In section 2.6 of Linde's Probability in Banach Spaces: Stable and Infinitely Divisible Distributions the author is pointing out that in infinite-dimensional Banach spaces the convergence of characteristic functions does not imply the weak convergence of measures.
I wonder what exactly is going wrong.
Let $E$ be a normed vector space and $\mathcal M(E)$ denote the space of finite signed measures on $\mathcal B(E)$ equipped with the total variation norm $\left\|\;\cdot\;\right\|$.
We know that if $(\mu_t)_{t\in I}\subseteq\mathcal M(E)$ is a bounded and tight$^1$ net and $\mu\in\mathcal M(E)$ is tight with$^2$ $$\forall g\in\mathcal C:(\mu_t)_{t\in I}\to\mu g\tag1$$ for some point-separating subalgebra $\mathcal C$ of $C_b(E)$ with $1\in\mathcal C$, then $(\mu_t)_{t\in I}\to\mu$ weakly.
Now, if $(\mu_n)_{n\in\mathbb N}\subseteq\mathcal M(E)$ and $\mu\in\mathcal M(E)$, the convergence of the characteristic functions of $(\mu_n)_{n\in\mathbb N}$ to the characteristic function of $\mu$ precisely means that $(1)$ holds for $$\mathcal C:=\left\{e^{{\rm i}\varphi}:\varphi\in E'\right\}.$$
Since $E'$ is point-separating, $\mathcal C$ is point-separating as well.
So, assuming that $(1)$ holds for this $\mathcal C$, shouldn't we be able to immediately conclude $(2)$?
(I would really appreciate if someone would share a reference which further elaborates on this topic. Maybe from a more functional-analytic point-of-view.)
$^1$ i.e. for any $\varepsilon>0$, there is a compact $K\subseteq E$ with $\sup_{t\in I}|\mu_t|(K^c)<\varepsilon$, where $|\mu_t$ denotes the total variation of $\mu_t$.
$^2$ As usual, $\mu g:=\int g\:{\rm d}\mu$.
