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I'm struggling to show that $ \int_0^1 \int_0^1 \log \left| x-y\right|\,dx\,dy >-\infty$ which means that $f(x,y):=\log| x-y|$ is in $L^1([0,1]^2,\lambda\otimes \lambda)$. The part "$<+\infty$" is not difficult.

The lower bound $1-\frac{1}{u} \leq \log u$ for all $u>0$, reduces the problem to $ \int_0^1 \int_0^1 \frac{1}{\left| x-y\right|}\,dx\,dy>-\infty$.

Any help/hint would be appreciated.

Michael Hardy
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anonymus
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  • The diagonal ${x=y}$ divides the square into two triangles, over which the integral is the same. On each one of them, I guess you can use Fubini. – Amitai Yuval Jun 13 '17 at 14:37

3 Answers3

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By setting $f(x,y)=\log\left|x-y\right|$ we have $f(x,y)=f(y,x)$ and $f\leq 0$ for any $(x,y)\in(0,1)^2$, hence

$$ \iint_{(0,1)^2}f(x,y)\,dx\,dy = 2\int_{0}^{1}\int_{0}^{x}f(x,y)\,dy\,dx \stackrel{y\mapsto xz}{=} 2\iint_{(0,1)^2}x\,f(x,xz)\,dz\,dx$$ and the original integral boils down to: $$ 2\int_{0}^{1}\int_{0}^{1}x \log(1-z)+x\log(x)\,dx\,dz = 2\int_{0}^{1}x(\log x-1)\,dx=\color{red}{-\frac{3}{2}}.$$

Jack D'Aurizio
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Proceeding we note from symmetry that we can write the integral of interest as

$$\begin{align} \int_0^1\int_0^1 \log(|x-y|)\,dx\,dy&=2\int_0^1\int_y^1 \log(x-y)\,dx\,dy\\\\ &=2\int_0^1 \left.\left((x-y)\log(x-y)-(x-y)\right)\right|_{x=y}^{x=1}\,dy\\\\ &=2\int_0^1\left( (1-y)\log(1-y)-(1-y)\right)\,dy\\\\ &=-\frac32 \end{align}$$

Mark Viola
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Showing $$ \int_0^1 \int_0^1 \log \left| x-y\right|dxdy>-\infty $$ is equivalent to showing $$ \int_0^1 \int_0^1 \log \frac1{\left| x-y\right|}dxdy<\infty. $$ Noting $$ x^{\frac12}\ln\frac1x\le C $$ for some constant, so one has $$ \int_0^1 \int_0^1 \log \frac1{\left| x-y\right|}dxdy\le C\int_0^1 \int_0^1 \frac1{\sqrt{\left| x-y\right|}}dxdy<\infty. $$

xpaul
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