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!

  • 3
    $\begingroup$ What is the norm (topology) on this space (of continuous integrable functions)? There doesn't seem to be an obvious candidate. $\endgroup$ – Christian Remling Sep 16 '14 at 17:06
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    $\begingroup$ Then your space is a dense subspace of $L^1$, so continuous functionals have unique extensions to elements of $(L^1)^*=L^{\infty}$. $\endgroup$ – Christian Remling Sep 16 '14 at 17:16
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    $\begingroup$ What is "boundedness of a measure"? Do you want to know when a measure is Radon? $\endgroup$ – Asaf Sep 16 '14 at 17:39
  • 5
    $\begingroup$ Can you formulate your question into a well posed question? I'm confused. Do you want a function space whose dual is your space of measures? Do you want to know the dual of the space of continuous integrable functions? (With a given norm, maybe the sum of the $L^1$ and $L^\infty$ norms.) If it comes naturally in some research context (as you mention in a comment), can you elaborate? $\endgroup$ – Joonas Ilmavirta Sep 16 '14 at 17:46
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    $\begingroup$ Voting to close until we get some idea of what the question means by "boundedness". $\endgroup$ – Nik Weaver Sep 16 '14 at 18:18

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.

  • $\begingroup$ Thanks Bazin for your answer. I am working on measures on $\mathbb{R}$. The examples that I give above all have infinite total mass. $\endgroup$ – Anand Sep 16 '14 at 19:31
  • $\begingroup$ I have added a complement to my answer above. $\endgroup$ – Bazin Sep 17 '14 at 15:45
  • $\begingroup$ Thanks Bazin for your help! It is a good idea to use the Fourier transform. Your condition is a bit too strong because the example $\mu=\sum_{m\in\mathbb{Z}}\delta_n$ fails. Am I right? Thanks! $\endgroup$ – Anand Sep 17 '14 at 16:01
  • $\begingroup$ In that case, $\hat \mu=\mu$ by Poisson summation formula. It is probably possible to weaken my requirement to $$\hat \mu\hat f\text{ is a measure with a finite total mass}.$$ If $f=e^{-\vert x\vert}$, you find $\hat f\hat \mu$ with finite mass since $\sum_{n\ge 1}\frac{1}{n^2}<+\infty$. $\endgroup$ – Bazin Sep 17 '14 at 19:31
  • $\begingroup$ Thanks Bazin. Do you mean $\hat{\mu}\hat{f}$ is a measure on complex plain with finite total mass? $\endgroup$ – Anand Sep 17 '14 at 22:00

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