Cauchy-Schwarz-like inequality of integrals

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.

• 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 $\frac{1}{48}$

• 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