# Understanding the relation of weak and weak star toplogy

I'm working with Eberlein- Smulian Theorem fromm the book "Topics in Banach Space Theory". During the proof I have seen that there is used a lot the concept of weak topology and weak star topology. For a given Banach space X according to the definitions the weak topology is defined on X and the weak star topology is defined on the dual space X* . While trying to understand the proof I see that for a set A in X there is considered the weak* closure W of A in X** . *I don't understand how we can speak about the weak and weak-star closure of a set A in X, since the weak-star closure, to me, makes sense only for sets in X **. I am really struggling with this issue! Is anybody who can tell me which is the clue here? The set A to which I am referring is used to prove that (iii) implies (i) in the last paragraph of the theorem. Any advice would be really appreciated!

• The author is thinking of $X$ embedded in $X^{\ast\ast}$ via the natural map $x \mapsto x^{\ast\ast}$, where $x^{\ast\ast}(f) = f(x)$ – Prahlad Vaidyanathan Oct 23 '13 at 17:46
• I have seen this embedding to be mentioned in several places. The nature of X and X** is different, so how is this embedding constructed ? Thank You! – Kristina Dedndreaj Oct 23 '13 at 20:41
• The embedding is precisely as @PrahladVaidyanathan gave it above. In more detail: If $x\in X$ then you embed that as $x^{**}\in X^{**}$ via the given formula. So $x^{**}$ needs to be a linear functional on $X^*$; you pick some $f\in X^*$ to apply $x^{**}$ to, and discover that now you have $x\in X$ and $f\in X^*$, and you combine them in the only way that makes sense: $f(x)$. Thus arriving at $x^{**}(f)=f(x)$. – Harald Hanche-Olsen Oct 23 '13 at 20:54

1. $X^{**}$ is a dual space, so it has a weak* topology.
2. $X$ can be considered a subspace of $X^{**}$.
As a side remark, the weak topology on $X$ coincides with the subscpace topology inherited from $X^{**}$ with the weak* topology, since both are derived from the functionals in $X^*$.