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Let $E$ be a Banach space (think of $E=\mathbb R^d$, if it's easier to understand the following for you), $(\mu_n)_{n\in\mathbb N}$ and $(\nu_n)_{n\in\mathbb N}$ be tight sequences of finite nonnegative measures on $\mathcal B(E)$.

I would like to conclude that the sequence $(\mu_n\ast\nu_n)_{n\in\mathbb N}$ of convolutions is tight as well.

Let $\varepsilon>0$. By the tightness assumption, there is a compact $K\subseteq E$ with $$\max\left(\sup_{n\in\mathbb N}\mu_n(K^c),\sup_{n\in\mathbb N}\nu_n(K^c)\right)<\varepsilon\tag1.$$

Now, $K+K$ is clearly compact and \begin{equation}\begin{split}(\mu_n\ast\nu_n)(K+K)&=(\mu_n\otimes\nu_n)(\{(x,y)\in E^2:x+y\in K+K\})\\&\ge(\mu_n\otimes\nu_n)(K\times K)=\mu_n(K)\nu_n(K)\end{split}\tag2\end{equation} for all $n\in\mathbb N$.

If I assume that $\mu_n,\nu_n$ are probability measures for all $n\in\mathbb N$, then $(2)$ yields $$(\mu_n\ast\nu_n)(K+K)\ge(1-\varepsilon)^2\tag3$$ and hence we obtain the claim.

But how can we show the claim in general?

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  • $\begingroup$ What exactly is your definition of tight? Depending on that, for a counterexample, it might be good to consider something like $\mu_n = n^{-1} \delta_n$ and $\nu_n = n \delta_0$ with the usual Dirac measure $\delta_x$ at $x$. $\endgroup$
    – PhoemueX
    Commented Dec 25, 2020 at 14:36
  • $\begingroup$ @PhoemueX A family $\mathcal F$ is tight if for all $\varepsilon>0$, there is a compact $K$ with $\sup_{\mu\in\mathcal F}|\mu|(K^c)<\varepsilon$. $\endgroup$
    – 0xbadf00d
    Commented Dec 25, 2020 at 14:43

1 Answer 1

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As I wrote in the comments, the claim is not true in general. To see this, let $\delta_x$ be the Dirac measure at $x$, meaning that $\delta_x(A) = 1$ if $x \in A$ and $\delta_x(A) = 0$ otherwise. Then define $\mu_n = n^{-1} \cdot \delta_n$ and $\nu_n = n \cdot \delta_0$.

Then choosing $K = \{0\}$, we see that $(\nu_n)_n$ is a tight sequence. Likewise, for each $N$, we can choose $K = \{1,\dots,N\}$ to get $\sup_n \mu_n(K^c) \leq 1/N$ for all $n \in \Bbb{N}$, showing that also $(\mu_n)_n$ is a tight sequence.

But $\mu_n \ast \nu_n = \delta_n$, which easily implies that $(\mu_n \ast \nu_n)_n$ is not a tight sequence.

Probably the answer is positive if one in addition assumes that $(|\mu_n|)_n, (|\nu_n|)_n$ are both uniformly bounded.


If one assumes the measures to be uniformly bounded, meaning that there is $C > 0$ with $\mu_n(E) \leq C$ and $\nu_n (E) \leq C$ for all $n$, then the claim is true.

To see this, simply note that what you already shows implies \begin{align*} (\mu_n \ast \nu_n)((K + K)^c) & = (\mu_n \ast \nu_n)(E) - (\mu_n \ast \nu_n)(K + K) \\ & \leq \mu_n(E) \nu_n(E) - \mu_n(K) \nu_n(K) \\ & = \mu_n(E) \cdot (\nu_n(E) - \nu_n(K)) + \nu_n(K) (\mu_n(E) - \mu_n(K)) \\ & \leq C \epsilon + C \epsilon . \end{align*}

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  • $\begingroup$ Thank you for the counterexample. Actually, in the situation I've got in mind, I always assume that $(\mu_n)_{n\in\mathbb N}$ and $(\nu_n)_{n\in\mathbb N}$ are (uniformly) bounded in total variation. So, if you know how we can obtain the result under this assumption, that would be amazing ;) $\endgroup$
    – 0xbadf00d
    Commented Dec 25, 2020 at 16:53
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    $\begingroup$ @0xbadf00d: I updated my answer to include the uniformly bounded case. $\endgroup$
    – PhoemueX
    Commented Dec 26, 2020 at 9:52
  • $\begingroup$ Thank you for the update. Please note that you have a typo on the rhs of your first equality: It should be $(\mu_n\ast\nu_n)(E)$ or $(\mu_n\otimes\nu_n)(E\times E)$. $\endgroup$
    – 0xbadf00d
    Commented Dec 26, 2020 at 11:19
  • $\begingroup$ Do you think we can extend the result even further to finite signed measures? Maybe it's possible to reduce to the nonnegative case. Is there a useful formula for the Hahn-Jordan decomposition of a convolution $\mu\ast\nu$? $\endgroup$
    – 0xbadf00d
    Commented Dec 26, 2020 at 11:20
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    $\begingroup$ @0xbadf00d: Should be possible. Maybe it is even enough to decompose each of the "input" measures and work with that. Let me think for a bit. $\endgroup$
    – PhoemueX
    Commented Dec 26, 2020 at 14:35

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