Images of the functional calculi of an operator

Question

Let $H$ a Hilbert space and $T$ a bounded operator of $H$.

1. Is the unitary subalgebra $\{f(T) \text{ s.t. } f: \Omega \supseteq \sigma(T) \to \mathbb{C} \text{ holomorphic}\}$ closed in $B(H)$ ? What is it precisely ? Is it equal to $B(H)$ ?
2. Suppose $T$ is normal. Is $\{f(T) \text{ s.t. } f: \Omega \supseteq \sigma(T) \to \mathbb{C} \text{ continuous}\}$ a sub-C*-algebra of $B(H)$ ? Is it equal to $B(H)$ ?
3. Suppose $T$ is selfadjoint. Is $\{f(T) \text{ s.t. } f: \Omega \supseteq \sigma(T) \to \mathbb{C} \text{ measurable}\}$ a sub-von Neumann algebra of $B(H)$ ? Is it equal to $B(H)$ ?

The Wikipedia pages about these three calculi don't indicate what these images are and I have no idea how to attack the problem. The following question gives only a partial answer: Image of Borel functional calculus of a bounded normal operator . Thanks in advance.

Answer 1

Since 2 and 3 were already answered by @pitariver in the comments, I'll only say something that partially answers 1.

Let's call $\mathrm{Hol}(T)$ to the image of the holomorphic functional calculus for $T$ in $B(H)$. In general $\mathrm{Hol}(T)$ is not all of $B(H)$ simply because $\mathrm{Hol}(T)$ will always be a commutative unital subalgebra of $B(H)$. Indeed, let $f, g$ be holomorphic functions in a neighborhood of $\sigma(T)$, then the function $fg=gf$ is also holomorphic and since the functional calculus is an algebra homomorphism we have $$ f(T)g(T)=(fg)(T)=(gf)(T)=g(T)f(T) $$

As for what exactly is the image in $B(H)$, we have to look at the Banach subalgebra of $B(H)$ generated by $T$, which is the smallest Banach subalgebra in $B(H)$ that contains $T$. Let's call it $\mathrm{Ban}(T)$, which is clearly a closed commutative subalgebra of $B(H)$.

We claim that $\mathrm{Hol}(T) \subseteq \mathrm{Ban}(T)$. To prove this, let $f$ be a holomorphic function on an open neighborhood of $\sigma(T)$, say $\Omega$. Using basic complex analysis we can write $\Omega$ as countable increasing sequence of compacts subsets and use Stone-Weierstrass to find polynomials $p_n$ such that $p_n \to f$ in the uniform convergence on compact subsets of $\Omega$. Then, by basic properties of the holomorphic functional calculus we have that $\|p_n(T)-f(T)\| \to 0$. Since each $p_n(T) \in \mathrm{Ban}(T)$, it follows that $f(T) \in \mathrm{Ban}(T)$. This proves the claim.

As far as I know this is the best result we can get. I wasn't able to quickly find a counter example to show that the inclusion $\mathrm{Hol}(T) \subseteq \mathrm{Ban}(T)$ is proper. I suspect $\mathrm{Hol}(T)$ might not even be closed in $\mathrm{Ban}(T)$ but I also don't have an example for this.