Relative Entropy and p-norm

Question

I asked this question on StackExchange but could not get any answer, therefore, I am posting it here.

I am currently reading the book "A Dynamical Approach to Random Matrix Theory". The authors introduce the notion of relative entropy and remark that relative entropy is a weaker measure of the distance between probability measures than the $L^p$ distance for any $p>1.$ The following inequality (authors call it 'elementary' and therefore leave the proof as an exercise) is mentioned as a justification for this claim: $$\int f\log f\;d\mu\le 2\left[\int |f-1|^p\; d\mu\right]^{1/p}+\frac{2}{p-1}\int |f-1|^p\; d\mu,$$ where $f\;d\mu$ and $\mu$ are a probability measure.

I have no idea how to get this inequality. I thought of using $$\int f\log f\;d\mu= \int (f-1)\log f\;d\mu+ \int \log f\;d\mu,$$ and then apply H"older's inequality to the first part but it doesn't seem to give me what I want. Any hint/help is appreciated.

Answer 1

The argument below is not very elegant,but it is, indeed, a standard exercise. Let $g=\max(f-1,0)$. We shall prove that $$ f\log f\le 2g+\frac 2{p-1}g^p\,. $$ The integration and Holder then give the result immediately

If $f<1$, there is nothing to prove ($LHS<0=RHS$). If $0\le g\le 1$, then $$ f\log f=(1+g)\log(1+g)\le (1+g)g\le 2g. $$ Now assume that $g>1$. Let $g=e^t$, $t>0$. Then $$ f\log f=(1+g)\log(1+g)\le 2g\log(2g)=2g(\log 2+\log g) \\ \le 2e^t(1+t)\le 2e^t+2te^t=2g+2\frac{(p-1)t}{p-1}e^t \\ \le 2g+2\frac{e^{(p-1)t}}{p-1}e^t=2g+2\frac{e^{pt}}{p-1}=2g+2\frac{g^p}{p-1}\,. $$ (we used that $e^u\ge u$ for all $u$ and, in particular, for $u=(p-1)t$).