Clarification on proof in Murphy's C*-algebras

in the proof theorem 4.3.4 in Murphy's C*-algebras and operator theory I stumbled upon something I don't quite get.

The Theorem states, that for every von Neumann algebra $A$ in $H_1$ and every weakly continuous *-homomorphism $\phi\colon A \to B(H_2)$ the range $\phi(A)$ is a von Neumann algebra.

I'll sum up the proof:

1. without loss of generality we assume $1_H\in A$. And we know already that $\phi(A)$ is a C*-algebra
2. He shows $\phi(\{a\in A\colon ||a|| < 1\})=\{b\in \phi(A)\colon ||b|| <1\}$
3. He states that $\phi( (A)_1 )$ is weakly compact
4. He states that the unit ball in $A$ is the weak closure of $R=\{a\in A\colon ||a|| < 1\}$
5. He shows that $\phi((A)_1) = (\phi(A))_1$
6. He shows with the Kaplansky density theorem that $\phi(A)$ is weakly closed

Why does he explicitly mention (4.)? I got the feeling that it's not needed in the proof.

I suppose one can only answer this question if one has a copy of the book at hand, but maybe someone is willing to have a look on page 132 and help me with my concerns.

I do believe that point 4. is superfluous. All you really need is that $(A)_1$ is weakly compact and that $R\subset (A)_1$. Since point 6. follows from point 5., let's analyze the proof of 5. sentence-by-sentence to see if it's necessary:

For in the proof that $(\phi(A))_1\subset\phi((A)_1)$:

Take $v\in\phi(A)$ with $\|v\|\leq1$, and a sequence $\varepsilon_n\in(0,1)$ converging to $1$.

Definitely not needed here.

Then for all $n$, $\|\varepsilon_nv\|<1$, so $\varepsilon_nv=\phi(u_n)$ for some $u_n\in R$

This is by point 2., so 4. is not needed.

Then $\varepsilon_n v$ is a sequence in $\phi((A)_1)$,

just because $R\subset(A)_1$

converging in norm to $v$, so $v\in\phi((A)_1)$.

Here since $(A)_1$ is weakly compact, $\phi((A_1))$ is as well, hence it is norm closed, so 4. is not necessary.

Now the other inclusion is not necessary to invoke 4. either, for if $v\in\phi((A)_1)$, then $v=\phi(u)$ for some $u\in A$ with $\|u\|\leq1$, and since $\phi$ is a $*$-homomorphism, $\|\phi\|\leq1$, so $\|v\|\leq1$ and thus $v\in(\phi(A))_1$.

So it appears you were correct.

• thanks, that was what I was thinking but I'm not used to feeling any comments in literature superfluous ... more of a confidence issue than a mathematical issue I suppose – VanillaThunder Mar 27 '18 at 13:27
• You're welcome. This actually happens more often than you'd think (not extremely common, but not unheard of). That you noticed it is a sign of mathematical maturity. – Aweygan Mar 27 '18 at 13:36