Convergence of mollified functions in weighted $L^p$ norm

Question

$ \newcommand{\bR}{\mathbb{R}} \newcommand{\bE}{\mathbb{E}} \newcommand{\supp}{\operatorname{supp}} $ Let $(\rho_n)_{n \geq 1}$ be a sequence of mollifiers on $\bR^d$, i.e., each $\rho_n$ is a probability density function such that $\rho_n \in C_c^{\infty} (\bR^d)$ and $\supp \rho_n \subset \overline{B(0,1 / n)}$. We have from Brezis' Functional Anlysis that

Theorem 4.22. Assume $f \in L^p (\bR^d)$ with $1 \le p<\infty$. Then $\| \rho_n * f -f \|_{L^p} \to 0$ as $n\to \infty$.

Above, $*$ denotes the mollification operation. Now we fix a Borel probability measure $\mu$ on $\bR^d$. Is the following statement true?

Assume $f \in L^p (\bR^d, \mu)$ with $1 \le p<\infty$. Then $\| \rho_n * f -f \|_{L^p (\mu)} \to 0$ as $n\to \infty$.

References are appreciated. Thank you for your elaboration.

Answer 1

It doesn't hold in general. Indeed, let $\mu = \delta_{0}$ and $p=1$. If it worked, we would have $\sup_{n \in \mathbb{N}} \|\rho_{n} \ast f\|_{L^{1}(\mu)} \leqslant C \|f\|_{L^{1}(\mu)} = C |f(0)|$ for some constant $C>0$, using the Uniform Boundedness Principle. We have (for nonnegative $f$ and assuming for simplicity that $\rho_n(-x) = \rho_{n}(x)$)

$ \|\rho_{n}\ast f\|_{L^{1}(\mu)} = \int_{\mathbb{R}} \rho_{n}(y) f(y) dy $.

This cannot be bounded by $C f(0)$, because we can assume that $f$ grows very fast around $0$, say, bigger than an arbitrary $M$ on a subset on which $\int \rho_{n}(y) dy \geqslant \frac{3}{4}$, while $f(0)\leqslant 1$, which would show that such a constant $C$ does not exist.

On the other hand, if $\mu$ has a density satisfying an extra condition, the convergence holds. Indeed, we want to bound

$ \int \left| \int \rho_n(x-y) f(y) dy \right|^p d\mu(x)\leqslant \int \int |f(y)|^p \rho_{n}(x-y) dy d\mu(x) $

and then it suffices to show the convergence on the dense set of smooth, compactly supported functions. We will first perform the integration over $x$, i.e.

$ \int \rho_{n}(x-y)d\mu(x) = \int \rho_{n}(x-y) g(x) dx$,

where $g$ is the density of $\mu$. By a change of variables, we obtain

$\int \rho_n(t) g(t+y) dt$

and if $g$ has the property that $g(t+y) \leqslant C g(y)$ for some $C>0$ and sufficiently small $t$ (uniformly over $y$), than we can estimate this by $Cg(y) \int \rho_n(t) dt = g(y)$; recall that $t \in \operatorname{supp}(\rho_n)$, so $|t| \leqslant \frac{1}{n}$. Coming back to our integral, we get

$ C\int |f(y)|^p(y) g(y) dy = C \int |f|^p d\mu(y) = C \|f\|^{p}_{L^{p}(\mu)}$.

This condition seems to be pretty restrictive, for example the Gaussian measure does not satisfy it, but the symmetric exponential distribution does.