If we consider the dual Banach space $V' :=\{f:V\rightarrow \mathbb{C}$ such that $f$ is linear and bounded$\}$. We know $V'$ also forms a Banach space in the norm topology where the norm is the general operator norm. But the open ball is not compact in the norm topology (V is not finite dimensional), but it is so by the weak* topology by the Banach Alaoglu Theorem.

My question is that are these two topologies i.e. the norm topology and the weak* topology comparable, i.e. is one of them weaker than the other?

The second question is, whether $V'$ is still a Banach space with the weak* topology?


2 Answers 2


A subbasic open set of the weak$^\ast$ topology is of the form $O(v,\epsilon):=\{f \in V': |f(v)| < \epsilon\}$ (for some $v \in V, \epsilon>0$ and so for every $f \in O(v,\epsilon)$, we have $f \in B_d(f, \epsilon) \subseteq O(v, \epsilon)$, where $d$ is the sup-operator norm and so every weak$^\ast$ open set is operator-norm open. So exactly as the name suggests, the weak$^\ast$ topology is weaker than the norm topology on $V'$ and strictly weaker for infinite-dimensional $V$, by the Banach-Alaoglu theorem (and the fact that a Banach space is locally compact iff it is finite-dimensional). So $V'$ will almost never be a Banach space in the weak$^\ast$ topology (only if $V \simeq V' \simeq \Bbb R^n$ for some $n \in \Bbb N$).

  • $\begingroup$ May I ask how to show $f \in B_d(f, \epsilon) \subseteq O(v, \epsilon)$? $\endgroup$
    – user760
    Commented Nov 8, 2022 at 10:38

Every $weak^{*}$ open set is open in the norm topology. These two topologies coincide iff the space is finite dimensional. $V'$ with $weak^{*}$ topology is a Banach space iff $V$ is finite dimensional. [This is an easy consequence of Banach Alaoglu Theorem].


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