Can we show that $(x_n)$ is relatively compact if and only if $(\langle x_n,\;\cdot\;\rangle)$ restricted to a closed ball is relatively compact?

Question

Let $E$ be a Banach space, $\iota:E\to E''$ denote the canonical injection of $E$ into $E''$, i.e. $$\iota x:=\langle x,\;\cdot\;\rangle\;\;\;\text{for }x\in E,$$ $\sigma_c(E',E)$ denote the topology of compact convergence on $E$ and $(x_n)_{n\in\mathbb N}$.

Are we able to show that $(x_n)_{n\in\mathbb N}$ is relatively compact if and only if there is a $r>0$ such that $(\left.\iota x_n\right|_{B^{E'}_r(0)})_{n\in\mathbb N}$ is relatively compact with respect to the topology of uniform convergence on $C(B^{E'}_r(0),\sigma_c(E',E))$?

The claim can be found in the proof of Linde's book Probability in Banach Spaces - Stable and Infinitely Distributions, Lemma 2.3.6:$^1$

He also seems to conclude (most probably using the Arzelà-Ascoli theorem) that "$(\left.\iota x_n\right|_{B^{E'}_r(0)})_{n\in\mathbb N}$ is relatively compact with respect to the topology of uniform convergence on $C(B^{E'}_r(0),\sigma_c(E',E))$" follows from "$(\left.\iota x_n\right|_{B^{E'}_r(0)})_{n\in\mathbb N}$ is $\sigma_c(E',E)$-equicontinuous". But this is only true, if $(\left.\iota x_n\right|_{B^{E'}_r(0)})_{n\in\mathbb N}$ is pointwise bounded, which should be the case if (and only if) $(x_n)_{n\in\mathbb N}$ is bounded. Why is that the case?

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$^1$ $V_\delta:=\{\varphi\in E':\left\|\varphi\right\|_{E'}\le\delta\}$ is endowed with the restriction of the topology $\sigma_c(E',E)$ and $\delta_x$ denotes the Dirac measure concentrated at $x$.

Answer 1

Given a set $E$, denote by $M(E)$ the set of all bounded functions on $E$ with the supremum norm topology. We will consider $M(V_{\delta})$, where $V_{\delta} = \{\psi \in E' \, \mid \, \|\psi\| \leq \delta\}$.

Claim: $(x_{n})_{n \in \mathbb{N}}$ is relatively compact in $E$ if and only if, for some $\delta > 0$, $(i x_{n})_{n \in \mathbb{N}}$ is relatively compact in $M(V_{\delta})$.

The claim follows from the following equivalence: given any subsequence $(n_{j})_{j \in \mathbb{N}}$ and any $\delta > 0$, $(x_{n_{j}})_{j \in \mathbb{N}}$ is a Cauchy sequence in $E$ if and only if $(ix_{n_{j}})_{j \in \mathbb{N}}$ is a Cauchy sequence in $M(V_{\delta})$.

Indeed, notice that, given $\delta > 0$, we have \begin{align*} \sup \left\{ |ix_{n_{j}}(\psi) - ix_{n_{k}}(\psi)| \, \mid \, \psi \in V_{\delta} \right\} &= \sup \left\{ |\psi(x_{n_{j}}) - \psi(x_{n_{k}})| \, \mid \, \psi \in V_{\delta} \right\} \\ &= \delta \|x_{n_{j}} - x_{n_{k}}\|. \end{align*}
Thus, by the definition of Cauchy sequence, $(x_{n_{j}})_{j \in \mathbb{N}}$ is Cauchy for the norm metric in $E$ if and only if it is Cauchy for the supremum norm metric in $M(V_{\delta})$.

Answer 2

Peter Morfe's answer answered my question, which is why I've accepted it. However, since the whole stuff with the $\sigma_c(E',E)$-topology on $E'$ confused me and it turned out that everything is way more elementary than I thought, I will reformulate the result, which can be proved in the way Peter described in his answer.

Let

• $\mathbb F=\mathbb R$ or $\mathbb F=\mathbb C$;
• $X$ be a normed $\mathbb F$-vector space;
• $\langle\;\cdot\;,\;\cdot\;\rangle$ denote the canonical duality pairing between $X$ and $X'$;
• $\iota$ denote the canonical embedding of $X$ into $X''$, i.e. $$\iota:X\to X''\;,\;\;\;x\mapsto\langle x,\;\cdot\;\rangle;$$
• $V_r:=\overline B^{X'}_r(0)$ for $r>0$.

Remember that if $V$ is any set, then $$B(V):=\left\{f:V\to\mathbb F\mid\left\|f\right\|_\infty<\infty\right\}$$ equipped with $\left\|\;\cdot\;\right\|_\infty$ is a $\mathbb F$-Banach space.

Proposition: Let $(x_n)_{n\in\mathbb N}\subseteq X$ and $$f_n:=\iota(x_n)\;\;\;\text{for }n\in\mathbb N.$$

1. If $(x_n)_{n\in\mathbb N}$ is relatively compact, then $(f_n)_{n\in\mathbb N}$ is relatively compact in $\iota(X)$. In particular, $\left(\left.f_n\right|_{V_r}\right)_{n\in\mathbb N}$ is relatively compact in $B(V_r)$ for all $r>0$.
2. If $X$ is complete and $\left(\left.f_n\right|_{V_r}\right)_{n\in\mathbb N}$ is relatively compact in $B(V_r)$ for some $r>0$, then $(x_n)_{n\in\mathbb N}$ is relatively compact.

Now the result asked for in the question is a simple corollary: If $r>0$ and $\tau$ is any compact topology on $V_r$, then $$C(V_r,\tau):=\left\{f:V_r\to\mathbb F\mid f\text{ is }\tau\text{-continuous}\right\}$$ is a closed subspace of $B(V_r)$. Thus, if $\iota(X)\subseteq C(V_r,\tau)$, then $\left(\left.f_n\right|_{V_r}\right)_{n\in\mathbb N}$ being relatively compact in $B(V_r)$ implies that $\left(\left.f_n\right|_{V_r}\right)_{n\in\mathbb N}$ is relatively compact in $C(V_r,\tau)$.