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Problem: The following is Exercise $4.29$(d) in Hugo Junghenn's Principles of Analysis:
Use Jensen's inequality to verify the following for a probability measure $\mu:$
$$\|f\|_1\log(\|f\|_1)\leq\log(\|f\log(f)\|_1\quad\text{where }f>0.$$ I tried the problem for a while and now I think that the claim is incorrect. I have cooked up the following counterexample.

Consider the probability space $((0,1),\mathcal B,\mu)$, where $\mathcal B$ is the Borel $\sigma$-field of $(0,1)$ and $\mu$ is the Lebesgue measure. Next, let $f(x)=x^2$ for $x\in(0,1)$. Then $f>0$ everywhere and we have $$\|f\|_1\log(\|f\|_1)=\frac{1}{3}\log\left(\frac{1}{3}\right).$$ On the other hand, using integration by parts and the fact that $x^3\log(x)\to0$ as $x\searrow0$, we see that $$\|f\log(f)\|_1=\int_0^1 x^2|\log(x^2)|\,dx=2\int_0^1x^2|\log(x)|\,dx=\frac{2}{9}.$$ But then $\log(2/9)<3^{-1}\log(3^{-1})$, hence the inequality does not hold.

However, I do believe that the author meant to ask to prove $$\|f\|_1\log(\|f\|_1)\leq\|f\log(f)\|_1,$$ which is an easy consequence of Jensen's inequality taking the convex function to be $x\log(x).$


My Question: Do you agree with my counterexample above? If not, I would like to ask if I am wrong and the inequality does hold, or if there is another, this time correct, counterexample.

Thank you very much for your time and appreciate all the feedback and help.

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    $\begingroup$ You are right. The first log should not have been there on the right side. $\endgroup$ – Kavi Rama Murthy Sep 16 at 5:22

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