Compact operator, functional calculus

Question

Let $H$ be a Hilbert space, T be a compact operator and $f$ be a bounded function on $\sigma(T)$. Now I want to show that the operator $f(T)$ (in the sense of functional calculus) is compact if $dim(H)<\infty$. I think the compactness of $f(T)$ follows directly. Under which assumptions do we have the other direction? Thanks for your help!

Answer 1

This is not true. If $T$ is a normal compact operator on $H$, and $f$ is continuous on $\sigma(T)$ and $f(0)=0$, then $f(T)$ lies in the $C^*$-algebra generated by $T$, and can be approximated by polynomials in $T,T^*$ with no constant term. The algebra of such polynomials in $T,T^*$ contains only compact operators, hence $f(T)$ is compact (regardless of the dimension of $H$).

EDIT

As your question currently stands, it is trivial. If $\dim H<\infty$, every operator on $H$ is compact, hence if $T\in B(H)$, then $f(T)\in B(H)$ is compact a priori.

Answer 2

As stated this is clearly not so. For example let $f(z)=z$; then $f(T)=T$ is compact whether $H$ has finite dimension or not. (Maybe you were supposed to show that if $f(T)$ is compact for all such $f$ then $H$ has finite dimension?)