 Celia Lucas

2022-06-19

Showing that ${\int }_{X}\mathrm{log}\left(f\right)d\mu \le \mu \left(X\right)\mathrm{log}\frac{1}{\mu \left(X\right)}$ 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 $f\in {L}_{0}^{+}\left(\mathrm{\Sigma }\right)$ satisfies ${\int }_{X}fd\mu ={\int }_{\left(0,\mathrm{\infty }\right)}\mu \left(\left\{f>t\right\}dm\left(t\right)\right)$
At first I thought I should use $\left\{|f|<\frac{1}{\mu \left(X\right)}\right\}$,$\left\{|f|\ge \frac{1}{\mu \left(X\right)}\right\}$ and show that if $\int \mathrm{log}fd\mu >\mu \left(X\right)\mathrm{log}\left(\frac{1}{\mu \left(X\right)}\right)$, then
$\int \mathrm{log}fd\mu \le \mu \left(\left\{|f|<\frac{1}{\mu \left(X\right)}\right\}\right)\mathrm{log}\frac{1}{\mu \left(X\right)}+{\int }_{\left\{|f|\ge \frac{1}{\mu \left(X\right)}\right\}}\mathrm{log}fd\mu$ and therefore
$\mathrm{log}\frac{1}{\mu \left(X\right)}\left(\mu \left(X\right)-\mu \left(\left\{|f|<\frac{1}{\mu \left(X\right)}\right\}\right)\right)=\mathrm{log}\frac{1}{\mu \left(X\right)}\left(\mu \left(\left\{|f|\ge \frac{1}{\mu \left(X\right)}\right\}\right)\right)<{\int }_{\mu \left(\left\{|f|\ge \right\}\right)}\mathrm{log}fd\mu$
which I can't seem to contradict. Any ideas how this can be shown, without using Jensen's Inequality? hopeloothab9m

Use that
$\mathrm{log}\left(x\right)\le \mathrm{log}\left(s\right)-1+\frac{x}{s}$
for any constant $s>0$. Then take $s=\mu \left(X{\right)}^{-1}$ 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.)

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