Is a function of a compact operator a compact operator

Question

Let $H$ be an infinite-dimensional Hilbert space and $T$ a compact operator on $H$.

1. Given a holomorphic function $f: \Omega \to \mathbb{C}$ where $\Omega$ is an open set containing the spectrum $\sigma(T)$ of $T$, is $f(T)$ also compact ?
2. Moreover, suppose that $T$ is normal. Given a continuous function $f : \Omega \to \mathbb{C}$ where $\Omega$ is an open set containing the spectrum $\sigma(T)$ of $T$, is $f(T)$ also compact ?
3. In the case where $T$ is selfadjoint, I also know that there exists a Borel functional calculus. So when is $f(T)$ bounded? And when is it compact?

The answer to these questions would probably proceed by approximating $f(T)$ by finite rank operators, but I have no clue how to do this.

Answer 1

(1) In the holomorphic case (which I believe is rather Independent from the continuous case, since $T$ needs not be normal), assuming that $f(0)=0$, one may write $f(z)= zg(z)$, for some other holomorphic function $g$, so $$ f(T)=Tg(T), $$ hence $f(T)$ is compact.

(2) In the continuous case, again assuming that $f(0)=0$, one may approximate $f$ uniformly by functions of the form $zg(z)$, by Stone Weierstrass, and the result follows similarly.

(3) In the Borel case, as noted in the comments, nothing can be said.