# Proof: Integral is finite (Fubini/Tonelli?)

1. Mar 22, 2013

### ChemEng1

1. The problem statement, all variables and given/known data
Let f:[0,1]→ℝ be non-negative and integrable. Prove that $\int_{[0,1]}\frac{f(y)}{|x-y|^{1/2}}dy$ is finite for ae x in [0,1]

2. Relevant equations
This looks like a Fubini/Tonelli's Theorem problem from the problem givens.

3. The attempt at a solution
I honestly don't know where to start with this problem. Any help or gentle nudges would be appreciated.

2. Mar 22, 2013

### haruspex

It doesn't seem to me to be true, so I must be missing something.
Given x in (0, 1), consider $f(y) = |x-y|^{-\frac{1}{2}}$ for y≠x, f(x)=0. Isn't that integrable? But $\int_0^1|x-y|^{-1}dy$ is not finite.

3. Mar 25, 2013

### ChemEng1

Here's what I am thinking.

Consider: $\int_{[0,1]}f(y)\left[\int_{[0,1]}\frac{1}{|x-y|^{1/2}}dx\right]dy=\int_{[0,1]}f(y)\left[2\left(\sqrt{1-y}-\sqrt{y}\right)\right]dy\leq\int_{[0,1]}f(y)\cdot 2<∞$. Therefore $\int_{[0,1]^{2}}\frac{f(y)}{|x-y|^{1/2}}<\infty$ by Tonelli's Theorem. Then $\int_{[0,1]}\frac{f(y)}{|x-y|^{1/2}}dy$ is finite for ae x in [0,1] again by Tonelli.

Think I still need to add some measurability and non-negative qualifiers, but this is close.

Thoughts?

Last edited: Mar 25, 2013