Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 4 / 22H / Solution

A bounded linear operator is compact if it maps the unit ball to a set with compact closure; equivalently, every bounded sequence has a subsequence whose images converge.

If $S,T$ are compact operators, select a subsequence on which $S{x}_n$ converges and then a further subsequence on which $T{x}_n$ converges. This proves compactness of any linear combination. For norm closure, suppose compact $T_j$ converge to $T$ in operator norm. Given $\epsilon >0$, choose $j$ with $\Vert T-T_j\Vert <\epsilon /3$. A finite $\epsilon /3$-net for $T_j$ of the unit ball is then a finite $2\epsilon /3$-net for its image under $T$. This image is totally bounded, and its closure is compact because $X$ is complete. Thus $B_0(X)$ is a closed linear subspace.

For bounded $S$ and compact $T$, the image of a bounded sequence under $S$ is bounded, so $TS$ is compact; and a convergent subsequence of $T{x}_n$ remains convergent after applying the continuous map $S$, so $ST$ is compact. Therefore the compact operators form a closed two-sided ideal in $B(X)$.

For the weighted backward shift, truncate its output after coordinate $N$, obtaining a finite-rank operator $T_N$. The exact norm estimate isHence $T$ is compact. Its iterates areThus for every nonzero complex $\lambda$ the Neumann seriesconverges absolutely in operator norm. Zero is an eigenvalue, since $T{e}_1=0$, whereas every nonzero value is in the resolvent set. Consequently