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Let $\mu$ be a positive measure on a measure space $\Omega$. How can I show that for all non-negative measurable $f$ $$\exp\left(\int_\Omega f \; d\mu \right) \leq \int_\Omega \exp(f) \; d\mu$$ the above inequality is valid if $\mu(\Omega) = 1$ and not valid if $\mu(\Omega) \neq 1.$

This is what I've been able to come up with.

Since $\exp$ is convex, if $\mu(\Omega) =1$, then the inequality is just Jensen's inequality. Do I have to say more? How do I show that the inequality is not valid if $\mu(\Omega) \neq 1$.

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What we have to show is that the following assertions are equivalent:

(i) For all $f$ measurable and non-negative, we have $\exp\left(\int_{\Omega} fd\mu\right)\leq \int_{\Omega}\exp(f)d\mu$;

(ii) $\mu(\Omega)=1$.

The direction $(ii)\Rightarrow (i)$ is as noticed a consequence of Jensen inequality. To see the converse, take $f:= 1$, which is non-negative and mesurable. Let $m:=\mu(\Omega)$. We have $\exp(m)\leq m\cdot e$. Let $h(t)=e^t-te$. Then $h'(t)=e^t-e$, so $h$ is non-negative and $h(t)=0$ if and only if $t=1$, so $m=1$.

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  • $\begingroup$ why do you show that $h(t) = 0$? $\endgroup$ – Joe Mar 15 '12 at 15:07
  • $\begingroup$ I show that $h(t)=0$ if and only if $t=0$. $\endgroup$ – Davide Giraudo Mar 15 '12 at 15:50
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Let $\Omega=[0,a]$ with standard Lebesgue measure and $f(x)=1$, then $$ \exp\left(\int_{\Omega}fd\mu\right)=\exp(a) $$ $$ \int_{\Omega}\exp(f)d\mu=a\exp(1) $$ You can easly check that $\exp(a)\geq a \exp(1)$ and equality holds only if $a=1$. So for this particular case inequality $$ \exp\left(\int_\Omega f d\mu \right) \leq \int_\Omega \exp(f) d\mu $$ holds iff $\mu(\Omega)=1$

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Since the inequality can be true for certain $f$, $\Omega$ and $\mu$ which are not covered by your assumptions you may provide a simple counterexample:

Take $\Omega=[0,2]$, $f(x) = 1$, $\mu = dx$. Then $$\exp\left(\int_0^2 f(x) \; dx\right) = \exp(2)$$ but $$\int_0^2\exp(f(x))dx = 2\exp(1)<\exp(2).$$

(Note that $f=0$ fulfills the inequality, also for $\Omega = [0,2]$.)

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