Past exam of the functional analysis course of the University of Oxford / 2011 / 1 / b / i

Let $Q_N$ be truncation to the first $N$ coordinates in ${\ell }^1$. For $x\in {\ell }^1$, boundedness of $T$ and absolute convergence giveApplying the difference to each $e_n$ with $n>N$ gives the reverse bound on the operator norm. HenceThe stated condition makes this tend to zero. Each $T{Q}_N$ has finite rank, so part (a)(ii) proves compactness. Thus the condition is sufficient on ${\ell }^1$.

For failure of necessity, set $Tx=({\sum }_{n\ge 1}{x}_n)e_1$. The sum defines a bounded linear functional on ${\ell }^1$, so $T$ is a bounded rank-one operator and therefore compact. But $T{e}_n=e_1$ for every $n$, so $\Vert T{e}_n{\Vert }_1=1$. Thus compactness need not make the basis images tend to zero on ${\ell }^1$.