$$ \mbox{If}\quad\mathrm{f}\left(x\right) = \int_{B\left(0,R\right)} \log\left(\frac{\left\vert\,x - y\,\right\vert} {\left\vert\,x - {R^{\large 2} \over \left\vert\left\vert\,y\,\right\vert\right\vert^{\large 2}} \,y\,\right\vert}\right)\,\mathrm{d}y $$ where $x\in B\left(0,R\right)$, how to find the minimum or bound of $\mathrm{f}\left(x\right)$ on $x \in B\left(0,R\right)$ only dependent on the radius $R$ ?.
UPDATE: Taking derivative will only lead to upper bound but not lower bound. Could anyone kindly help? Thanks.
We have the following Lemma, which is related to the properties of the Poisson kernel: $$ \text{if }a\in(0,1),\qquad \int_{0}^{2\pi}\log\left|a-e^{it}\right|\,dt = 0,$$ $$ \text{if }a>1,\qquad \int_{0}^{2\pi}\log\left|a-e^{it}\right|\,dt = 2\pi\log a.$$ Let us focus on $\iint_{B(0,R)}\log|x-y|\,d\mu$.
By symmetry we may clearly assume that $x\in(0,R)$. With such assumption we have $$ \iint_{B(0,R)}\log|x-y|\,d\mu = \int_{0}^{2\pi}\int_{0}^{R}\rho \log|x-\rho e^{i\theta}|\,d\rho\,d\theta\\=\int_{0}^{R}2\pi\rho\log\rho+2\pi\rho\max\left(0,\log\frac{x}{\rho}\right)d\rho $$ hence the LHS equals $$ \frac{\pi R^2}{2}(2\log R-1)+2\pi\int_{0}^{x}\rho\log\frac{x}{\rho}\,d\rho=\frac{\pi R^2}{2}(2\log R-1)+\frac{\pi x^2}{2}. $$ Similarly, $$ \iint_{B(0,R)}\log\left|x-\frac{R^2 y}{\|y\|^2}\right|\,d\mu=\int_{0}^{R}\int_{0}^{2\pi}\rho \log\left|x-\frac{R^2}{\rho} e^{i\theta}\right|\,d\theta\,d\rho =\frac{\pi R^2}{2}(2\log R+1)$$ does not really depend on $x$. It follows that for any $R>0$ the minimum of the given integral, $f(x)=-\pi R^2+\frac{\pi}{2}\|x\|^2$, is attained at the origin, and it equals $$ \iint_{B(0,R)}\log\frac{\|y\|^2}{R^2}\,d\mu = -\pi R^2. $$