Is there any counterexamples for that? or there is quick proof?
Is the dual space of a Banach space Hausdorff?
2
$\begingroup$
general-topology
functional-analysis
-
2In what topology? – 2012-10-22
3 Answers
10
The dual of Banach space is a Banach space, hence Hausdorff.
2
Here:
normed => metric => Hausdorff
2
For $X^*$ to be Hausdorff in the weak topology, or even the weak* toplogy, it is enougn to know that the elements of $X$ separate points of $X^*$. But that is just the definition of equality for functions.
Given two distinct points $f,g \in X^*$, there is $x \in X$ so that $f(x) \ne g(x)$, and from this we get two disjoint open neighborhoods $\{k \in X^*: k(x) > r\}$ and $\{k \in X^*: k(x) < r\}$ where $r$ is chosen between $f(x)$ and $g(x)$.