I have proved using Green's theorem the following equality:$$\frac{1}{2\pi i}\left(\frac{1}{\zeta}-\bar{\zeta}\right)\int_{\partial \mathbb{D}}f(z+\zeta)\log\left(\frac{z}{z-\zeta}\right)dz=f(\zeta)+\frac{\zeta}{\pi}\iint_{\mathbb{D}}\frac{\bar{z}-\bar{\zeta}}{1-\zeta \bar{z}}f(z+\zeta)dxdy $$ For some $\zeta$ inside the disc and for $f:\Omega\to \mathbb{C}$ , $\mathbb{D} \subset \Omega$. Now I would like to estimate the integral in the left side. I consider a $\zeta$ such that $0<|\zeta|<1$ Now i would like to estimate $f(\zeta)$. After calculating the contour integral with residues, what i get is: $$f(\zeta)=\zeta f(2\zeta)\left(\frac{1-|\zeta|^2}{\zeta}\right)+\frac{\zeta}{\pi}\iint_{\mathbb{D}}\frac{\bar{z}-\bar{\zeta}}{1-\zeta \bar{z}}f(z+\zeta)dxdy$$ and after using the triangle inequality I get: $$|f(\zeta)|\le (1-|\zeta|^2)|f(2\zeta)|+\frac{|\zeta|}{\pi}\iint_{\mathbb{D}}|f(z+\zeta)|dxdy.$$ . After substituting $f(z+\zeta)$ with $(f^{'}(z+\zeta))^2$ one has a lower bound for the Area of the function inside $\mathbb{D}$. Does this estimation have a meaning? Some calculations give a good estimation but I would like a full answer.
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