Given positive $\epsilon$ and $c$, find a density $\phi$ such that $t\phi(\epsilon/t) \ge c \|\phi'\|_\infty$ for all positive $t$

Question

A nice density (on $\mathbb R$) is function $\phi:\mathbb R \to \mathbb R$ such that

• (1) $\phi(x) \ge 0$ for all $x \in \mathbb R$,
• (2) $\int_{-\infty}^\infty \phi(x) \mathrm{d}x = 1$,
• (3) $\phi$ is even, i.e $\phi(-x) = \phi(x)$ for all $x \in \mathbb R$, and
• (4) $\phi$ is differentiable and $\|\phi'\|_\infty := \sup_{x \in \mathbb R}|\phi'(x)|$ is finite.

The 3rd requirement may be dropped in case it courses trouble.

Question 1. Given positive $\epsilon$ and $c$, is it possible to find a nice density $\phi$ (which may depend on $\epsilon$ and $c$) such that

• $t\phi(\epsilon/t) \ge c \|\phi'\|_\infty$ for all positive $t$ ?

Question 2. In case Question 1 has an affirmative answer, can $\phi$ be chosen in the family of (centered) Gaussian densities given by $\phi_\sigma(x) \propto e^{-x^2/(2\sigma^2)}$, for some positive $\sigma$ ?

Answer 1

Of course, not.

Indeed, suppose that $t\phi(\epsilon/t) \ge c \|\phi'\|_\infty$ for all positive $t$. Then, letting $s:=\epsilon/t$, we get $\phi(s)\ge Cs$ for $C:=c \|\phi'\|_\infty/\epsilon>0$ and all real $s>0$. So, $\int_{\mathbb R}\phi=\infty$ and hence $\phi$ cannot be a pdf.