I want to prove the following inequality: $$\int_{-\infty}^{\infty} f^2\log \left(\frac{|f|}{\|f\|_2}\right) dx \leq \int_{-\infty}^{\infty} f^2\left(\log \frac{|f|}{\|f\|_2}\right)^2 dx$$
Note that this inequality remains the same if we multiply $f$ by a scalar $c$. Hence, without loss of generality, we may assume that $\|f\|_2=1$.
Partial solution: By Holder's inequality, and by recalling that $\|f\|_2=1$, we have that $$\int_{-\infty}^{\infty} f^2\log \left(\frac{|f|}{\|f\|_2}\right) dx \leq \left[\int_{-\infty}^{\infty} f^2\left(\log \frac{|f|}{\|f\|_2}\right)^2 dx\right]^{1/2}$$
Hence, if $\int_{-\infty}^{\infty} f^2\left(\log \frac{|f|}{\|f\|_2}\right)^2 dx\geq 1$, then this implies the inequality that we want. However, I'm unable to prove this inequality if $0\leq \int_{-\infty}^{\infty} f^2\left(\log \frac{|f|}{\|f\|_2}\right)^2 dx<1$.
I am probably missing something quite elementary.
Best Answer
You are unable to prove the inequality because it is not true in general. Indeed, let's consider the family of normalized functions $$ f_a(x):=\left(\frac{a}{\pi}\right)^{1/4}e^{-\frac{1}{2}ax^2}\quad (a>0). \tag{1} $$ Since $\|f_a\|_2=1$, the LHS of the inequality is equal to \begin{align} F_L(a):&=\int_{-\infty}^{\infty}\left(\frac{a}{\pi}\right)^{1/2}e^{-ax^2} \log\left(\left(\frac{a}{\pi}\right)^{1/4}e^{-\frac{1}{2}ax^2}\right)dx \\ &=\int_{-\infty}^{\infty}\left(\frac{a}{\pi}\right)^{1/2}e^{-ax^2} \left(\frac{1}{4}\log\left(\frac{a}{\pi}\right)-\frac{1}{2}ax^2\right)dx \\ &=\frac{1}{4}\log\left(\frac{a}{\pi}\right)-\frac{1}{4}, \tag{2} \end{align} whereas its RHS is equal to \begin{align} F_R(a):&=\int_{-\infty}^{\infty}\left(\frac{a}{\pi}\right)^{1/2}e^{-ax^2} \left(\log\left(\left(\frac{a}{\pi}\right)^{1/4}e^{-\frac{1}{2}ax^2}\right)\right)^2dx \\ &=\int_{-\infty}^{\infty}\left(\frac{a}{\pi}\right)^{1/2}e^{-ax^2} \left(\frac{1}{16}\left(\log\left(\frac{a}{\pi}\right)\right)^2 -\frac{1}{4}\log\left(\frac{a}{\pi}\right)ax^2 +\frac{1}{4}a^2x^4\right)dx \\ &=\frac{1}{16}\left(\log\left(\frac{a}{\pi}\right)\right)^2 -\frac{1}{8}\log\left(\frac{a}{\pi}\right)+\frac{3}{16}. \tag{3} \end{align} Now I'll show that there is a range of values of $a$ such that $F_L(a)>F_R(a)$. Indeed, let $\lambda:=\frac{1}{4}\log\left(\frac{a}{\pi}\right)$; then \begin{align} F_L(a)>F_R(a) &\iff \lambda-\frac{1}{4} > \lambda^2-\frac{1}{2}\lambda+\frac{3}{16} \\ &\iff \lambda^2 -\frac{3}{2}\lambda +\frac{7}{16} < 0 \\ &\iff \frac{3-\sqrt{2}}{4}<\lambda<\frac{3+\sqrt{2}}{4} \\ &\iff \pi e^{3-\sqrt{2}}<a< \pi e^{3+\sqrt{2}}. \phantom{\frac{7}{16}} \tag{4} \end{align} In particular, $(2)$ and $(3)$ yield $$ F_L(\pi e^3)-F_R(\pi e^3)=\left(\frac{3}{4}-\frac{1}{4}\right)-\left(\frac{9}{16} -\frac{3}{8}+\frac{3}{16}\right)=\frac{1}{8}>0. \tag{5} $$