I am asking how to prove that:
If $f$ is nonconstant and analytic on $B(0,R)$, then
$I(r):= \frac{1}{2\pi}\int_0^{2\pi}\rvert f(re^{i\theta})\rvert d\theta$
is a strictly increasing function (and I believe this should be: $I: (0,R)\to \mathbb{R}$).
(using fairly elementary stuff such as Schwarz Lemma and Max Modulus, nothing about harmonic functions though).
What I have so far:
so far I have the following (part of which I found from another MSE post, but part of the answer I found in that other MSE post is either not correct or at least very unclear):
Let $B:=B(0,R)$. $f$ is not identically zero, so the set of circles $\{\{re^{i\theta}\mid \theta\in \mathbb{R}\}\mid r\in (0,R)\}$ which contain at least one zero of $f$, must be finite by the identity principle. Pick an $r\in (0,R)$ s.t. $f$ is nonzero everywhere on the circle of center origin and radius $r$ and let $\phi(\theta):= \frac{\rvert f(re^{i\theta}\rvert}{f(re^{i\theta})}$. Then $\phi:[0,2\pi]\to \mathbb{C}$ is continuous and consequently (By Morera's theorem for triangles) $F(z):= \int_0^{2\pi}f(ze^{i\theta})\phi(\theta)d\theta$ is analytic on $B$. Note also that for $t\in (0,R)$,
$max_{\rvert z\rvert = t}\rvert F(z)\rvert\leq max_{\rvert z\rvert = t}\int_0^{2\pi}\rvert f(ze^{i\theta})\rvert \rvert\phi(\theta)\rvert d\theta = max_{x\in \mathbb{R}}\int_0^{2\pi}\rvert f(ze^{i(x+\theta)})\rvert \rvert\phi(\theta)\rvert d\theta = \int_0^{2\pi}\rvert f(te^{i(\theta)})\rvert d\theta = 2\pi I(t)$.
As in: $max_{\rvert z\rvert = t}\rvert F(z)\rvert\leq 2\pi I(t)$.
If I can conclude that $F$ is nonconstant (e.g. somehow BWOC using the fact that $f$ is nonconstant), then I think I can figure the rest out using the above inequality and the Max Modulus Theorem, but I really cannot figure out to save my life why $F$ should be nonconstant (I have tried a couple arguments but not getting anything yet).