# Banach-Alaoglu theorem and coarseness of weak star topology

Let $X$ be a normed space and let $X^\ast$ denote its continuous dual. There is a norm on $X^\ast$ defined by $\|\varphi\|=\sup_{\|x\|=1}|\varphi(x)|$. The weak star topology on $X^\ast$ is defined to be the weakest (=coarsest) topology such that the maps $e_x: \varphi \mapsto \varphi(x)$ are continuous.

The Banach-Alaoglu theorem states that $D=\{\varphi \in X^\ast : \|\varphi\|\le1\}$ is compact in the weak star topology.

I hope everything I wrote so far is correct. I believe the diagram here should apply to $X^\ast$ in that the norm topology on $X^\ast$ is finer than the weak star topology. But if it was so then Banach-Alaoglu would follow immediately since of course $D$ is compact in the norm topology and sets that are compact in a stronger topology are automatically compact in a weaker topology, that is, if $C$ is compact in $T$ and $T'\subseteq T$ then $C$ is also compact in $T'$ where $T,T'$ are topologies.

Therefore the weak star topology cannot be weaker than the norm topology. Is this correct and if so could someone please provide an example of a set that is open in the weak star topology but not open in the norm topology?

• If $X$ is infinite dimensional, then $D$ is not compact in the norm topology. (If $X$ is finite dimensional, then the norm and weak-star topologies are the same.) – Nate Eldredge Apr 7 '14 at 16:30

of course $D$ is compact in the norm topology
That is your error. When $X$ is infinite dimensional, $D$ is not compact in the norm topology.
The norm topology on $X^*$ is always at least as fine (strong) as the weak-* topology. When $X$ is finite dimensional, both topologies are the same, and when $X$ is infinite dimensional, the norm topology is strictly finer.