says "if $(h,k)$ moves towards $(0,0)$ in such a way that $h/k$ has not zero as limit, $h/\theta k$ will not have zero for a limit", so that the RHS above goes to $f_{xy}$, and dually also to $f_{yx}$. But it seems to me that $\theta$ depends on $h$ and $k$, so there is no reason it might not blow up as $(h,k)\to 0$ even if $h/k$ has a finite nonzero limit. Is there some reason why this works?

Answer 1

The result is correct. Note that they are not saying just that the second partial derivatives exist. They are saying that the derivative map $$Df(x,y) = \begin{bmatrix} \frac{\partial f}{\partial x}(x,y) & \frac{\partial f}{\partial y}(x,y)\end{bmatrix}$$ is differentiable as a function $U\to L(\Bbb R^2,\Bbb R)$, where $U$ is a neighborhood of $(a,b)$ and $L(E,F)$ denotes the vector space of linear maps from $E$ to $F$.

This implies that the second derivative gives you a symmetric bilinear map $\Bbb R^2\times\Bbb R^2\to\Bbb R$, i.e., that the mixed partials are equal.

This is proved, for example, in Dieudonné's multi-volume Analysis text.