weak topology has less open set than strong topology (in Banach spaces). Why?

Question

Let $E$ a Banach spaces of infinite dimension. The weak topologie is the thickest that makes functional continuous. Let denote $\mathcal T_W$ the weak topology on $E$.

1) I call "Dual topological" the element of $E^*=\mathcal L(E,\mathbb R)$ that are continuous (wrt the strong topology that I denote $\mathcal T$, i.e. induced by the norm of $E$). I denote it $E'$. I know that $E'\subsetneq E^*$, i.e. there are linear functional that are not continuous.

2) In the book "Analyse fonctionnelle : Théorie et application" of Haim Brezis it's written that all open of the weak topology are also open in the strong topology.

3) Then it's written that the weak topology is strictly thicker than the strong topology in the sense that there are less open.

Question : For me, 1) is not coherent with 2) and 3). If $E'\subset E^*$ and that $\mathcal T_W$ makes all element of $E^*$ continuous, there are more continuous function with refer to the weak topology, and thus, it should has more open set no ? If we consider $E$ with a topology $T$ and $\mathcal F(E,\mathbb R)$ the set of function $f:E\to \mathbb R$, a priori, thiner is $T$ and more element of $\mathcal F(E,\mathbb R)$ are going to be continuous, and thus, more open there are no ? So with this argument, weak topology should has more open than strong topology. Could someone give me more explications ?

Answer 1

I suspect that you refer to Remark 2 in page 32 of the said book:

Les ouverts de la topologie faible sont aussi ouverts pour la topologie forte. Lorsque $E$ est de dimension infinie la topologie faible est strictement moins fine que la topologie forte i.e. il existe des ouverts pour la topologie forte qui ne sont pas ouverts pour la topologie faible.

In the English version (page 59):

Open sets in the weak topology are always open in the strong topology. In any infinite-dimensional space the weak topology is strictly coarser than the strong topology; i.e., there exist open sets in the strong topology that are not open in the weak topology.

Therefore, contrary to what you assume at the beginning of your question, it is not said that "all open of the strong topology are open in the weak topology".

Answer 2

You make confusions :

• Indeed, the topology defined as "the weakest topology s.t. elements of $E^*$ are continuous" is much thiner than the strong topology, and thus not very interesting. I recall that in normed vectors spaces we are looking for topologies that has few opens sets. The reason for that is that such topology has more compacts sets and such sets are very important for existences theorems.

• The weak topology is the weakest topology s.t. continuous linear form are continuous. So you don't consider all linear forms, but only continuous linear form. And such a topology is by definition thicker than the strong topology.