# Question from Rudin: Jensen?

I came across this question from Rudin's Real and Complex Analysis, 3rd Edition (p.75 # 25)

"Suppose that $\mu$ is a positive measure on $X$ and $f:X\rightarrow (0,\infty)$ satisfies $\int_X f d\mu = 1$. Prove for every $E \subset X$ with $0 < \mu(E) < \infty$ that $\int_E(\log{f})d\mu \leq \mu(E)\log{\frac{1}{\mu(E)}}$

Also, when $0<p<1$, we have $\int_E{f^p}d\mu \leq \mu(E)^{1-p}$."

My first thought was that since log is a concave function, we can use Jensen's inequality (in the opposite direction), but that is not giving me what I want. Any suggestions?

Further Addendum : Jensen's inequality only works on a set of measure 1 (or by redefining an interval to get a measure 1 set) so this is clearly not the correct approach.

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Rudin has several analysis texts. Is this from Real and Complex Analysis? If so, what edition? My edition gets up to a problem 21 on page 74 and starts a new chapter on page 75. – Henry T. Horton Aug 19 '12 at 4:00
This is from Rudin Real and Complex Analysis 3rd edition. This is from the exercises on the Lp space chapter. – James Aug 19 '12 at 4:07

## 2 Answers

First note that since we are only interested in $f$ on $E$, the condition $\int_X f\,d\mu=1$ translates to $\int_E f\,d\mu\le1$.

Let $\bar{f_E}=\int_E f\,d\mu/\mu(E)\le 1/\mu(E)$ denote the average value of $f$ on $E$. We then get $$\int_E \ln f\,d\mu \le\int_E \bar{f_E}\,d\mu =\mu(E)\ln\bar{f_E} \le\mu(E)\ln\frac{1}{\mu(E)}.$$

Replace the logarithm with power $p$, and the second result follows in exactly the same manner.

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If you want to use Jensen's Inequality, then you must check the measure 1 condition first, as you say.

Here,since $\mu(E)$ may not be 1,you can change the measure $\mu$ to $\frac{1}{\mu(E)}\mu$. That is, you can apply the Jensen's inequality(for concave function) as follows:

$$\frac{1}{\mu(E)}\int_E\rm{log} \,f\,\rm{d}\mu\le \rm{log}\left(\frac{1}{\mu(E)}\int_E \,f\,\rm{d}\mu\right)\le\rm{log}\frac{1}{\mu(E)}.$$ You can do this similarly by change $\rm{log} f$ to $f^p$.

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