Let $f:\mathbb{R}^2 \to \mathbb{R}$ be twice differentiable in $a$. Suppose that $\frac{\partial^2f}{\partial x \partial y}$ is continuous in $a$. Is it possible that $\frac{\partial^2 f}{\partial y \partial x}$ is discontinuous in $a$?

The question is motivated by Schwarz' Theorem - as this observation would show that it would not help at merely computing the derivatives because mostly we would not know whether both second derivates are continuous. And that's needed to apply Schwarz (at least in the way I know the theorem).

  • $\begingroup$ I remember seeing this statement somewhere ... $\endgroup$ May 15, 2018 at 21:08
  • $\begingroup$ Very nice question! $\endgroup$
    – user525755
    May 15, 2018 at 21:14
  • 1
    $\begingroup$ See my answer here. $\endgroup$ May 15, 2018 at 21:16

1 Answer 1


Theorem 9.41 in Baby Rudin:

Suppose $\frac{\partial f}{\partial x}$, $\frac{\partial^2f}{\partial x \partial y}$ and $\frac{\partial f}{\partial y}$ exist on a neighborhood of $a$ and $\frac{\partial^2f}{\partial x \partial y}$ is continuous at $a$.

Then $\frac{\partial^2f}{\partial y \partial x}$ exists at $a$ and $$\frac{\partial^2f}{\partial x \partial y}(a)=\frac{\partial^2f}{\partial y \partial x}(a)$$

Here the existence of $\frac{\partial f}{\partial x}$, $\frac{\partial^2f}{\partial x \partial y}$ and $\frac{\partial f}{\partial y}$ is guaranteed by the twice differentiability assumption.

  • $\begingroup$ Perfect! Thanks. $\endgroup$
    – Qi Zhu
    May 15, 2018 at 22:00

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