Consider $\mathbb{R}^n$ and the standard topology. Assume $A$ and $B$ are two subsets of $\mathbb{R}^n$ with $A$ closed such that $A\cap B=\emptyset$. Assume that the function $f:\mathbb{R}^n\to\mathbb{R}_{\geq 0}$ is positive semidefinite and positive definite in $B$ with respect to $A$, that's to say:
- $f(x)\geq 0$ for all $x\in\mathbb{R}^n$
- $f(x)=0$ for all $x\in A$
- $f(x)>0$ for all $x\in B$
My question is whether I can find a scalar function $\rho:\mathbb{R}_{\geq 0}\to\mathbb{R}_{\geq 0}$ which is positive definite on $\bar B=\{|x|_A: x\in B\}$ with respect to $\{0\}$ such that: $$ f(x)\geq\rho(|x|_A) $$ where $|x|_A:=\inf_{y\in A}|x-y|$ is the distance to the closed set $A$. I have the following counter example:
- $A=\{(x_1,x_2): x_2=0, x_1\geq 0\}$
- $B=\{(x_1,x_2): x_1x_2\geq 1, x_1\geq 0\}$
- $f(x)=\exp(-x_1)|x_2|^2$
Apparently, we can not lower bound $f(x)$ by any p.d. function of $|x|_A=|x_2|$, is that right? So in that case, can we impose further assumptions on $A$ and $B$ to obtain the lower bound needed?