1
$\begingroup$

Let $A$ be an unital commutative C*-subalgebra of $B(H)$, and $\Omega$ be its character space. By spectral theorem $$\phi: B_\infty(\Omega)\to B(H);~~~~~f\to \int f \, dP$$ is a $*-$ homomorphism where $B_\infty(\Omega)$ is the space of bounded Borel-measurable functions and $P$ is the spectral measure correspondence $A$.

If $x\in B(H)$ is normal, put $A=C^*(x,1)$ (C*-algebra generated by $x,1$). Could we show that $\phi(B_\infty (\sigma(x))) = vn(x,1)$ (von Neumann algebra generated by $x,1$)? Thanks in advance.

$\endgroup$
1
$\begingroup$

After 6 hours, I think the answer is yes, if for every $\xi\in H$, $x\xi\neq 0$. In this case $\phi(B_\infty(\Omega))$ is nondegenerate. Now it's sufficient to show that $\phi(B_\infty(\Omega))$ is wot-closed. Easily we can show that if $f_n\to f$ (pointwise- bounded) then $\phi(f_n)\to \phi(f)$ (wot), and it's the key point of solution.

For $g\in \phi(B_\infty(\Omega))$, there is $f\in B_\infty(\sigma(x))$ such that $g=\phi(f)$. Also there is a sequence $\{f_n\}\subset B_\infty(\Omega)$ such that $f_n\to f$ (p.b), so $\phi(f_n)\to g$ (wot). It's desired result.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.