Past exam of the functional Analysis and Banach and C*-Algebras course of the University of Oxford / 2007 / A / 4

A bounded linear operator is compact when the closure of the image of its unit ball is compact. Equivalently every bounded sequence has a subsequence whose images converge; on a Banach target it also suffices that the image of its unit ball be totally bounded.

For a finite-rank operator that image is bounded in a finite-dimensional subspace, whose closed bounded sets are compact. Thus every finite-rank operator is compact.

Given a bounded sequence, first pass to a subsequence on which the images under $T_1$ converge. From that subsequence extract another on which the images under $T_2$ converge. Both image sequences, and hence their sum, converge on the latter subsequence. Thus $T_1+T_2$ is compact.

Extract a subsequence whose $T$ images converge, and apply continuity of the bounded operator $S$ to those images. This proves $ST$ is compact. The reverse composition $TS$ is compact as well, since $S$ sends a bounded sequence to a bounded sequence to which compactness of $T$ applies.

Suppose $T_n$ are compact and $\Vert T_n-T\Vert \to 0$. For any $\epsilon >0$ take $n$ with $\Vert T_n-T\Vert <\epsilon$. A finite $\epsilon$-net of $T_n$ of the unit ball is then a finite $2\epsilon$-net of its $T$ image. That image is totally bounded, so $T$ is compact by the stated criterion. Hence $\mathcal{K}(X)$ is norm closed in $B(X)$.

Compact. Let $P_N$ keep the first $N$ coordinates. The operator $P_{N}T$ has finite rank and $\Vert T-P_{N}T\Vert \le 1/(N+1)$, since the remaining diagonal coefficients are at most that number. Norm-closedness of compact operators proves the claim.

Not compact. For coordinate unit vectors $e_j$ with $j\ge 2$, the images are $e_{j-1}$. Distinct such images have ${\ell }^1$ distance two, so no image subsequence converges.

Compact. This operator is $BD$, where $D$ is the compact diagonal operator in (a) and $B$ is the bounded backward shift in (b). The composition property applies. Equivalently its tails have the same vanishing-coefficient finite-rank approximation.

Compact. Its column indexed by $m$ has entries $n!/m!$ for $1\le n<m$, and zero otherwise. The sum of that column is $c_m=(1!+\cdots +(m-1)!)/m!\le 2/m$ for $m\ge 2$, by the elementary induction $1!+\cdots +k!\le 2k!$. For operators from ${\ell }^1$ the supremum of the absolute column sums is the operator norm: one inequality is the triangle inequality, and coordinate unit vectors give the reverse. Keeping just the first $N$ input columns is finite rank and its remainder norm is at most ${\sup }_{m>N}{c}_m\le 2/(N+1)$. Hence it converges to $T$ in norm, proving compactness.