Let $\ f:[0,1]\rightarrow\ [1,\infty$) be Lebesgue measurable. Which quantity is greater:

$\int_{0}^{1} f(x)\log f(x)dx$


$\int_{0}^{1} f(y)dy\int_{0}^{1}\log f(w)dw$ ?

Prove your answer, for all such $f.$

Is the Hölder inequality sufficient to prove this?

  • 1
    $\begingroup$ You should show how you tried to apply the Holder inequality. This questions smells like homework and homework tends to only get help when the student shows what they have tried already. Why do you think it's not sufficient, so forth. $\endgroup$ – Alec Teal Nov 27 '13 at 16:22
  • $\begingroup$ Is $f^2$ integrable on $[0,1]$ ? $\endgroup$ – derivative Nov 27 '13 at 16:44
  • $\begingroup$ @derivative: Does your problem state explicitly that $f^2$ is integrable? It needs not be, of course. Take for example the function $f(x)=\begin{cases}1/\sqrt{x} & x\in (0, 1]\\ 0 & x=0\end{cases}$. But maybe you omitted something... (By all means, I don't think that this problem can be solved by a straightforward application of Hölder's inequality. You need to think more) $\endgroup$ – Giuseppe Negro Nov 27 '13 at 17:41

$$ \int_0^1 f(x) \log f(x) dx = \int_0^1 \psi (f(x))dx $$ where $\psi (t) =t \log(t)$ is a convex function. So, by Jensen's inequality $$ \int_0^1 f(x) \log f(x) dx \ge \psi\left(\int_0 ^1 f(x)dx\right)=\left(\int_0 ^1 f(x)dx\right)\; \log\left(\int_0 ^1 f(x)dx\right) $$ If we can show that $$ \log\left(\int_0 ^1 f(x)dx\right) \ge \int_0 ^1 \log f(x)dx $$ we are done. Call $g=\log f(x)$. The previous inequality is equivantly to $$ \int_0^1\exp{g(x)}dx \ge \exp\left( \int_0^1 g(x) dx\right) $$ which is true, again, thanks to Jensen's inequality.

  • $\begingroup$ Can we not directly reverse the Inequality, without the step with exp-function, if we have a concave function ? $\endgroup$ – derivative Dec 1 '13 at 17:47

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