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The following is taken from Peter D. Lax book Functional Analysis. Exercise 5 page 165.

Let $X,U$ be Banach spaces. I want to define the weak* topology on the set of bounded linear maps from $X$ to the dual of $U$ denoted $\mathcal{L}(X,U')$.

Am i correct in assuming that a sub basis for this topology is given by the sets

$$\{M\in \mathcal{L}(X,U')\mid |(Mx)u-(M_0x)u|<r\}$$

where $x\in X$$u\in U$, $M_0\in \mathcal{L}(X,U')$ and $r>0$?


EDIT: If my understanding of this is correct, we want the map $T:\mathcal{L}(X,U')\rightarrow \mathbb{R}$ defined by $T(M)=(Mx)u$ to be continuous, for any $x\in X$, $u\in U$. Therefore the inverse image of every open set $(a,b)$ must be open in $\mathcal{L}(X,U')$.

Suppose that $(M_0x)u\in (a,b)$. Choose $r$ such that $B_r[(M_0x)u]\subseteq (a,b)$. This denotes the ball of radius $r$ centred at $(M_0x)u$. Then

$$\{M\in \mathcal{L}(X,U')\mid |(Mx)u-(M_0x)u|<r\}\subseteq T^{-1}\Big((a,b)\Big).$$

On the other hand every set of the above type must be open so this is indeed the smallest topology which makes the above function continuous?

OgvRubin
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