For what nonnegative measures $\mu$ does $\mu*e^{-|\cdot|}\in L^{\infty}$?

Question

I am trying to characterize all measures on $\mathbb{R}$ such that

$$ \sup_{x\in\mathbb{R}} \: (\mu*f)(x)<+\infty, $$

where $f(x)$ is some specific integrable functions, such as $f(x)=e^{-|x|}$, and "$*$" denotes the convolution. This suggests that find whether the measures do not have any growing tails (hence the title of the post).

The space of Radon measures is the dual of continuous functions with compact support. It may have arbitrary large growing tails. For example, $\mu(d x)=e^{|x|}d x$ and $\mu=\sum_{n\in\mathbb{Z}}|n|\: \delta_n$.

Measures which fall in our class include the following examples:

1. Absolutely continuous measures with bounded density function, such as the Lebesgue measure.
2. $\mu=\sum_{n\in\mathbb{Z}}\delta_n(x).$

Does any one come across this kind of measures? Is this set of measures studied somewhere?

Thanks for any references and remarks!

Answer 1

Let $(X,\mathcal M)$ be a measurable space and let $\lambda$ be a complex measure on $(X,\mathcal M)$. The total variation $\vert \lambda\vert$ is a positive measure on $(X,\mathcal M)$ with a finite total variation, i.e. such that $\vert \lambda\vert(X)<+\infty$. We define $$ \vert \lambda\vert(X)=\sup_{\substack{{\text{$E_k$ pairwise disjoint}}\\{\text{with union $E$, $E_k\in \mathcal M$}}} }\sum_{k\in \mathbb N}\vert\lambda(E_k)\vert. $$ The mapping $\lambda\mapsto\vert \lambda\vert(X)$ is a norm on the vector space $\mathscr M(X,\mathcal M)$ of complex measures on $(X,\mathcal M)$ and make it a Banach space. So what you call the space of bounded measures is $\mathscr M(X,\mathcal M)$. A fact should be noted: a positive measure $\mu$ on $(X,\mathcal M)$ is not always a complex measure in particular since $\mu(X)$ may be $+\infty$.

Regarding your question, you may require a somewhat stronger property which would be $$ \hat \mu\hat f\in L^1(\mathbb R).\tag{$\sharp$} $$ This implies $\mu\ast f\in L^\infty$ with a norm smaller than the $L^1$ norm of $\hat \mu\hat f$. However, you have to make sure that the Fourier transform of $f$ makes sense as well as the product. For instance when $f(x)=e^{-\vert x\vert}$, this would require $$ \hat\mu(\xi)=(1+\xi^2) g(\xi),\quad\text{with $g\in L^1(\mathbb R)$,} $$ providing examples. A weaker requirement than $(\sharp)$ would be $$ \hat \mu\hat f\text{ is a bounded measure}.\tag{$\flat$} $$ As noted in my comment below that would contain the case $f=e^{-\vert x\vert}$ and $\mu$ the Dirac comb.