I'm studying Takesaki's Theory of operator algebras book by myself. The following is a theorem from that book:
I have several questions about this proof:
1- He claims, in the first line of proof, "we may assume that $A$ is a C*-algebra", while I do not why.
2- He claims, in the last line of first paragraph of proof, "we may assume that $A$ contains the identity", but before it he just shows that identity belongs to the strong* clouser of $A$, which is clearly is different from $A$.
3- In the middle of proof, he claims $fog(x)=x=gof(x)$ for $x\in S$. I accept this equation satisfies for normal elements of unit ball, but why is correct for all of them?
4- finally in the last paragraph he says, the unit ball of $A$ with respect to the uniform topology is not dense in the unit ball of $M$, while I know that $\|.\|-cl (A_{\|.\|\leq 1}) = wot-cl (A_{\|.\|\leq 1})$, also by the kapalansky density theorem $wot-cl (A_{\|.\|\leq 1})= M_{\|.\|\leq 1}$ and therefore $\|.\| - cl (A_{\|.\|\leq 1}) = M_{\|.\|\leq 1}$. So I'm confused really about that claim.
Please help me. Thanks in advance.
Because the unit ball of the C$^*$-algebra generated by $A$ sits in between the unit ball of $A$ and the unit ball of the weak closure of $A$; so if the former is dense in the latter, so is the unit ball of $C^*(A)$.
Because $A$ is non-degenerate, the identity is in its weak closure. So if the unit ball of $A$, together with the identity, is dense in the unit ball of the weak closure of $A$, so is the unit ball of $A$.
The equality $f\circ g=g\circ f$ is just an algebraic equality. And the functions are designed so that they apply to any element, not just normal ones.
What you say is not true: as Takesaki says, the norm closure is usually not equal to the weak closure. As a trivial example, if you consider $C[0,1]$ as multiplication operators on $L^2[0,1]$, it is norm closed and its weak closure is $L^\infty[0,1]$