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!