I have asked this question on Mathoverflow, but it did not receive much attention there.

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?

  • $\begingroup$ If $f$ is $k$ times differentiable (with continuity or not), then the $k$-th partials all coincide. Differential Calculus by Cartan or Foundations of Modern Analysis by Dieudonne. $\endgroup$ – Will M. Mar 8 '17 at 16:57
  • $\begingroup$ Also, another known result: $f$ is $k$ times differentiable with continuity on some open set iff all the $k$-th ordre partials exists and are continuous. I know this result is false if you don't assume continuity. Also, I am not sure if your claim "For example if $\partial_i f,$ $\partial_j f$ exist on $U$ and they are both differentiable (in the sense of Fréchet) at some point $a \in U$ then $\partial_i \partial_j f(a) = \partial_j \partial_i f(a)$" is true, it seems false without continuity. $\endgroup$ – Will M. Mar 8 '17 at 17:01
  • $\begingroup$ @WillM. The problem is that if $f$ is 3 times differentiable at $a$, then it must be twice differentiable around $a$. Hence the 1st partials of $f$ are differentiable around $a$, which is what I am asking if it is possible to avoid. Also about your 2nd comment, the claim in the question is true and you can find a proof (of a little different version, although essentially the same) in Pugh's analysis book. $\endgroup$ – Mohammad Safdari Mar 8 '17 at 17:08
  • $\begingroup$ I checked Pugh's, a quick look. in chapter 5, sects. 2 and 3. I couldn't find the claim. I'd be very interesting in knowing it because is one of these things I have wondered for a long time. $\endgroup$ – Will M. Mar 8 '17 at 17:23
  • $\begingroup$ @WillM. It is theorem 16 of chapter 5 (2nd edition), although as I said you need to modify it a little bit. $\endgroup$ – Mohammad Safdari Mar 9 '17 at 5:41

All of this still works if $U$ is restricted to any neighborhood of $a$. So it isn't that you need differentiability on all of $U$, just on some neighborhood of $a$. $U$ is simply acting as that neighborhood.

But, yes, you do need the full force of differentiability of the first partials for this to work, and for a very simple reason: If the the first partials are not differentiable on some neighborhood of $a$, then the second partials are not defined on a neighborhood of $a$, and thus the third partials at $a$, which are defined from the behavior of the second partials on a neighborhood of $a$, do not exist.

So the reason for requiring the full differentiability of the first partials is so that the two expressions being compared will exist in the first place.

  • $\begingroup$ The question is about functions that have 2nd partial derivatives on a neighborhood of the point $a$, and for this we certainly do not need the 1st partials to be differentiable in the sense of Frechet. $\endgroup$ – Mohammad Safdari Mar 8 '17 at 11:28

This is more of an expanded comment than an answer.

Let $\mathrm{U}$ be an open subset of $\Bbb R^2.$ If $f:\mathrm{U} \to \Bbb R$ is such that $\mathbf{D}_1 f$ and $\mathbf{D}_2 f$ exist on $\mathrm{U}$ and are differentiable, then $f$ is twice differentiable on $\mathrm{U}.$ To see this, write $$\mathbf{D} f(x, y) = \mathbf{D}_1 f(x, y) \circ \mathrm{pr}_1 + \mathbf{D}_2 f(x, y) \circ \mathrm{pr}_2,$$ then the order one expansion for $\mathbf{D}f$ is clear provided those for $\mathbf{D}_1 f$ and $\mathbf{D}_2 f.$

If $f:\mathrm{A} \subset \Bbb R^d \to \Bbb R$ is such that $\mathbf{D}_i f$ and $\mathbf{D}_j f$ exist on $\mathrm{A}$ and are differentiable there, then $\mathbf{D}_{i,j} f = \mathbf{D}_{j,i} f$ on $\mathrm{A}.$ Just define $g(x_i, x_j) = f(x_1, \ldots, x_d)$ and apply previous result.

So, in some sense, for the partials of higher order to be symmetric, you need higher order differentiation, at least in the involved variables. That is, if you want to permute the partials with respect to the variables $x_i,$ $x_j$ and $x_l,$ then you need to define the function $g(x_i, x_j, x_l) = f(x_1, \ldots, x_d)$ and assume $f$ is such that $g$ is as many times differentiable as needed and wanted.


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