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I am currently reading Jesse Peterson's lecture notes on von Neumann algebras. I'm confused by lemma 4.4.2. In particular, it seems to me that Hahn Banach theorem here can only conclude the map $\phi$ is in the X** not in X. Without this it seems the subsequent proof of uniqueness of predual of von Neumann algebra cannot go through. Can somebody help? Thank you very much.

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    $\begingroup$ I haven't read Peterson's notes, but I can tell you that Sakai's original paper proving uniqueness of the predual is pretty readable. $\endgroup$
    – Nik Weaver
    Commented Jan 18, 2020 at 13:24

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The proof is using the version of the Hahn-Banach theorem which works for locally convex spaces. We have a $C^*$-algebra $A$ and a Banach space $X$ with $X^*=A$. We then equip $A$ with the weak$^*$-topology, that is, $\sigma(A,X)$. As the proof notes, $(A)_1 \cap A_+$ is $\sigma(A,X)$ compact, so if $a<0$ we can find a $\sigma(A,X)$-continuous linear functional which separates. But $\sigma(A,X)$-continuous linear functionals are simply elements of $X$.

I agree that this is a little unclear in the proof. What's a good book for the Hahn-Banach theorem at this level of generality? You could look at Rudin, Functional Analysis. Or I really like Conway's Functional Analysis. I think Theorem 3.9 of Chapter IV in the 2nd edition is enough. I'm afraid I don't know a good online resource (can anyone suggest one?)

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    $\begingroup$ I don't know an online resource either, but these statements can be found in Schaefer's Topological Vector Spaces as II.9.2 and IV.1.2, and also somewhere in the first volume of Dunford and Schwartz. The statement that $\sigma(A,X)$-continuous linear functionals all come from elements of $X$ is itself not trivial unless you know to use the definition of continuity of linear maps in terms of neighbourhoods of $0$. $\endgroup$
    – Robert Furber
    Commented Jan 18, 2020 at 10:59
  • $\begingroup$ Thank you very much for the help. I understand the proof now. $\endgroup$
    – user151245
    Commented Jan 18, 2020 at 20:28

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