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I am aware there are other proofs of line of this statement. But I am interested in the argument outlined here on page 62-63

Corollary II.2.2.9 Let $A$ and $B$ be $C^*$ algebras, $\phi:A \rightarrow B$ be injective $*$-homomoprhism. Then $\phi$ is isometric, i.e. $||\phi(x) || = ||x||$ for all $x \in A$.

The proof goes as follows: Wlog we may assume $A,B$ are commutative (I got this ), and it is obvious from II.2.2.4. (as below).

Theorem II.2.2.4 If $A$ is a commutative $C^*$ algebra, then the Gelfand trasform is an isometric $*$-isomorphism from $A$ onto $C_0(\hat{A})$.

How does II.2.2.4 imply 2.2.9?

Bryan Shih
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1 Answers1

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That would be II.2.2.4, the Gelfand Transform, a few theorems back. If you are new to C$^*$-algebras, reading Blackadar (with almost no proofs and a general point of view) is probably not the best idea.

Answer to the edit: as Blackadar says, the C$^*$-identity lets you restrict the problem to $A,B$ abelian. So you have maps $$ C_0(\hat A)\xrightarrow{\ \ \ \ \ } A\xrightarrow{\ \ \phi\ \ } B\xrightarrow{\ \ \ \ \ } C_0(\hat B), $$ where the three maps are injective ($\phi$ by hypothesis and the other two by II.2.2.4). Now when restricted to algebras of continuous functions you can use II.2.2.5/II.2.2.7: say $\psi:C(X)\to C(Y)$ is an injective homomorphism. Then $\breve\psi:Y\to X$ is surjective. Now, for $f\in C(X)$, you have (I do the compact case to avoid a few epsilons) $$ \|f\|=|f(x_1)|=|f(\breve\psi (y_1))|=|(\psi f)(y_1)|\leq \|\psi f\|. $$ Also, $$ \|\psi f\|=|(\psi f)(y_0)|=|f(\breve\psi(y_0))|\leq \|f\|. $$ Thus $\|f\|=\|\psi f\|$.

Martin Argerami
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  • That's the only part of the proof that Blackadar did write! – Martin Argerami Apr 12 '19 at 18:01
  • No, he says "For the first part, it remains only to show that $\Gamma$ is isometric (then the range will be closed, hence all of $C_0(\tilde A)$ by the Stone-Weierstrass Theorem). By the C$^*$-axiom and II.1.6.6., it suffices to show that if $x = x^∗$, then $|\hat x|\geq|x|$. By II.2.1.2. there is $λ \in \sigma_A(x)$ with $|λ| = |x|$, so there is a φ with $|\tilde x(φ)| = |x|$." – Martin Argerami Apr 12 '19 at 18:05
  • Edited. The "theorem" in the "obvious" statement is II.2.2.8 and not II.2.2.4. – Martin Argerami Apr 12 '19 at 18:57
  • I mean - how did you get the last $\le$ for both of your lines? – Bryan Shih Apr 12 '19 at 21:05
  • They are points. The norm of a function $f$ is $|f|=\max{|f(t)|: t}$. So you always have $|f(t)|\leq |f|$. – Martin Argerami Apr 12 '19 at 21:07
  • I see ! Thanks a lot. I thought you meant operator, now everything makes sense. – Bryan Shih Apr 12 '19 at 21:09
  • Glad I could help. That was the whole point of the move to C$^*$-algebras of continuous functions. – Martin Argerami Apr 12 '19 at 21:13
  • Hi Martin, I wonder if you know a reference/ proof of the more general statement for the post here on ess. s.a. operators. https://math.stackexchange.com/questions/3183960/decomposition-for-essentially-self-adjoint-operators – Bryan Shih Apr 12 '19 at 22:38
  • I know very little about unbounded operators, so I'm not sure I can help. – Martin Argerami Apr 12 '19 at 22:55
  • Ok, you have helped a lot already. I also have one, that requires little no knowledge : i just want to apply functional calculus but do not get the desired conclusion. I wonder if you can have a look at where I got wrong. https://math.stackexchange.com/questions/3185681/applying-functional-calculus-to-the-bounded-operator-t-pm-ii-1 – Bryan Shih Apr 12 '19 at 23:39
  • Why can you focus on commutative $C^$-algebras? How does the $C^$-identity allow restriction? – MyWorld Jan 20 '22 at 18:25
  • Since $|a|=|a^a|^{1/2}$, it is enough to prove that $\phi$ is isometric on positive elements. So one considers the restriction of $\phi$ to $C^(a)$ for $a\geq0$, which is abelian. And the image of an abelian C$^$-algebra through a $$-homomorphism is abelian. – Martin Argerami Jan 20 '22 at 21:47