Does there exist $f=f(x)$ satisfying $f(x)\ge0$ for $x\in\mathbb{R}$, $f(x)=f(-x)$ for $x\in\mathbb{R}$ (i.e. $f$ is even), $\int_{\mathbb{R}} f(x)\,dx<\infty$, and $\int_{\mathbb{R}} x^2\,f(x)\,dx<\infty$ such that for some $C>0$ (independent of $a$ and $b$), we have
\begin{equation} a\,f\Big(\frac{x}{a}\Big)-b\,f\Big(\frac{x}{b}\Big)\ge C\,(a-b)\,f\Big(\frac{x}{a-b}\Big) \end{equation} for all $a>b>0$ and $x\in\mathbb{R}$. My idea is to use the Jesen inequality to show that the above inequality holds for some class of $f$. However, I failed to do so. Is it possible to find an example of such $f$? Any comment, idea, or suggestion is welcome. Thanks!
Here is an example: $$f(x)=\begin{cases}1, |x|\le 1\\0, |x|>1\end{cases}$$ The integrals are finite. Since $a>b>0$, you get $a>a-b>0$. There are two cases, $b\ge a-b$ and $b<a-b$.
If $|x|>a$, $f(x/a)=f(x/b)=f(x/(a-b))=0$, so your condition is verified for any $C$. If $|x|<b$ and $|x|<a-b$, then $f(x/a)=f(x/b)=f(x/(a-b))=1$, and your condition is verified for any $C<1$. What you are left with is $|x|\le a$ and
(a)$|x|>b, |x|>a-b$,
(b)$|x|>b, |x|<a-b$,
(c)$|x|<b, |x|>a-b$.
For (a) you get $a-b\cdot 0\ge C(a-b)\cdot 0$, which is always true. For (b) you have $a-b\cdot 0\ge C(a-b)$ which once again is true if $C<1$. For (c) you get $a-b\ge C(a-b)\cdot 0$, true for any C