Let $X$ a vector space. We define weak topology as the weakest topology s.t. continuous linear form are continuous. In my course they only define weak topology for Banach space. I was wondering : if $X$ is not a Banach space could we define a weak topology ? And if not why ? Because using this definition (i.e. weakest topology s.t. linear form are continuous), I don't really see in what the fact that it's Banach is important.

  • $\begingroup$ A weak topology can be defined an a topological vector space. The results are richer in Banach spaces, of course. $\endgroup$ – copper.hat Aug 23 '18 at 23:54
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    $\begingroup$ A "weak" topology may be defined for any space $X$ and any collection of maps $f_a: X \rightarrow Y_a$, where $Y_a$ are spaces. See Brezis's functional analysis text for more details. $\endgroup$ – RandomWalker Aug 24 '18 at 2:56

The assumption "X Banach" is unnecessary in order to define the weak topology; even some of the results of the "Banach case" can be rearranged in a more general contest. If for example X is "only" a topological vector space, the Banach Alaoglu theorem, that states that the closure of the ball in the dual space is weak-* compact, is still valid. Of course here there is no distance, so given $V$ nbd of $0$ in $X$ you define the polar set $V°=\{f \in X^* | f(x) \leq 1 \forall x \in V\}$. With the same proof of the Banach case, you can prove that $V°$ is weak-* compact. (Sometimes in fact the theorem is formulated in the more general case).


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