Let $\Omega\subseteq\mathbb C$ be an open set and let $\phi:\Omega\to\mathbb R_{\geq 0}$ be an exhausting (i.e. proper) smooth subharmonic function. Fix $p\in\Omega$. Does there exist a harmonic function $h:\Omega\to\mathbb R$ such that $h(q)\leq\phi(q)$ with equality iff $q=p$?
1 Answer
The answer is no. The reason is that a subharmonic function does not have to be continuous. For example, there is a subharmonic function $u$ in the unit disk, $u(0)=1$ and $u(z_k)=0$ for a sequence $z_k\to 0$. It is not difficult to make it proper. Now evidently there is no harmonic support function at $0$ because harmonic functions are continuous.
To construct $u$ take $$v(z)=1+\sum a_k\log1z/z_k,$$ where $z_k\to 0$ and $a_k>0$ are so small that the sum converges and $v(0)=1$. To make it positive, take $v_1=\max\{ v,0\}$. To make it proper take $u(z)=\max\{ u,\logz+M\},$ where $M$ is very large.
EDIT. Requiring that $u$ is continuous or smooth will not help, as seen from the following argument. Let $u_n\geq u$ be a decreasing sequence of smooth subharmonic functions, $u_n(z)\to u(z)$ pointwise. Such a sequence can be constructed by taking convolutions of $u$ with smooth positive "hat" functions. Suppose that we have harmonic functions $v_n\leq u_n$ with $v_n(0)=u_n(0)$. As our $u_n$ are uniformly bounded from avove, $v_n$ are also uniformly bounded from above. So you can find a subsequence on which $v_n$ converge uniformly on compact subsets. Then the limit function $v$ will be harmonic, $v(0)=u(0)$ and $v\leq u$. Contradiction.
EDIT 2. I don't know why you need this, but there is a theorem that can perhaps substitute: Consider functions of the form $\logf$ where $f$ is analytic. Then allow linear combinations with positive coefficients and taking pointwise $\max$, and limits of decreasing sequences. Then you obtain all subharmonic functions. This is due to Harnack.

$\begingroup$ OK, yes. I guess I should have added the assumption that $\phi$ is smooth, since that's the case I'm interested in. Do you know if it holds in that case? $\endgroup$ Commented May 4, 2014 at 3:17