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$\newcommand{\R}{\mathbb R}$Let $M$ denote the set of all finite signed measures on a separable Banach space $B$. For any $\mu\in M$, let \begin{equation*} \exp^*(\mu):=\sum_{k=0}^\infty\frac{\mu^{*k}}{k!}. \end{equation*} The following question arose during a discussion on this page:

Suppose that $(\mu_n)$ is sequence in $M$ such that $\mu_n\to\mu$ (weakly) for some $\mu\in M$ and $\|\mu_n\|\to c$ for some $c\in(0,\infty)$, where $\|\cdot\|$ is the total variation norm. Does it then follow that the sequence $(\nu_n)$ with \begin{equation*} \nu_n:=\frac{\exp^*(\mu_n)}{\|\exp^*(\mu_n)\|} \end{equation*} converges?

This question will be answered, negatively, below.

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$\newcommand{\R}{\mathbb R}\newcommand{\la}{\lambda}\newcommand{\de}{\delta}\newcommand{\ka}{\rho}$Let $B=\R$.

For each natural $n$, let \begin{equation*} \la_n:=2\de_{\pi/n}-\de_{1/n},\tag{1} \end{equation*} where $\de_a$ is the Dirac probability measure at point $a$. Then \begin{equation*} \la_n\to\mu:=\de_0 \tag{2} \end{equation*} and hence, by dominated convergence or using characteristic functions,
\begin{equation*} \exp^*(\la_n)\to\exp^*(\mu)=\exp^*(\de_0)=e\de_0. \tag{3} \end{equation*} Also, for each $k=0,1,\dots$ we have $$\la_n^{*k}=\sum_{j=0}^k\binom kj2^j(-1)^{k-j}\de_{j\pi/n-(k-j)/n};$$ therefore and because the values of $j\pi/n-(k-j)/n=j\big(\pi-(-1)\big)/n-k/n$ are distinct for distinct values of $j\in\{0,\dots,k\}$, we have
\begin{equation*} \|\la_n^{*k}\|=\sum_{j=0}^k\binom kj2^j=3^k. \tag{4} \end{equation*} Since $\pi$ is irrational, it is easy to see that for each $n$ the signed measures $\la_n^{*0},\la_n^{*1},\la_n^{*2},\dots$ are mutually singular, whence \begin{equation*} \|\exp^*(\la_n)\|=\sum_{k=0}^\infty\frac{\|\la_n^{*k}\|}{k!}=\sum_{k=0}^\infty\frac{3^k}{k!}=e^3. \tag{5} \end{equation*} Thus, by (3), \begin{equation*} \frac{\exp^*(\la_n)}{\|\exp^*(\la_n)\|}\to\frac{e\de_0}{e^3}. \tag{6} \end{equation*}

Somewhat similarly to (1), for each natural $n$, let $$\ka_n:=2\de_{1/n}-\de_0.$$ Then, quite similarly to (2)--(4), we have \begin{equation*} \ka_n\to\mu=\de_0,\tag{7} \end{equation*} \begin{equation*} \exp^*(\ka_n)\to\exp^*(\mu)=e\de_0,\tag{8} \end{equation*} \begin{equation*} \|\ka_n^{*k}\|=3^k. \end{equation*}

However, (5) and (6) do not hold with $\ka_n$ in place of $\la_n$. Indeed, \begin{equation*} \|\ka_n^{*0}+\ka_n^{*1}\|=\|\de_0+(2\de_{1/n}-\de_0)\|=2, \end{equation*} so that \begin{equation*} b:=\|\exp^*(\ka_n)\|\le\|\ka_n^{*0}+\ka_n^{*1}\| +\sum_{k=2}^\infty\frac{\|\ka_n\|^k}{k!}=2+(e^3-1-3)=e^3-2<e^3 \end{equation*} and hence, in view of (8), \begin{equation*} \frac{\exp^*(\ka_n)}{\|\exp^*(\ka_n)\|}\to\frac{e\de_0}b,\quad\text{and}\quad b\ne e^3; \tag{9} \end{equation*} (since $\exp^*(\ka_n)(A)=\exp^*(\ka_1)(nA)$ for all $A\subseteq\mathbb R$, we see that $b=\|\exp^*(\ka_n)\|$ does not depend on $n$).

Let now $\mu_n:=\la_n$ if $n$ is odd and $\mu_n:=\ka_n$ if $n$ is even. Then, in view of (2), (7), (3), (8), \begin{equation*} \mu_n\to\mu \end{equation*} and \begin{equation*} \exp^*(\mu_n)\to\exp^*(\mu). \end{equation*} Also, \begin{equation} \|\mu_n\|=3\to3. \end{equation} However, in view of (6) and (9), \begin{equation*} \frac{\exp^*(\mu_n)}{\|\exp^*(\mu_n)\|} \end{equation*} does not converge.

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