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 $q=p$?



Harmonic function osculating a given subharmonic function
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 $$v(z)=1+\sum a\_k\log|1-z/z\_k|,$$
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 
$u(z)=\max\{ u,\log|z|+M\},$ 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 $\log|f|$ 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.


