The following general definition of subharmonic function comes from the classical text book [elliptic partial differential equations of second order] by Gilbarg and Trudinger.

We call a function $u$ subharmonic in $\Omega$ if $u \in C^0(\Omega)$ and for every ball $B \subset \subset \Omega$ and every function $h$ harmonic in $B$ satisfying $u \le h$ on $\partial B$, we also have $u \le h$ in $B$.

It is known by Aleksandrov's theorem that convex function has second derivatives almost everywhere, and convex function is subharmonic, so I wonder whether one can prove that a subharmonic function also has second derivatives almost everywhere. Notice that a subharmonic function need not to be convex, for example in $\mathbb{R}^2$, consider $u(z)=log|z|$.

If it is impossible to prove the existence of second derivative, what if we add more conditions on the subharmonic function, for example, we require $u$ to be $W^{1,2}$? The motivation to ask this question is that, in this case, $\lambda:=\Delta u$ would be a positive Radon measure, then I can prove that for almost every $r>0$ such that $B_r \subset \subset \Omega$, $$\int_{B_r} d\lambda = \int_{\partial B_r} \nabla u \cdot \nu $$where $\nu$ is the unit outer normal. The formula above looks very like the trace theorem for BV functions if thinking $\nabla u$ as a BV vector. In philosophy, if trace theorem is true for a function $u$, then $u$ must have one more derivative in some sense.

Disregarding the further condition for $u$, I think the first claim should be provable by adapting the proof of Aleksandrov's theorem, but If it is a known result, I would like to just accept it without doing by myself.

Any comments or ideas would be really appreciated.

  • $\begingroup$ Even the first derivative of a subharmonic function is not defined pointwise. But of course all derivatives exist as Schwartz distributions. $\endgroup$ – Alexandre Eremenko Aug 12 '15 at 1:03
  • $\begingroup$ @Alexandre Eremenko, sorry I didn't say it clearly in the previous title. I'm asking whether subharmonic function has second derivatives almost everywhere. If not true, does it have some fine properties? $\endgroup$ – student Aug 12 '15 at 1:25
  • $\begingroup$ The answer is no. The distributional second derivative can be ARBITRARY positive measure. BTW, a subharmonic function can be discontinuous everywhere. $\endgroup$ – Alexandre Eremenko Aug 12 '15 at 18:59
  • $\begingroup$ @Alexandre Eremenko, I guess a subharmonic function can be discontinuous quasi-everywhere (not everywhere). $\endgroup$ – user111 May 23 '18 at 9:19

Almost everywhere is too strong, but a Lusin-like theorem holds true :

Let $u$ be a subharmonic function in a domain $D\subset\mathbb{R}^n$, $K\subset D$ a compact set, and $\epsilon>0$. There exist an open set $G_\epsilon$ whose Lebesgue measure is less than $\epsilon$ and a twice continuously differentiable function $f$ such that the restriction of $f$ to $K\setminus G_\epsilon$ coincides with $u$.

This was proved in

S. A. Imomkulov, Twice differentiability of subharmonic functions, Russian Acad. Sci. Izv. Math. 41 (1993), no. 1, 157–167

By the way, the above result was also recently mentioned (as an example of an unknown theorem) here:



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.