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Takesaki in his operator theory says

A C*-algebra $M$ of operators on Hilbert space $H$ means a nondegenerate ( $\text {cl} (MH) = H$) $*-$ subalgebra of $B(H)$ which is closed under the uniform topology.

Suppose $M$ is a C*-algebra. Clearly $M$ is a $*-$ Banach algebra and $\text{cl}(MH)\subset H$. Just show that $H\subset \text{cl}(MH)$ . Let $\{u_i\}$ be a approximate identity of $M$. Thus $u_i\to1$(sot) and $\xi = \lim u_i\xi \in \text{cl} (MH)$ for $\xi\in H$.

For converse direction I know that $\|x\|^2=\|xx^*\|$ in general. but I would like to know the relation between nondegenerate and C*-property. Please give me a hint. Thanks.

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  • $\begingroup$ What are you trying to prove, exactly? If you don't assume anything, there's no reason for a $C^*$-subalgebra of $B(H)$ to be nondegenerate. For example, scalar multiples of a given orthogonal projection look like a decent $C^*$ algebra to me, though they're very far from being nondegenerate. On the other hand, I think it should be true that for a $C^*$-subalgebra of $B(H)$, the restriction to its maximal invariant (closed) subspace should be faithful. But the quote from Takesaki looks like a definition, not a theorem. $\endgroup$
    – tomasz
    Commented Jan 10, 2015 at 7:45
  • $\begingroup$ No, it isn't, if it is not closed. And degeneracy is irrelevant here: any closed $*$-subalgebra of $B(H)$ (or any $C^*$-algebra, really) is a $C^*$-algebra. $\endgroup$
    – tomasz
    Commented Jan 10, 2015 at 8:04
  • $\begingroup$ Sorry,It was a typo. I mean every $*-$ Banach alg is also C*-alg so why does Takesaki emphasis on it? $\endgroup$
    – niki
    Commented Jan 10, 2015 at 8:08
  • $\begingroup$ I don't have the book, so I can't tell for sure, but as I have said, it looks to me like he's defining the term "$C^*$-algebra of operators on $H$". $\endgroup$
    – tomasz
    Commented Jan 10, 2015 at 8:10
  • $\begingroup$ @Tomasz: Thanks $\endgroup$
    – niki
    Commented Jan 10, 2015 at 8:11

1 Answer 1

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Equality $\| x\|^2=\| x x^*\|$ holds for any bounded operator $x$ on $H$. Namely, let $\xi \in H$ be of norm $1$. Then $$ \| x\xi\|^2=\langle \xi,x^*x\xi\rangle\leq \| \xi\| \| x^* x\xi\|\leq \| x^* x\|\leq \| x^*\| \| x\|.$$ Since $x^{**}=x$ it follows easily that $\| x^*\|=\| x\|$. Now one has $$ \| x\|^2=\sup\{ \|x \xi\|^2; \xi\in H, \| \xi\|\leq 1\}\leq \sup\{ \| x^* x\xi\|; \xi\in H, \| \xi\|\leq 1\}=\| x^* x\|\leq \| x^*\| \| x\|=\| x\|^2.$$ Hence $\| x^* x\|=\| x\|^2$.

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  • $\begingroup$ I know it, I would like know the relation between C*-algebra and nongenerate property. $\endgroup$
    – niki
    Commented Jan 10, 2015 at 7:33
  • $\begingroup$ @niki Can you explain this more explicitly, please? $\endgroup$
    – Janko Bracic
    Commented Jan 10, 2015 at 7:37
  • $\begingroup$ I know that every $*-$ Banach algebra is not a C*-algebra. Also $\|x\|^2=\|x^*x\|$ is the only property of a C*-algebra that does not satisfy in any $*-$ Banach algebra. Now I would like prove its property using nondegenerate property, based on Takesaki's claim $\endgroup$
    – niki
    Commented Jan 10, 2015 at 7:40
  • $\begingroup$ @niki But it is said that $M$ is $\ast$-algebra of operators on a Hilbert space. If $M$ is just an abstract $\ast$-Banach algebra which acts on $H$ via some representation and one has $cl(MH)=H$ it does not follow that $M$ is $C^*$ algebra, I guess. $\endgroup$
    – Janko Bracic
    Commented Jan 10, 2015 at 7:45
  • $\begingroup$ Why $-1$ for my answer? $\endgroup$
    – Janko Bracic
    Commented Jan 10, 2015 at 7:48

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