Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f:[0,1] \to [1,\infty)$ be such that $$ \int_0^1f(x)\ln f(x) dx < \infty. \tag{1} $$ Show $$ \int_0^1 f(x) dx \int_0^1 \ln f(x) dx \le \int_0^1 f(x) \ln f(x) dx. \tag{2} $$

From (1), $f$ and $\ln f$ are integrable on $[0,1]$.
Using this post, $\ln x \le x \ln x$, when $x >0$.
I can't go further.

share|cite|improve this question
up vote 4 down vote accepted

Here's a cute trick I learned in a probability course. You can apply Fubini's theorem to show that the inequality stated is equivalent to the positivity of the following double integral:

$ \int_{[0,1]^2} \left( f(x) - f(y) \right) \left(\ln f(x) - \ln f(y) \right) dx dy$

That's the cute part. Notice that there's a kind of 'evenness' here for us to exploit: decompose [0,1]^2 as $\{(x,y) : f(x) \geq f(y)\} =: A$ and its complement. Now on $A$ the integrand is positive by the monotonicity of $\ln$ and the assumption $f \geq 1$. On $A^c$ the integrand is the product of two negatives, hence also positive. Therefore the integral is positive and the inequality holds.

I'll also mention that the above procedure can be applied to a very wide variety of situations (but in order to apply it you must have a condition analogous to (1) to use Fubini!).

share|cite|improve this answer
(+1) That was cute. (Welcome to the stackexchange, btw.) – Cameron Buie Mar 7 '13 at 16:24

$g(x):=x\log(x)$ is convex: $g'(x)=\log(x)+1$, $g''(x)= \frac{1}{x}\geq 0$, the second derivative implying convexity. Then, by Jenson's inequality:

$\int f(x)\ln f(x)dx \geq g(\int_0^1f(x)dx)=\left(\int_0^1f(x)dx\right) \log(\int_0^1f(x)dx)\geq \int_0^1f(x)dx\int_0^1\log f(x)dx$

where in the last step we use concavity of $\log(x)$ and Jenson's inequality again for concave functions this time.

share|cite|improve this answer

Your Answer


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.