Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
Can you prove the case of simple function first? –  ziyuang Oct 6 '12 at 0:26

2 Answers 2

up vote 21 down vote accepted
+50

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.

share|improve this answer
    
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
1  
@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.

share|improve this answer
    
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
1  
Thanks for your insight. BTW, Chebychev's inequality hasn't been introduced in the text so far. –  PeterM Oct 6 '12 at 1:38

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.