2
$\begingroup$

Let $f(x,y)$ be a a real valued function on an open subset of $\mathbf{R}^2$ with continuous partial derivatives $\frac{\partial^2 f}{\partial x^2}$ and $\frac{\partial^2}{\partial y^2}$. Is $f$ twice differentiable?

$\endgroup$

2 Answers 2

4
$\begingroup$

Here is an explicit counterexample in $B_1$, $$f(x,y) = x\, y \, \log(-\log(x^2+y^2))$$

Both $\partial_{xx} f$ and $\partial_{yy} f$ are continuous but $f$ is not twice differentiable at the origin.

$\endgroup$
1
  • $\begingroup$ Upon a quick check it looks just fine. Nice idea to take the double logarithm of the troublesome term in the usual counterexample for the insufficiency of single partial differentiability to ensure the total one. My curiosity came from the observation that over the p-adic numbers this criterion oddly seems sufficient. Thank you! $\endgroup$
    – Tiffy
    Nov 12, 2012 at 19:32
0
$\begingroup$

For very general open sets, this is not true.

However, assume true if the boundary is sufficiently smooth and the boundary conditions are nice enough (say, Dirichlet). In that case, your question boils down to asking whether a continuous function whose Laplacian is continuous, too, is also in C^2. This is surely true if instead of continuity you assume Hölder continuity: i.e., if your aim is to pass from $\Delta u\in C^{0,\alpha}$ to $u\in C^{2,\alpha}$: this is exactly the assertion of Schauder's estimates; or else if you content yourself with weak differentiability instead of classical one: in that case, see e.g. §6.3.2 in Evans' Partial Differential Equations.

$\endgroup$
1
  • $\begingroup$ The are well known counterexamples of functions for which $\Delta f$ is continuous, but f is not $C^2$. The question asks what happens under the stronger condition that both $f_{xx}$ and $f_{yy}$ are continuous. $\endgroup$ Sep 25, 2012 at 20:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.