I am trying to solve the following problem:
Let $f:[0,1] \rightarrow [1,\infty)$ be Lebesgue measurable. Which quantity is greater:
$\int_0^1 \! f(x)\log f(x) \, \mathrm{d}x$
or
$\int_0^1 \! f(y) \, \mathrm{d}y \int_0^1 \! \log f(z) \, \mathrm{d}z. $
I was thinking of using Fubini-Tonelli in some way, but I am not getting anywhere. I would be very grateful for some help. Thanks!
On
The functions $u:x\mapsto x$ and $v:x\mapsto\log x$ are both nondecreasing on $[1,+\infty)$ hence they are positively associated. Thus, for every function $f$ with values in $[1,+\infty)$, $$\int_0^1u(f)\cdot\int_0^1v(f)\leqslant\int_0^1u(f)v(f). $$ The simplest approach to prove this might be to expand the LHS of the inequality $$ \int_0^1\!\!\int_0^1\left[u(f(x))-u(f(y))\right]\cdot\left[v(f(x))-v(f(y))\right]\cdot\mathrm dx\mathrm dy\geqslant0. $$
Did originally posted this answer, with a small omission, but changed to another. I think it is worth having this answer, so I will repeat it.
The function $u(x)=x\log(x)$ is convex on $(0,\infty)$: $u''(x)=\frac1x\gt0$.
By Jensen's Inequality, we have $$ \int_0^1f(x)\,\mathrm{d}x\cdot\log\left(\int_0^1f(x)\,\mathrm{d}x\right) \le\int_0^1f(x)\log(f(x))\,\mathrm{d}x\tag{1} $$ Furthermore, since $\log(x)$ is concave on $(0,\infty)$ we have $$ \int_0^1\log(f(x))\,\mathrm{d}x\le\log\left(\int_0^1f(x)\,\mathrm{d}x\right)\tag{2} $$ Since $\int_0^1f(x)\,\mathrm{d}x\ge0$, $(1)$ and $(2)$ imply that $$ \int_0^1f(x)\,\mathrm{d}x\cdot\int_0^1\log(f(x))\,\mathrm{d}x \le\int_0^1f(x)\log(f(x))\,\mathrm{d}x\tag{3} $$