Let $F(z)$ be an analytic function in an open sector $\Sigma_{\gamma}=\{0<\arg z<\gamma<\frac{\pi}{2}\}$, and continuous to the boundary. Then $\log{|F(re^{i\theta})|}$, $z=re^{i\theta}\in \Sigma_{\gamma}$, is sub-harmonic.
My question is for each fixed $r>0$, is $\log{|F(re^{i\theta})|}$ a convex function of $0<\theta<\gamma$? i.e., do we have $$ |F(re^{i\theta})|\leq |F(re^{i\gamma})|^{\frac{\theta}{\gamma}}|F(r)|^{1-\frac{\theta}{\gamma}},~~0<\theta<\gamma~~? $$
The answer should be no. Take $r=1, \gamma = \pi/4$, and $\theta = \pi/6$. Let $F(z) = 1/(z-2+0.1i).$ This is analytic in the open sector since it's isolated from the pole at $z=2-0.1i.$ But you can check numerically that
$$ \vert F(e^{i\pi/6}) \vert > \vert F(e^{i\pi/4}) \vert^{2/3} \cdot \vert F(1)\vert^{1/3} $$
I'm sure someone can come up with something more elegant, but I got the idea just by drawing a picture.