Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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, $\|x^{-1/3}f(x)\|_1\leq \|x^{-1/3}\|_2\|f(x)\|_2 < \infty$.


share|cite|improve this question

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 $L^{2}([0,1])$.

share|cite|improve this answer
Thanks! That's what I just figured out! – Ferenc Nov 17 '12 at 16:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.