Take a weak solution $u$ of the Poisson equation on $\mathbb{R}^d$ $$ \Delta u = f $$ By standard elliptic regularity theory we have (for some $\alpha\in (0,1]$) $f\in C^{0, \alpha}_{\text{loc}}(\mathbb{R}^d)$, then $u \in C^{2, \alpha}_{\text{loc}}(\mathbb{R}^d)$. Now my (naive) question is:

Are there some weaker assumptions that on $f$ that ensure that $u \in C^2(\mathbb{R}^d)$?

By weaker assumptions I mean something like $f\in C^0(\mathbb{R}^d)$ plus some condition which is not of Hölder type (I am aware that $f\in C^0(\mathbb{R}^d)$ is not sufficient).

  • 3
    $\begingroup$ There is a standard counterexample that shows the existence of $f \in C^0$ such that $u\not\in C^2$ (which I guess you are aware of); so I guess you are looking for an assumption of the form $C^0$ + "something not measured on the Holder scale"? $\endgroup$ Jul 12 '19 at 16:30
  • $\begingroup$ @WillieWong Exactly. I will edit such that it becomes more transparent. $\endgroup$ Jul 12 '19 at 16:52

Dini continuity may be what you are looking for: if $f = \Delta u$ is Dini continuous, that is, $$\int_0^1 \frac{\omega_f(t)}{t} \, dt < \infty,$$ then $u$ is $C^2$. This is a rather old result, but I do not know the reference. A quick Google search leads to Poisson's equation by T. Gantumur, see Theorem 42 and Corollary 43 there.

  • $\begingroup$ Thanks. I'll check it out. $\endgroup$ Jul 12 '19 at 18:04
  • 2
    $\begingroup$ Additional references: the Laplacian version seems to have been due to Shapiro, and there is a version with more general elliptic operators due to Burch. $\endgroup$ Jul 15 '19 at 13:49
  • $\begingroup$ @WillieWong: Interesting, thanks! (I learned about the result from a colleague of mine. When I meet him, I will ask if he knows any relevant references.) $\endgroup$ Jul 15 '19 at 19:41
  • $\begingroup$ @MateuszKwaśnicki: and I learned about the result from you :-) I just looked up the keywords on MathSciNet for those references. $\endgroup$ Jul 15 '19 at 21:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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