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