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

A linear operator $T:X\to Y$ is a compact operator if the image of the closed unit ball has compact closure in $Y$. Equivalently, every bounded sequence $(x_n)$ in $X$ has a subsequence whose images converge in the norm of $Y$. This implies boundedness of $T$.

The Schauder theorem for compact operators says that for a bounded linear operator between Banach spaces,where the dual map of bounded linear operator is $T^{\ast }f=f\circ T$. The forward implication also holds for normed spaces; completeness of the target gives the usual converse. The compactness definition requires the image subsequence's limit to belong to $Y$, so that completeness qualification matters when stating the theorem beyond Banach spaces.

Let $A_n$ be finite-rank bounded operators with $\Vert T-A_n\Vert \to 0$ in operator norm. Fix $ϵ>0$ and choose $n$ with $\Vert T-A_n\Vert <ϵ/3$. The set $A_n(B_X)$ is bounded in a finite-dimensional subspace, hence has a finite $ϵ/3$-net $y_1,\dots ,y_m$.

For each $x\in B_X$, some $j$ satisfiesThus $T(B_X)$ is totally bounded. Its closure is totally bounded and, as a closed subset of the Banach space $Y$, complete. A complete totally bounded metric space is compact. Consequently $T$ is compact.

The operator-norm limit is bounded, since $\Vert T\Vert \le \Vert A_n\Vert +\Vert T-A_n\Vert$ for any sufficiently good approximation. This proves directly that an approximable operator into a Banach space is compact; no assertion that every compact operator on an arbitrary Banach space is approximable is needed.

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 ${\ell }^1$.

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.