Let $\ f:[0,1]\rightarrow\ [1,\infty$) be Lebesgue measurable. Which quantity is greater:
$\int_{0}^{1} f(x)\log f(x)dx$
or
$\int_{0}^{1} f(y)dy\int_{0}^{1}\log f(w)dw$ ?
Prove your answer, for all such $f.$
Is the Hölder inequality sufficient to prove this?
$$ \int_0^1 f(x) \log f(x) dx = \int_0^1 \psi (f(x))dx $$ where $\psi (t) =t \log(t)$ is a convex function. So, by Jensen's inequality $$ \int_0^1 f(x) \log f(x) dx \ge \psi\left(\int_0 ^1 f(x)dx\right)=\left(\int_0 ^1 f(x)dx\right)\; \log\left(\int_0 ^1 f(x)dx\right) $$ If we can show that $$ \log\left(\int_0 ^1 f(x)dx\right) \ge \int_0 ^1 \log f(x)dx $$ we are done. Call $g=\log f(x)$. The previous inequality is equivantly to $$ \int_0^1\exp{g(x)}dx \ge \exp\left( \int_0^1 g(x) dx\right) $$ which is true, again, thanks to Jensen's inequality.