$\sum^{\infty}_{n=0}\frac{1}{2^n}\log |z-\frac{1}{2^n}|$ defines a subharmonic function over $\mathbb{C}$

Question

I am preparing for my complex analysis qualifying exam. I encountered the following question that I don't really know how to solve.
The question has two parts and I am able to solve the first part which is $$\frac{1}{2\pi}\int^{2\pi}_0 \log|re^{it}-\rho|dt=\max\{\log |r|,\log |\rho|\}$$ for real numbers $r>0$ and $\rho$ such that $|r|\neq \rho$.
The second part is to show that $u(z)=\sum^{\infty}_{n=0}\frac{1}{2^n}\log |z-\frac{1}{2^n}|$ defines a subharmonic function over $\mathbb{C}$.
My thought is to show that for $\overline{B(a,r)}\subset \mathbb{C}$, we have $$u(a)\le \frac{1}{2\pi}\int^{2\pi}_0 u(a+re^{it})dt$$ Notice that when $a=\frac{1}{2^n}$, the left hand side is $-\infty$ so we don't need to do anything. However, when $a$ is not one of $\frac{1}{2^n}$, I don't know how to do them. My feeling is that if $\overline{B(a,r)}$ doesn't contain any $\frac{1}{2^n}$ by using dominated convergence theorem we may be able to show that the equality holds.

Thank you.

Answer 1

The comment by kobe pretty much gives you how to proceed via your idea. So with that hint in the bag, the only hurdle left is justifying interchanging an integral and the summation defining $u$. Let $\overline{B}(z, r)$, $r > 0$ be a closed disc in the complex plane. From the triangle inequality and the fact that the real logarithm is strictly increasing, $$\log\left|z - \frac{1}{2^n} + re^{it}\right| \leq \log\left(\left|z - \frac{1}{2^n}\right| + r\right)$$ for any $t \in [0, 2\pi]$ and $n = 0, 1, 2, \dots$. So by continuity, for any $n = 0, 1, 2, \dots$ and $t \in [0, 2\pi]$, $$\left|\frac{1}{2^n}\log\left|z - \frac{1}{2^n} + re^{it}\right|\right| \leq \frac{1}{2^n} \sup_{x \in [0, 1]} \left|\log(|z - x| + r)\right|.$$ Call $M_{z, r} = \sup\limits_{x \in [0, 1]} |\log(|z - x| + r)|$ and note that $M_{z,r} > 0$. Since $\sum\limits_{n = 0}^\infty \frac{M_{z, r}}{2^n}$ converges, the function $$t \mapsto \sum_{n = 0}^\infty \frac{1}{2^n}\log\left|z - \frac{1}{2^n} + re^{it}\right|$$ converges uniformly on $[0, 2\pi]$ by the Weierstrass M-test. Hence, we may interchange the integral and summation to obtain \begin{align*} \frac{1}{2\pi}\int_0^{2\pi} u(z+re^{it})dt ={} & \frac{1}{2\pi}\int_0^{2\pi} \sum_{n = 0}^\infty \frac{1}{2^n}\log\left|z - \frac{1}{2^n} + re^{it}\right| dt \\[5pt] ={} & \sum_{n = 0}^\infty \frac{1}{2^n} \frac{1}{2\pi}\int_0^{2\pi} \log\left|z - \frac{1}{2^n} + re^{it}\right| dt \\[5pt] ={} & \sum_{n = 0}^\infty \frac{1}{2^n}\max\left\{\log\left|z - \frac{1}{2^n}\right|, \log r\right\} \\[5pt] \geq{} & \sum_{n = 0}^\infty \frac{1}{2^n}\log\left|z - \frac{1}{2^n}\right| \\[5pt] ={} & u(z). \end{align*} As $\overline{B}(z, r)$ was an arbitrary closed disc, it follows that $u(z) = \sum\limits_{n = 0}^\infty \frac{1}{2^n}\log\left|z - \frac{1}{2^n}\right|$ defines a subharmonic function on the complex plane.