TITLE: Big Rudin Exercise 3.26 - Which integral is larger
QUESTION [15 upvotes]: This is exercise 3.26 in Rudin's Real & Complex Analysis:

If $f$ is a positive measurable function on $[0,1]$, which is larger,
  $$\int_0^1 f(x) \log f(x) \, dx$$
  or
  $$\int_0^1 f(s) \, ds \int_0^1 \log f(t) \, dt$$

I tried a bunch of functions and always got the first to be larger, which suggests that Hölder's inequality won't help here (at least not a direct application). I couldn't find an example that made the second larger. I'm stuck otherwise.
(This is self-study, not homework)
Clarification: The integral here is the Lebesgue integral. The only answer so far is only applicable to Riemann integrable functions.

REPLY [25 votes]: The function $x\mapsto x\log x$ is convex on $(0,\infty)$, as its second derivative is positive. Thus by Jensen's inequality,
$$\int_0^1 f(t)\log f(t) dt \geq \int_0^1 f(t) dt \log\left( \int_0^1 f(t) dt \right) .$$
The function $x\mapsto \log x$ is concave, so another application of Jensen's inequality yields
$$\log\left( \int_0^1 f(t) dt \right) \geq \int_0^1\log f(t) dt .$$
Combining these two inequalities proves the result.