The following is left unproven in a monograph. The author refers to it as "elementary exercise" but I am unable to prove it. Any insight is appreciated. $$ \int f \log f d\mu \le 2 \left[\int|f-1|^p d\mu\right]^{1/p}+\frac{2}{p-1}\int |f-1|^p d\mu,\quad p>1 $$ where $f$ is a probability density.
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4$\begingroup$ You really mean $f$ a probablity density? not $\mu$? Also please quote the monograph. $\endgroup$– YCorCommented Sep 5, 2020 at 7:03
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$\begingroup$ @YCor I rather expect that $\mu$ is probability measure. Because the inequality is not homogeneous with respect to $\mu$. $\endgroup$– Fedor PetrovCommented Sep 5, 2020 at 9:04
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8$\begingroup$ This follows from $(1+x) \log (1+x)\le 2x +2/(p-1)x^p$ for $x \ge 0$ writing $0 \le f=1+g$ and integrating only where $g \ge 0$. $\endgroup$– Giorgio MetafuneCommented Sep 5, 2020 at 10:38
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$\begingroup$ $\mu$ is indeed a probability measure in the context of the monograph. f is probability density with respect to $\mu$. But seems that it is true in general by Metafune's proof? The inequality is from (13.6) in "Dynamic to random matrix" by Erdos and Yau. $\endgroup$– Daniel LiCommented Sep 5, 2020 at 17:20
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1$\begingroup$ I used that $\mu$ is a probability measure to have $\|g\|_1 \le \|g\|_p$ and $f \ge 0$. $\endgroup$– Giorgio MetafuneCommented Sep 5, 2020 at 17:37
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