I suppose that for $f(x) \geq 0$, $$ \left(\int_\Omega f\,dx\right)^2 \geq C\int_\Omega f^2\,dx $$ because $(a+b)^2 \geq a^2 + b^2$ for $a,b \geq 0$.

Is this inequality true? How can I prove it?

  • $\begingroup$ If $\Omega$ has measure $1$, then you have the reverse inequality with $C=1$ (by Jensen's inequality). $\endgroup$ – Prahlad Vaidyanathan May 11 '14 at 7:16

No, the result is not true, unless $C = 0$. Consider $$f_n = n \chi_{[0,1/n]}$$

on $\Omega = [0,1]$ (or $\Omega = \mathbb R)$. Then $$\int_{\Omega} f_n dx = 1$$

for every $n$, while

$$\int_{\Omega} f_n^2 dx = n$$

Essentially, the question is equivalent to asking whether

$$\|f\|_{L^2} \lesssim \|f\|_{L^1}$$

after taking square roots; this is quite false without (perhaps) some sort of uniform boundedness on $f$.


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