# 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. Aug 12, 2015 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? Aug 12, 2015 at 1:25
• The answer is no. The distributional second derivative can be ARBITRARY positive measure. BTW, a subharmonic function can be discontinuous everywhere. Aug 12, 2015 at 18:59
• @Alexandre Eremenko, I guess a subharmonic function can be discontinuous quasi-everywhere (not everywhere). May 23, 2018 at 9:19
• @AlexandreEremenko In fact a weaker version of Aleksandrov's differentiability is true, see my answer. Feb 29, 2020 at 16:29

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

• I am just curious. Did you learn about Imomkulov's theorem from my post or you knew it before? Feb 29, 2020 at 16:29
• Hi, that's some time ago..I was interested in regularity of potentials, like quasi-continuity. That may be the reason why I found the paper of Imomkulov. There is also another related paper : A. Sadullaev and R. Madrakhimov, Smoothness of subharmonic functions, Math. USSR Sb. 69 (1991). Feb 29, 2020 at 16:55

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 $$1\leq p ($$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

• One of the MO questions linked here is deleted at the moment. Of course, 10k+ users can still see it - for others, there is at least a snapshot in the Wayback Machine. Nov 10 at 8:35
• @MartinSleziak I undeleted the question. I deleted it temporarily since I had my solution and I was in process ot typying it. Nov 11 at 15:22