Celia Lucas

2022-06-19

Showing that ${\int}_{X}\mathrm{log}(f)d\mu \le \mu (X)\mathrm{log}\frac{1}{\mu (X)}$ given that ${\int}_{X}fd\mu =1$ for positive $f$ without using Jensen's inequality.

We weren't presented with Jensen's inequality, but with Fubini's Theorem and another theorem which is:

in a finite measure space $f\in {L}_{0}^{+}(\mathrm{\Sigma})$ satisfies ${\int}_{X}fd\mu ={\int}_{(0,\mathrm{\infty})}\mu (\{f>t\}dm(t))$

At first I thought I should use $\{|f|<\frac{1}{\mu (X)}\}$,$\{|f|\ge \frac{1}{\mu (X)}\}$ and show that if $\int \mathrm{log}fd\mu >\mu (X)\mathrm{log}(\frac{1}{\mu (X)})$, then

$\int \mathrm{log}fd\mu \le \mu (\{|f|<\frac{1}{\mu (X)}\})\mathrm{log}\frac{1}{\mu (X)}+{\int}_{\{|f|\ge \frac{1}{\mu (X)}\}}\mathrm{log}fd\mu $ and therefore

$\mathrm{log}\frac{1}{\mu (X)}(\mu (X)-\mu (\{|f|<\frac{1}{\mu (X)}\}))=\mathrm{log}\frac{1}{\mu (X)}(\mu (\{|f|\ge \frac{1}{\mu (X)}\}))<{\int}_{\mu (\{|f|\ge \})}\mathrm{log}fd\mu $

which I can't seem to contradict. Any ideas how this can be shown, without using Jensen's Inequality?

We weren't presented with Jensen's inequality, but with Fubini's Theorem and another theorem which is:

in a finite measure space $f\in {L}_{0}^{+}(\mathrm{\Sigma})$ satisfies ${\int}_{X}fd\mu ={\int}_{(0,\mathrm{\infty})}\mu (\{f>t\}dm(t))$

At first I thought I should use $\{|f|<\frac{1}{\mu (X)}\}$,$\{|f|\ge \frac{1}{\mu (X)}\}$ and show that if $\int \mathrm{log}fd\mu >\mu (X)\mathrm{log}(\frac{1}{\mu (X)})$, then

$\int \mathrm{log}fd\mu \le \mu (\{|f|<\frac{1}{\mu (X)}\})\mathrm{log}\frac{1}{\mu (X)}+{\int}_{\{|f|\ge \frac{1}{\mu (X)}\}}\mathrm{log}fd\mu $ and therefore

$\mathrm{log}\frac{1}{\mu (X)}(\mu (X)-\mu (\{|f|<\frac{1}{\mu (X)}\}))=\mathrm{log}\frac{1}{\mu (X)}(\mu (\{|f|\ge \frac{1}{\mu (X)}\}))<{\int}_{\mu (\{|f|\ge \})}\mathrm{log}fd\mu $

which I can't seem to contradict. Any ideas how this can be shown, without using Jensen's Inequality?

hopeloothab9m

Beginner2022-06-20Added 25 answers

Use that

$\mathrm{log}(x)\le \mathrm{log}(s)-1+\frac{x}{s}$

for any constant $s>0$. Then take $s=\mu (X{)}^{-1}$ and use this bound to get an upperbound for the integrand and hence also for the integral. (This is, all said and done, just Jensen’s inequality.)

$\mathrm{log}(x)\le \mathrm{log}(s)-1+\frac{x}{s}$

for any constant $s>0$. Then take $s=\mu (X{)}^{-1}$ and use this bound to get an upperbound for the integrand and hence also for the integral. (This is, all said and done, just Jensen’s inequality.)