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

Let $1<p<\infty$ and $q=p/(p-1)$. The dual of ${\ell }^p$ is ${\ell }^q$. For every $a\in {\ell }^q$, $⟨a,e_n⟩=a_n\to 0$, so $e_n\rightharpoonup 0$ weakly. A compact operator sends a weakly convergent sequence to a norm-convergent sequence with the corresponding limit. Indeed, any failure of $\Vert T{e}_n\Vert \to 0$ would give a subsequence bounded away from zero, with a further norm-convergent subsequence by compactness; weak continuity of $T$ forces that norm limit to be zero, a contradiction. The condition is therefore necessary.

For failure of sufficiency, partition the positive integers into consecutive blocksand define the block averaging compactness counterexampleBy Hölder's inequality, $|(Tx{)}_k{|}^p\le {\sum }_{j\in B_k}|x_j{|}^p$. Summing over disjoint blocks shows $\Vert Tx{\Vert }_p\le \Vert x{\Vert }_p$, so $T$ is bounded. If $j\in B_k$, then $T{e}_j=k^{-1/q}{e}_k$; as $j\to \infty$, also $k\to \infty$, so $\Vert T{e}_j{\Vert }_p\to 0$.

However, the vectors $x^{(k)}=k^{-1/p}{1}_{B_k}$ have norm one and satisfy $T{x}^{(k)}=e_k$. Distinct output vectors have distance $2^{1/p}$, so this bounded sequence of inputs has no norm-convergent image subsequence. Thus $T$ is not compact.

To connect with the hint, define $(Uy{)}_j=k^{-1/q}{y}_k$ for $j\in B_k$. Then $U:{\ell }^q\to {\ell }^q$ is an isometry, and the dual pairing calculation gives $T=U^{\ast }$. The vectors $U{e}_k$ have disjoint supports and norm one. The direct estimates above establish the counterexample without relying only on the hint.