# Convolution of tight measures is tight

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 $$$$\begin{split}(\mu_n\ast\nu_n)(K+K)&=(\mu_n\otimes\nu_n)(\{(x,y)\in E^2:x+y\in K+K\})\\&\ge(\mu_n\otimes\nu_n)(K\times K)=\mu_n(K)\nu_n(K)\end{split}\tag2$$$$ for all $$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?

• 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 $x$. Commented Dec 25, 2020 at 14:36
• @PhoemueX A family $\mathcal F$ is tight if for all $\varepsilon>0$, there is a compact $K$ with $\sup_{\mu\in\mathcal F}|\mu|(K^c)<\varepsilon$. Commented Dec 25, 2020 at 14:43

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*}

• 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 ;) Commented Dec 25, 2020 at 16:53
• @0xbadf00d: I updated my answer to include the uniformly bounded case. Commented Dec 26, 2020 at 9:52
• 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 $(\mu_n\otimes\nu_n)(E\times E)$. Commented Dec 26, 2020 at 11:19
• 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 $\mu\ast\nu$? Commented Dec 26, 2020 at 11:20
• @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. Commented Dec 26, 2020 at 14:35