I'm trying to understand a result given in Theorem 2.3.7 (see below) of Linde's Probability in Banach spaces: Stable and Infinite Divisible Distributions.

Let $(\mu_n)_{n\in\mathbb N}$ be a sequence of Radon probability measures on the Borel $\sigma$-algebra $\mathcal B(E)$ of a Banach space $E$ such that there is a sequence $(x_n)_{n\in\mathbb N}\subseteq E$ such that $(\mu_n\ast\delta_{x_n})_{n\in\mathbb N}$ is relatively compact (denoted by "w.r.c." below) with respect to the topology of weak convergence of measures.

If we denote the topology of compact convergence on $E'$ by $\tau_c$, we can show that the family $\{\hat\mu:\mu\in\mathcal F\}$ of characteristic functions of a bounded (in total variation norm) and tight family $\mathcal F$ of finite signed measures on $\mathcal B(E)$ is uniformly $\tau_c$-continuous.

Assume now that there is a $\delta>0$ such that $C_\delta:=\left\{\left.\hat\mu_n\right|_{V_\delta}:n\in\mathbb N\right\}$ is $\tau_c$-equicontinuous at $0$ (which is actually equivalent to being uniformly $\tau_c$-equicontinuous, which in turn is equivalent to being relatively compact with respect to the uniform topology on $C(V_\delta)$), where $V_\delta:=\{\varphi\in E':\left\|\varphi\right\|_{E'}\le\delta\}$.

In conclusion, $C_\delta$ and $\left\{\left.\widehat{\mu_n\ast\delta_{x_n}}\right|_{V_\delta}:n\in\mathbb N\right\}$ are both $\tau_c$-equicontinuous at $0$.

This should yield the existence of $\rho>0$ and a compact $K\subseteq E$ such that $$\left|1-\hat\mu_n(\varphi)\right\|<\frac\varepsilon2\tag1$$ and $$\left|1-e^{{\rm i}\langle x_n,\:\varphi\rangle}\hat\mu_n(\varphi)\right\|<\frac\varepsilon2\tag2$$ for all $\varphi\in V_\delta$ with $p_K(\varphi):=\sup_{x\in K}|\langle x,\varphi\rangle|<\rho$ and $n\in\mathbb N$.

However, as you can see in the proof of Linde below, it seems like he is assuming $\rho=1$. That doesn't make sense to me. Why should $(1)$ and $(2)$ be smaller than $\varepsilon/2$ for an arbitrary $\varepsilon$ as long as $p_K(\varphi)\le1$?

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If $K$ is a compact subset of $E$, then so is $RK = \{Rx \, \mid \, x \in K\}$, no matter the choice of $R > 0$. Further, given $a \in E'$, $\sup \left\{ |\langle a,x \rangle| \, \mid \, x \in RK \right\} = R \sup \left\{ |\langle a,x' \rangle| \, \mid \, x' \in K \right\}$. Hence, by dilation, we can replace $\rho$ by $1$.

  • $\begingroup$ I thought on this problem in a more general setting of continuous functions and completely missed that we are considering linear continuous functions here. So, I simply can replace $K$ by $K/\rho$ ... $\endgroup$
    – 0xbadf00d
    Jan 13 '21 at 7:56

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