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 $\left\|\;\cdot\;\right\|$.



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> 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](https://en.wikipedia.org/wiki/Convergence_of_measures#Weak_convergence_of_measures), then $\left(\frac{\mu\_n}{\left\|\mu\_n\right\|}\right)\_{n\in\mathbb N}$ is convergent as well.
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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 $\mu\in\mathcal M(E)$.



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



If $(\exp(\mu_n))_{n\in\mathbb N}$ is weakly convergent, is the normalized sequence convergent as well?
$\newcommand\R{\mathbb R}$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 $\exp^\*(\mu\_n)\to\exp^\*(\mu)$.


