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Suppose $U$ is an open subset of $\mathbb{R}^n$, and $f:U\to \mathbb{R}$. When $f$ is $C^2$ we know that the mixed partial derivatives are symmetric, i.e. $\partial_i\partial_jf= \partial_j\partial_if.$ But as it is famous the continuity of the 2nd order partial derivatives is not necessary for this to happen. For example if $\partial_if$, $\partial_jf$ exist on $U$ and they are both differentiable (in the sense of Fréchet) at some point $a\in U$ then $$\partial_i\partial_jf(a)= \partial_j\partial_if(a).$$

Now for the 3rd order partial derivatives we can obtain the symmetry if we assume that the 1st order partial derivatives of $f$ are differentiable on $U$ and its 2nd order partial derivatives are differentiable at $a$. Let me explain the proof for the particular case $$\partial_3\partial_2\partial_1f(a)= \partial_2\partial_1 \partial_3f(a).\tag{$\star$}$$ First as $\partial_1 f$ has 1st order partial derivatives in $U$ and they are differentiable at $a$ we have $$\partial_3\partial_2\partial_1f(a)= \partial_2\partial_3 \partial_1f(a).\tag{1}$$ Then since the 1st order partial derivatives of $f$ are differentiable in $U$ we have $\partial_3\partial_1f(x)= \partial_1\partial_3 f(x)$ for all $x\in U$. Hence we can differentiate to obtain $$\partial_2\partial_3\partial_1f(a)= \partial_2\partial_1 \partial_3f(a).\tag{2}$$ By combining (1) and (2) we get ($\star$).

As you can see the full force of differentiability of the 1st order partial derivatives of $f$ on all of $U$ is only used for the equality of the 3rd order partial derivatives appeared in (2). So my question is

Question: Can we prove the symmetry of 3rd order mixed partial derivatives of $f$ at $a$ by merely assuming that the 1st and 2nd order partial derivatives of $f$ exist on $U$ and they are all differentiable at $a$? If not, can you provide a counterexample? Finally, if the answer is positive, can we generalize it to higher order mixed partial derivatives?

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    $\begingroup$ The most useful situation is for weak or distributional derivatives of distributions (in the sense of generalized functions). In that case, partials always commute. $\endgroup$ – Deane Yang Feb 25 '17 at 20:53
  • $\begingroup$ @DeaneYang : I am aware of that, but here I am looking for a pointwise equality. $\endgroup$ – Mohammad Safdari Feb 27 '17 at 5:05
  • $\begingroup$ Very interesting question, especially because some respectable authors omitted the assumption of 2-times differentiability on U! $\endgroup$ – Alexander Kuleshov Apr 18 '19 at 22:22
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For the accurate statement and rigorous proof see Theorem 3 of the following paper https://www.mdpi.com/2227-7390/8/11/1946/htm

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  • $\begingroup$ Thanks for your answer. $\endgroup$ – Mohammad Safdari Nov 19 '20 at 18:36
  • $\begingroup$ To clarify the answer: the pointwise differentiability assumption is sufficient to deduce the symmetry of higher order mixed partial derivatives. For the proof, instead of switching pairs of partial differentiation operators, the idea is to use a signed sum of the values of $f$ at the vertices of a higher-dimensional rectangle, similarly to the proof for the symmetry of 2nd order partial derivatives. $\endgroup$ – Mohammad Safdari Nov 19 '20 at 18:45

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