subharmonic on punctured disk but extends continuously to origin Suppose that $f$ is bounded and subharmonic on $\Omega = D(0,1) - \{0\}$.  Suppose also that $L = \sup_{z \in \Omega} f(z)$ and $\lim_{z \to 0}f(z) = L$. How can I prove that $f$ extends subharmonically to all of $D(0, 1)$ with $f(0) = L$?
 A: This follows from Theorem 5.18 in the book "Subharmonic functions", Vol. 1, by Hayman and Kennedy, London Math Society Monographs, number 9, Academic Press, 1976. (They prove much more, take a look.) 
A: One can proceed similarly as in How to prove Liouville's theorem for subharmonic functions:
If $f$ is subharmonic in the punctured unit disk $\Bbb D \setminus \{0\}$
 then
$$ \tag{*}
 f(z) \le \frac{\log r_2 - \log |z|}{\log r_2 - \log r_1} M(r_1, f)
 + \frac{\log |z| - \log r_1}{\log r_2 - \log r_1} M(r_2, f)
$$
for $0 < r_1 < |z| < r_2 < 1$, where
$$
  M(r, f) := \max \{ f(z) : |z| = r \} 
$$
(the "Hadamard three-circle theorem" for subharmonic functions, see below
for a proof).
Since $f$ is bounded above by $L$ it follows that
$$ 
 f(z) \le \frac{\log r_2 - \log |z|}{\log r_2 - \log r_1} L
 + \frac{\log |z| - \log r_1}{\log r_2 - \log r_1} M(r_2, f)
$$
for $0 < r_1 < |z| < r_2 < 1$. For $r_1 \to 0$ we conclude that
$$
 f(z) \le M(r_2, f)
$$
for $0 < |z| < r_2 < 1$, and for $z \to 0$ it follows that
$$
 L = \lim_{z \to 0} f(z) \le M(r_2, f) \le L \, .
$$
So $M(r_2, f) = L$, which means that $f$ attains its maximum in
the domain and therefore is constant.

Proof of $(*)$: Define
$$
 u(z) = \frac{\log r_2 - \log |z|}{\log r_2 - \log r_1} M(r_1, f)
 + \frac{\log |z| - \log r_1}{\log r_2 - \log r_1} M(r_2, f)
$$
for $z \in A = \{ z \mid r_1 \le \vert z \vert \le r_2 \}$.
$u$ is continuous on the annulus $A$ and harmonic in its interior.
Therefore $v := f - u$ is upper semicontinuous in $A$ and subharmonic
in its interior. Also $v \le 0$ on the boundary of $A$. 
The maximum principle for subharmonic functions implies that $v \le 0$
in $A$.
