Second order differentiability of subharmonic function almost everywhere?

Question

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.

Answer 1

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:

https://mathoverflow.net/q/296845

Answer 2

In fact a weaker version of Aleksandrov's theorem is true for subharmonic functions. Since $\Delta u$ is a Radon measure, the following result follows from Proposition 4.4 in 1.

Theorem. If $u:\Omega\to\mathbb{R}$ is subharmonic and locally integrable, then for almost all $x\in\Omega$ there exist a vector $Du(x)$ and a matrix $D^2u(x)$ such that for all $1\leq p<n/(n-2)$ ($1\leq p<\infty$, if $n=2$) $$ \left(\frac{1}{|B(x,r)|}\int_{B(x,r)}|u(y)-u(x)-Du(x)(y-x)-\frac{1}{2}(y-x)^TD^2u(x)(y-x)|^p\right)^{1/p}\\=o(r^2). $$

Although it is not stated in [AG] that the matrix $D^2u(x)$ is symmetric, you can always assume that since it appears in a quadratic form and in a quadratic form you can always replace a matrix by a symmetric one since $$ \langle Ax,x\rangle=\big\langle\frac{1}{2}(A+A^T)x,x\big\rangle. $$

As a corollary, the authors prove Imomkulov's theorem (see the answer of user111), but they do it independently (21 years later!) since they do not quote the work of Imomkulov.

If you want to compare it with the statement of Aleksandrov's theorem, see Second order differentiability of convex functions and Aleksandrov's proof of the second order differentiability of convex functions.

[AG] G. Alberti, S. Bianchini, C. G. Stefano, On the $L^p$-differentiability of certain classes of functions . Rev. Mat. Iberoam. 30 (2014), no. 1, 349–367. MathSciNet