Characterizing compact operators from normal operators Assume H is a Hilbert space and T is normal operator that is bounded with spectral measure E. Is it possible to say something about T being compact iff spectral measure has finite rank when applied to elements of the spectrum ?
 A: Let $dP$ denote the spectral measure of $T$. Then one has the following statement:

A normal operator $T$ is compact iff $dP(\Bbb C- B_\epsilon(0))$ has finite rank for all $\epsilon>0$.

In order to show it first assume the statement on the right-hand side is true. Then as a consequence
$$T_\epsilon := \int_{\Bbb C-B_\epsilon(0)}\lambda \ dP(\lambda)$$
has finite rank. However
$$\|T-T_{\epsilon}\| =\left\|\int_{B_\epsilon(0)}\lambda\ dP(\lambda)\right\|≤ \epsilon \left\|\int_{B_\epsilon(0)}dP(\lambda)\right\|=\epsilon$$
and you get that $T$ is then a norm limit of finite rank operators, in particular $T$ is compact.
Now for the other direction assume $T$ is compact. Suppose there is some $R$ so that $dP(\Bbb C-B_R(0))$ does not have finite rank. Let $V$ be the space onto which $dP(\Bbb C-B_R(0))$ projects, this space is infinite dimensional. Then for all $v\in V$ you have:
$$\|Tv\| = \left\|\int_{B_R(0)}\lambda\ dP(\lambda)v + \int_{\Bbb C - B_R(0)}\lambda \ dP(\lambda)v\right\| = \left\|\int_{\Bbb C-B_R(0)}\lambda \ dP(\lambda)v\right\|≥ R\|v\|.$$
Now if $T_n$ is finite you must have that $\ker(T_n)$ intersects $V$ (since $V$ is infinite dimensional and $\ker(T_n)$ has finite codimension). As such for any $v\in \ker(T_n)\cap V$ you have
$$\|T-T_n\|≥ \|(T-T_n)v\| = \|Tv\| ≥ R\|v\|$$
implying $\|T\|≥R$. Hence $T$ is not approximated by finite rank operators, contradicting compactness. Thus our assumption that there is some $R$ with $dP(\Bbb C- B_R(0))$ not having finite rank must have been false.
