# Find minimum or lower bound of $f(x)=\int_{B(0,R)}\log(\frac{|x-y|}{|x-\frac{R^2}{||y||^2}y|})dy$

$$\mbox{If}\quad\mathrm{f}\left(x\right) = \int_{B\left(0,R\right)} \log\left(\frac{\left\vert\,x - y\,\right\vert} {\left\vert\,x - {R^{\large 2} \over \left\vert\left\vert\,y\,\right\vert\right\vert^{\large 2}} \,y\,\right\vert}\right)\,\mathrm{d}y$$ 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.

• Consider what the power of a point theorem tells you about $$\oint_{\Gamma}\log|x-y|\,dy$$ where $\Gamma$ is a circle enclosing $x$. Additionally, it is pretty clear that $f(x)$ only depends on $\|x\|$ and there are not many candidate stationary points in $B(0,R)$. – Jack D'Aurizio Mar 1 '18 at 19:22
• @JackD'Aurizio Thank you for your hint. I don't quite understand your idea. Can you give me more details? – Sherry Mar 1 '18 at 21:11

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 $$\iint_{B(0,R)}\log\frac{\|y\|^2}{R^2}\,d\mu = -\pi R^2.$$