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We defined the weak limit as:

Let $X$ be a Banach space. $x_n \in X$ converges weakly to $x_0 \in X$, if $\: \:\forall _{\phi \in X^{*}}\:\phi \left(x_n\right)\rightarrow \phi \left(x_0\right)$

But why are we even bothering to introduce the weak limit in the first place? What's the motivation and use behind that?

I am in an Operator Theory course, and we only got a short introduction into functional analysis (we never had functional analysis before), and all I know is that it's important for the Riesz theorem (I assume) and Banch-Alaoglu theorem (due to weak topology) (which I also don't understand the use of it, especially regarding Operator Theory)

anon
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1 Answers1

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Your question basically is: why care about the weak topology? The reason is simple: given a topological vector space $(V, \tau)$, the weak topology $(V, \tau_w)$ has the same continuous dual: $$(V,\tau)^* = (V, \tau_w)^*.$$ The Hahn-Banach separation theorem implies that if $C\subseteq V$ is a convex set, then $$\overline{C}^{\tau}= \overline{C}^{\tau_w}.$$

I can't emphasise enough how important this result is. To show that an element $x$ is in $\overline{C}^\tau$, it suffices to check the (formally) weaker condition $x \in \overline{C}^{\tau_w}.$


Very often this result is applied in the following situation:

Let $V$ be a normed space and let $C\subseteq V$ be a closed convex subset and $v \in V$. If you can find a sequence (or more generally a net) $\{c_\lambda\}_{\lambda \in \Lambda}\subseteq C$ such that $\lim_{\Lambda }\omega(c_\lambda)= \omega(v)$ for all $\omega \in V^*$, then $v\in C$.

Try to show this without topological vector space techniques!

J. De Ro
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