A <compact operator> is a <bounded linear operator> carrying every bounded set into a relatively <compact set>. Equivalently, for every bounded <sequence> $(x_n)$, the <sequence> $(Tx_n)$ has a norm-convergent <subsequence>. It is enough to require this for the <closed unit ball>, because any bounded set lies in a scalar multiple of that ball.

Solution

Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 4 / 22H / b / Solution

 Solution (source code)