In reading a paper, I've come across an "obvious" statment that looks funky. To make the question applicable to a wider audience let me simplify it a bit:
Suppose $h$ is subharmonic on the annulus $A = \{ z \in \mathbb{C}: c_1 < |z| < c_2\}$. Define $f(z) = \frac{1}{2 \pi}\int_0^{2\pi} h(|z|e^{i\theta}) d \theta$. Is $f$ subharmonic on $A$?
Ideas or references are appreciated.
Essentially, this follows from sub-mean value property and using the correct reparametrization.
First note that $ f(w) = \frac{1}{2 \pi} \int_0^{2\pi} h(|w| e^{i\theta}) d \theta = \frac{1}{2 \pi} \int_0^{2\pi} h(w e^{i\theta}) d \theta$ for each $w \in A$.
Now consider an arbitrary point $p \in A$. Consider the circle $|z|=|p|$. Then for some small $c>0$ we have for each $z$, $h(z) \le \frac{1}{2 \pi} \int_0^{2\pi} h(z+ ce^{it}) dt = \frac{1}{2 \pi} \int_0^{2\pi} h(z+ ce^{i(t+\theta)}) dt$.
Integrating both sides over the circle |z|=|p| yields: \begin{align*} \frac{1}{2 \pi} \int_0^{2 \pi} h(|p|e^{i\theta}) d \theta &\le \left( \frac{1}{2 \pi} \right)^2 \int_0^{2 \pi} \int_0^{2 \pi} h(|p|e^{i\theta} + ce^{i(t+\theta)}) dt d\theta \\ &= \left( \frac{1}{2 \pi} \right)^2 \int_0^{2 \pi} \int_0^{2 \pi} h((|p|+ce^{it})e^{i\theta}) d\theta dt\\ &= \frac{1}{2 \pi} \int_0^{2 \pi} f(|p|+ce^{it}) d t \end{align*}
Thus $f$ is subharmonic at $p$.