I know there are examples of functions which are twice differentiable, but the mixed partials $f_{xy}\neq f_{yx}$ at the origin $(0,0)$.

Are there any examples of functions where the mixed partials exist, but are equal nowhere? I know a necessary condition is that the mixed partials are discontinuous. My question can be formally stated in a three parts:

For both of these cases, assume the second partial derivatives exist.

$1)$ let $f\in C^1(\mathbb{R}^2)$. Do there exists subsets $E \subset\mathbb{R}^2$ where $\mu(E)> 0$ where $f_{xy}\neq f_{yx}$, (where $\mu(E)$ denotes the Lebesgue measure of $E$). Note that having mixed partial $f_{xy}\neq f_{yx}$ at the origin answers this question in the affirmative for $\mu(E)=0$, since if $E=\{(0,0)\}$, then $\mu(E)=0$, and we know of many examples where this occurs.

$2)$ let $f\in C^1(\mathbb{R}^2)$. Is it possible for $f_{xy}$ to be continuous on a set $\mu(E)\geq 0, $ and $f_{xy}\neq f_{yx}$ everywhere or at least almost everywhere?

$3)$ would answers to these questions be interesting enough to publish as counterexamples or theorems? I was unable to find information about these two cases, let alone generalizations to functions with $f\in C^1(\mathbb{R}^n)$, leading me to believe these could be interesting research questions.

  • 1
    $\begingroup$ As functions $f_{xy}$ and $f_{yx}$ may be different, but as distributions they are the same. I would guess that $f_{xy}$ would have to equal $f_{yx}$ almost everywhere. $\endgroup$ Jun 24, 2017 at 3:55
  • $\begingroup$ As a preliminary thought, I wonder if the weak partial derivatives $f_{xy}$ and $f_{yx}$ must be equal a.e. Since I am assuming the strong partial derivatives exist, the weak derivatives would need to be equal to the strong ones. $\endgroup$ Jun 24, 2017 at 4:04

1 Answer 1


If $f_{xy}$ and $f_{yx}$ are assumed to exist everywhere, then we can make $f_{xy}-f_{yx}$ to be the indicator of a Cantor set with positive measure.

If $f_{xy}$ and $f_{yx}$ are only assumed to exist almost everywhere, then we can make $f_{xy}\ne f_{yx}$ almost everywhere.

At any point where $f_{xy}$ is continuous and $f_y$ is defined, we have $f_{xy}=f_{yx}$.

I have posted the proofs of these results on my blog: Part 1 and Part 2.

There is a comment saying that as distributions we have $f_{xy}=f_{yx}$. Indeed, if they are assumed to be $L^2$ then they are equal almost everywhere.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .