Let $u$ be a real harmonic function on $\Bbb C-\{0\}$. I'm trying to show that if $u$ is bounded above, then $u$ must be a constant.
This is my attempt: It is known that there exists constants $a,b\in \Bbb R$ such that $$\frac{1}{2\pi}\int_0^{2\pi} u(re^{i\theta})~d\theta=a \log r+b$$ for all $r>0$. Since $u$ is bounded above, we must have $a=0$. Then it seems that we should have $u \equiv b$, but I have no idea here.
Any hints?
I would suggest a direct proof as follows; first, by taking $u-M$ where $M$ is an upper bound for $u$, we can assume $u \le 0$.
Then consider the two simple connected domains $D_1, D_2$ obtained by cutting out the negative real axis and the positive imaginary axis respectively, so $D_1 \cap D_2$ has two components, $U_1, U_2$, with $U_2$ being the second quadrant and $U_1$ the rest of the plane minus the two rays (union of the first, third, fourth quadrants including the positive real axis ray and the negative imaginary ray)
Let $f_1$ be the unique analytic extension of $u$ on $D_1$, $f_1(1)=u(1)$, so $f_1=u+iv_1, v_1(1)=0$ and similarly $f_2$ on $D_2$, $f_2=u+iv_2, f_2(1)=u(1), v_2(1)=0$
Clearly $f_1-f_2$ is analytic and with real part $0$ on $D_1 \cap D_2=U_1 \cup U_2$, so $f_1-f_2$ is an imaginary constant (possible different) on $U_1$ and $U_2$ respectively. But by our choice $1 \in U_1$ so $f_1(1)=f_2(1)$ hence $f_1=f_2$ there. If $f_1=f_2$ on $U_2$, $f(z)=f_1(z), z \in D_1, f(z)=f_2(z), z \in D_2$ is well defined and analytic in the punctured plane and then $g=e^{-f}$ is bounded in the punctured plane, hence in particular it extends to an entire bounded function, so it is constant, hence $f$, hence $u$ is constant.
If $f_1(z)-f_2(z)= ic \ne 0, z \in U_2, c \in \mathbb{R}$, we define $g(z)=e^{-\frac{2\pi}{|c|}f_1(z)}, z \in D_1, g(z)=e^{-\frac{2\pi}{|c|}f_2(z)}, z \in D_2$ then clearly $g$ is well defined and bounded in the punctured plane, hence extends to an entire bounded function, hence is constant, hence $u$ constant