Second order differentiability of subharmonic function almost everywhere?

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.

• Even the first derivative of a subharmonic function is not defined pointwise. But of course all derivatives exist as Schwartz distributions. – Alexandre Eremenko Aug 12 '15 at 1:03
• @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? – student Aug 12 '15 at 1:25
• The answer is no. The distributional second derivative can be ARBITRARY positive measure. BTW, a subharmonic function can be discontinuous everywhere. – Alexandre Eremenko Aug 12 '15 at 18:59
• @Alexandre Eremenko, I guess a subharmonic function can be discontinuous quasi-everywhere (not everywhere). – user111 May 23 '18 at 9:19

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 $u$.