# Image of a commutative C*-algebra

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.

## 1 Answer

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.