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Proof: Integral is finite (Fubini/Tonelli?)

  1. Mar 22, 2013 #1
    1. The problem statement, all variables and given/known data
    Let f:[0,1]→ℝ be non-negative and integrable. Prove that [itex]\int_{[0,1]}\frac{f(y)}{|x-y|^{1/2}}dy[/itex] 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. jcsd
  3. Mar 22, 2013 #2


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    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.
  4. Mar 25, 2013 #3
    Here's what I am thinking.

    Consider: [itex]\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<∞[/itex]. Therefore [itex]\int_{[0,1]^{2}}\frac{f(y)}{|x-y|^{1/2}}<\infty[/itex] by Tonelli's Theorem. Then [itex]\int_{[0,1]}\frac{f(y)}{|x-y|^{1/2}}dy[/itex] 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.

    Last edited: Mar 25, 2013
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