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

Let $f,g,$ be integrable on $[a,b]$. Prove that $$\int_a^b(fg)^2\le\int_a^bf^2\int_a^bg^2$$

I know that from Cauchy-Schwarz we have $$\left(\int_a^bfg\right)^2\le\int_a^bf^2\int_a^bg^2$$

so if we showed that $$\int_a^b(fg)^2\le\left(\int_a^bfg\right)^2$$ we would be done. But I don't think this is even true in general, so this method doesn't seem to lead anywhere. Is there another approach to this problem that I'm missing?

Edit: The inequality seems to be false. Perhaps the inequality was given incorrectly.

share|cite|improve this question
Indeed, the last inequality does not hold in general: the converse holds by Jensen's inequality. – Clement C. Feb 3 '14 at 1:17
What is true is that this holds for (real-valued) square-integrable functions, for which the proof is a "simple" matter of proving that $\int_a^b fg$ is in fact an inner product. – Eric Stucky Feb 23 '14 at 22:13

This doesn't seem to be true in general. For example:

Take $f(x)=1$ for $x \in [0,\frac{1}{2}]$

$g(x)=x$ (same interval.)

Then, your LHS= $\frac{1}{24}$, RHS is $\frac{1}{48}$

share|cite|improve this answer
That seems correct. I'll have to ask about this problem. Perhaps it was written wrong and was meant to be the CSI itself. – ant11 Feb 3 '14 at 1:21
@ant11: Yes please verify the correct question. – voldemort Feb 3 '14 at 1:22

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.