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

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.