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I asked a previous question, but did a mistake in it. If $f\in L^2([0,1])$, is it true that $f/x^{1/3}$ will be in $L^1([0,1])$?

Edit: After thinking about it, I think the answer is yes. By Holder's inequality, $\|x^{-1/3}f(x)\|_1\leq \|x^{-1/3}\|_2\|f(x)\|_2 < \infty$.

Thanks!

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By Holder's inequality, for measurable functions $f,g$, we have $||fg||_{1}\leq||f||_{2}||g||_{2}$. Take $g(x)=x^{-1/3}$ and see if it is in $L^{2}([0,1])$.

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Thanks! That's what I just figured out! –  Ferenc Nov 17 '12 at 16:07
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