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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?2012-10-06

2 Answers 2

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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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    @ChristianBlatter - you have not caused any inconvenience. Requesting details of an argument improves the site.2013-11-09
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$\int_0^1f(t)d(t)=\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n f\left(\frac{i}{n}\right)$ 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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    Thanks for your insight. BTW, Chebychev's inequality hasn't been introduced in the text so far.2012-10-06