Let $f(z)$ be a function that is complex analytic in an open region containing the unit disk. Suppose we are given an upper bound of $|f(z)|$ on the unit circle, i.e. $|f(e^{i\theta})|\leq g(\theta), 0\leq\theta\leq 2\pi$. Then what is the maximal possible value of $|f(0)|$?
I feel this should be related to Cauchy's integral technique, since we have $$f(0)=\frac{1}{2\pi i}\oint_{|z|=1}f(z)\frac{h(z)}{z}dz, $$ where $h(z)$ is analytic in the closed unit disk and $h(0)=1$. Therefore, $$|f(0)|\leq \max_{0\leq\theta\leq2\pi}g(\theta)|h(e^{i\theta})|.$$ Hopefully by choosing an optimal $h(z)$ we could find a tight upper bound of $|f(0)|$.
Is this a well-researched problem? More generally, if we are given an upper bound of the absolute value of an analytic function $f(z)$ in a region or a closed curve, how could we get an optimal upper bound on $|f(z)|$ in the interior of this region or curve? Any useful hints or references are highly welcomed. Thanks in advance!
Assuming that $g$ is continuous and strictly positive, you can explicitly construct an extremal function $f$ maximizing $|f(0)|$ as follows: Find a harmonic function $u$ in the unit disk with boundary values $\log g$, let $v$ be its harmonic conjugate, and define $f=e^{u+iv}$. Then $f$ is analytic in the unit disk, with $|f|=e^u = g$ on the boundary, and $$ |f(0)| = e^{u(0)} = \exp\left[\frac{1}{2\pi} \int_0^{2\pi} \log g(\theta) \, d\theta\right]. $$ On the other hand, if you have any complex analytic function $\tilde{f}$ satisfying the given estimate, then $\tilde{u} = \log |\tilde{f}|$ is subharmonic in the unit disk, so you get $$ \log |\tilde{f}(0)| = \tilde{u}(0) \le \frac{1}{2\pi} \int_0^{2\pi} \log g(\theta) \, d\theta = \log |f(0)|. $$ I am pretty sure you can weaken the assumptions significantly to allow $g$ to have zeros and discontinuities, as long as it does not get too wild.
P.S.: I just realized that you wanted $f$ to extend as a complex analytic function to a larger disk. The extremal function constructed above does not even necessarily extend continuously to the closed unit disk, so you will have to rescale and look at functions $(1-\epsilon)f((1-\delta)z)$ with small $\delta,\epsilon > 0$ to get arbitrarily close to the bound given above.