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I'm self studying Rudin's Functional Analysis by myself. In page 68, he wants to show that $(X^*,w^*)^*=X$ where X is a topological vector space and whose dual is $X^*$.

He says every $x\in X$ induces a linear functional $f_x$ on $X^*$, defined by $f_x(\phi)=\phi x$ for every $\phi\in X^*$. then he says $X=\{f_x, x\in X\}$ . I can not understand why $X=\{f_x, x\in X\}$? I think just there is a linear bijection between $X$ and $\{f_x, x\in X\}$ and we can say there is a linear bijection between $X$ and $(X^*,w^*)^*$. Please help me.

Thanks in advance.

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  • $\begingroup$ When you work with vector spaces V,W , if there is a linear bijection L between them, they are isomorphic , so one often just says tha the two are equal unless there is some additional structure in consideration that is not preserved by L . $\endgroup$ – user99680 Jul 10 '14 at 10:03
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In the weak$^*$-topology of $X^*$, you can linearly and continuously inject $X$ into $X^{**}$. Rudin considers this injection an identification, which eases notation. In this relation, see page 95 for a more thorough discussion of the special case of Banach spaces.

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