If $(\exp(\mu_n))_{n\in\mathbb N}$ is weakly convergent, is the normalized sequence convergent as well?

Question

Let $E$ be a metric space and $\mathcal M(E)$ denoote the space of finite signed measures on $\mathcal B(E)$ equipped with the total variation norm $\left\|\;\cdot\;\right\|$.

I would like to know whether we can show that if $(\mu_n)_{n\in\mathbb N}$ is convergent with respect to the topology of weak convergence of measures, then $\left(\frac{\mu_n}{\left\|\mu_n\right\|}\right)_{n\in\mathbb N}$ is convergent as well.

This is at least true when $\mu_n$ is nonnegative for all $n\in\mathbb N$, but since the proof of this result utilizes the fact that $\left\|\mu_n\right\|=\mu_n(E)$, this doesn't seem to extend to the general case.

However, I'm especially interested in the following special case: Assume $E$ is a normed vector space and let $$\exp(\mu):=\sum_{k\in\mathbb N_0}\frac{\mu^{\ast k}}{k!},$$ where $\mu^{\ast k}$ denotes the $k$-fold convolution power, for $\mu\in\mathcal M(E)$.

I would like to know whether we can show that if $(\exp(\mu_n))_{n\in\mathbb N}$ is convergent, then $\left(\frac{\exp(\mu_n)}{\left\|\exp(\mu_n)\right\|}\right)_{n\in\mathbb N}$ is convergent as well.

Note that $\left\|\exp(\mu)\right\|=e^{\left\|\mu\right\|}$.

Answer 1

$\newcommand\R{\mathbb R}$The answer to the second question (and hence to the first one) is no.

Indeed, let $E:=\R$.

For all odd natural $n$, let $\mu_n:=\mu$, where $\mu$ is the uniform distribution on the interval $[2,3]$, so that $\mu_n(dx)=\mu(dx)=1(2\le x\le3)\,dx$, $\exp^*(\mu_n)=\exp^*(\mu)$, $\|\exp^*(\mu_n)\|=\|\exp^*(\mu)\|=e$, whence $$\frac{\exp^*(\mu_{2k+1})}{\|\exp^*(\mu_{2k+1})\|}=\frac{\exp^*(\mu)}e\to\frac{\exp^*(\mu)}e;$$ everywhere here, the convergence is for (natural) $k\to\infty$. In particular, letting $$\nu_n:=\frac{\exp^*(\mu_n)}{\|\exp^*(\mu_n)\|}\quad\text{and}\quad \nu:=\frac{\exp^*(\mu)}{\|\exp^*(\mu)\|}=\frac{\exp^*(\mu)}e,\tag{0}$$ we have $$\int1\,d\nu_{2k+1}=1\to1.\tag{1}$$

For all even natural $n$, let $\mu_n(dx):=\mu(dx)(1+2\pi\cos nx)$. Then, by the Riemann–Lebesgue lemma, $\mu_{2k}\to\mu$. Hence, by dominated convergence or using characteristic functions, $$\exp^*(\mu_{2k})\to\exp^*(\mu);$$ in particular, $$\int1\,d\exp^*(\mu_{2k})\to\int1\,d\exp^*(\mu)=e.\tag{2}$$ On the other hand, $$\|\mu_{2k}\|=\int_2^3 dx\,|1+2\pi\cos 2kx|\ge\int_2^3 dx\,(2\pi|\cos 2kx|-1)\to3.$$ The crucial point is that, for each $j\in\{0\}\cup\{2,3,\dots\}$, the support set of the measure $|\mu_{2k}|$ (which is the interval $[2,3]$) is disjoint from the support set of the measure $|\mu_{2k}|^{*j}$ (which is the interval $[2j,3j]$). Therefore, $$\liminf_k\|\exp^*(\mu_{2k})\|\ge\liminf_k\|\mu_{2k}\|\ge3$$ and hence, in view of (2), $$\limsup_k\int1\,d\nu_{2k} =\limsup_k\frac{\int1\,d\exp^*(\mu_{2k})}{\|\exp^*(\mu_{2k})\|} \le\frac e3<1.$$

Thus, in view of (0) and (1), the sequence $$\Big(\frac{\exp^*(\mu_n)}{\|\exp^*(\mu_n)\|}\Big)$$ does not converge, even though $\exp^*(\mu_n)\to\exp^*(\mu)$.

Answer 2

Let $E$ be a metric space.

Lemma 1: Let $(\mu_t)_{t\in I}$ be a net in $\mathcal M(E)$ with $$c:=\lim_{t\in I}\left\|\mu_t\right\|\ne0\tag2$$ and $$\nu_t:=\frac{\mu_t}{\left\|\mu_t\right\|}\;\;\;\text{for }t\in I.$$ Then,

1. $\exists\mu\in\mathcal M(E)$ with $(\mu_t)_{t\in I}\to\mu$ weakly;
2. $\exists\nu\in\mathcal M(E)$ with $(\nu_t)_{t\in I}\to\nu$ weakly

are equivalent. In that case, $$\nu=\frac\mu c.\tag2$$

Proof: "$\Rightarrow$": $(\mu_t)_{t\in I}\to\mu$ weakly $\Rightarrow$ $$\nu_tf=\frac1{\underbrace{\left\|\mu_t\right\|}_{\to\:c}}\underbrace{\mu_tf}_{\to\:\mu f}\xrightarrow{t\in I}\frac\mu cf\;\;\;\text{for all }f\in C_b(E)\tag3.$$

"$\Leftarrow$": $(\nu_t)_{t\in I}\to\nu$ weakly $\Rightarrow$ $$\mu_tf=\underbrace{\left\|\mu_t\right\|}_{\to\:c}\underbrace{\nu_tf}_{\to\:\nu f}\xrightarrow{t\in I}c\nu f\;\;\;\text{for all }f\in C_b(E).\tag4$$

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Now assume $(E,d)$ is a metric $\mathbb R$-vector space and let $$\operatorname{CPoi}(\lambda):=\frac{\operatorname{exp}(\lambda)}{\left\|\operatorname{exp}(\lambda)\right\|}\;\;\;\text{for }\lambda\in\mathcal M(E).$$

Corollary 2: Let $(\lambda_t)_{t\in I}\subseteq\mathcal M(E)$. If $(\left\|\lambda_t\right\|)_{t\in I}$ is convergent and $$\alpha:=\lim_{t\in I}\left\|\lambda_t\right\|,$$ then

1. $\exists\mu\in\mathcal M(E)$ with $(\operatorname{exp}(\lambda_t))_{t\in I}\to\mu$ weakly;
2. $\exists\nu\in\mathcal M(E)$ with $(\operatorname{CPoi}(\lambda))_{t\in I}\to\nu$ weakly

are equivalent. In that case, $$\nu=\frac\mu{e^\alpha}.\tag5$$

Proof: Let $$\mu_t:=\operatorname{exp}(\lambda_t)$$ and $$\nu_t:=\operatorname{CPoi}(\lambda_t)=\frac{\mu_t}{\left\|\mu_t\right\|}$$ for $t\in I$. By definition of $\alpha$, $$c:=\lim_{t\in I}\left\|\mu_t\right\|=\lim_{t\in I}e^{\left\|\lambda_t\right\|}=e^\alpha>0\tag6$$ and hence the claim follows readily from Lemma 1.