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We know that $A$ embeds into $A$** (the double dual space of $A$ ). Is the following true? If $\Psi$ is in $A$** and weak* continuous, is there an element $a \in A$ such that $ \Psi$ is the evaluation functional at $a$? That is to say, $\Psi(f)=f(a)$ for any $ f \in A^{*}$?

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up vote 7 down vote accepted

This has nothing to do with operator algebras as it is well-known that if $B$ is a linear topological space (local convexity and Hausdorff-ness hypothesis included), the topological dual of $B^{\ast}$ endowed with the wk*-topology is $B$ itself, that is, every wk*-continuous functional on $B^{\ast}$ is an evaluation functional. For the proof, see for example the first section of chapter V in J. B. Conway's Functional Analysis textbook.

Regards, G. Rodrigues

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