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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.

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Can you prove the case of simple function first? – ziyuang Oct 6 '12 at 0:26
up vote 22 down vote accepted

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.

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I don't see how your first inequality follows from the convexity of $x\mapsto x\log x$ and Jensen's inequality. – Christian Blatter Oct 9 '12 at 20:11
Let $\phi(x) = x\log x$. The left side is the integral of $\phi\circ f$, the right side is $\phi$ of the integral of $f$. – user15464 Oct 10 '12 at 0:08
Can you explain why my first inequality does not follow from Jensen's inequality? It still looks to me like it is a direct application of the statement of the inequality. – user15464 Oct 10 '12 at 12:26
You are right. Sorry for the inconvenience I have caused. – Christian Blatter Oct 10 '12 at 13:42
@ChristianBlatter - you have not caused any inconvenience. Requesting details of an argument improves the site. – Ben Blum-Smith Nov 9 '13 at 23:38

$$\int_0^1f(t)d(t)=\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n f\left(\frac{i}{n}\right)$$ So in order to prove the inequality $$ \int_0^1 f(x) \log f(x) dx \geq \int_0^1f(s)ds \int_0^1\log f(t)dt $$ it is adequate to show $$ \frac{1}{n}\sum_{i=1}^n f\left(\frac{i}{n}\right) \log f\left(\frac{i}{n}\right) \geq \frac{1}{n}\sum_{j=1}^n f\left(\frac{j}{n}\right) \cdot \frac{1}{n}\sum_{k=1}^n\log f\left(\frac{k}{n}\right) $$

Since $\log f(x)$ increases as $f(x)$ increases, we can apply Chebychev's inequality to give $$ \sum_{j=1}^n f\left(\frac{j}{n}\right) \cdot \sum_{k=1}^n\log f\left(\frac{k}{n}\right) \leq n \sum_{i=1}^n f\left(\frac{i}{n}\right) \log f\left(\frac{i}{n}\right) $$ from which the required result follows immediately.

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This assumes the function is Riemann integrable, no? The exercise is about measurable functions and Lebesgue integral. – PeterM Oct 4 '12 at 8:39
Hmm, interesting. I'm trying to think what sort of non-Riemann-integrable function would make the inequality come out the opposite way. I need to think about this some more. But I'm sure that somewhere in the middle of it all, you'll use Chebychev's inequality or something equivalent. I might replace this answer with something different if I manage to work it out. – user22805 Oct 4 '12 at 8:52
I guess what my (partial) answer does tell us is that if there IS a counterexample to this inequality, then it's not almost-everywhere equal to any Riemann-integrable function. Which means we have to look for something REALLY peculiar. I think the right way to go might be to think in terms of finite linear combinations of indicator functions and apply Chebychev's inequality to the co-efficients. But I'm not sure how to use the fact that $\log x$ is monotonically non-decreasing (which surely we must have to use). – user22805 Oct 4 '12 at 18:57
Thanks for your insight. BTW, Chebychev's inequality hasn't been introduced in the text so far. – PeterM Oct 6 '12 at 1:38

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