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?
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?
Copyright © 2021 JogjaFile Inc.
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 $f$.