Approximation property controls finite rank density in compact operators

Let $X$ be a normed space and let $Y$ be a Banach space with the approximation property (Approximation property and bounded approximation property). Then every compact operator $T:X\to Y$ (Compact linear operator) is approximable (Approximable operator), that is, $T$ lies in the operator-norm closure of the bounded finite-rank operators $X\to Y$.

The argument is short and worth recording. The set $C:=\overline{T({\stackrel{‾}{B}}_X)}\subseteq Y$ is compact, because $T$ is compact. For a real $\epsilon >0$ the approximation property of $Y$ supplies a bounded finite-rank operator $R:Y\to Y$ with ${\sup }_{c\in C}\mathrm{\parallel }Rc-c\mathrm{\parallel }<\epsilon$. Then $R\circ T$ is a bounded operator whose range lies in the finite-dimensional range of $R$, hence $RT$ is finite rank, and for every $x\in X$ with $\mathrm{\parallel }x\mathrm{\parallel }\le 1$ the point $Tx$ lies in $C$, so $\mathrm{\parallel }RTx-Tx\mathrm{\parallel }<\epsilon$; taking the supremum over the closed unit ball gives $\mathrm{\parallel }RT-T\mathrm{\parallel }\le \epsilon$ (The operator norm as the least bound and as the unit-sphere or unit-ball supremum, A bounded linear operator between normed spaces). Since $\epsilon >0$ was arbitrary, $T$ is in the norm closure of the bounded finite-rank operators, which is exactly approximability.

This is not a universal finite-rank approximation theorem for arbitrary Banach targets. The argument uses the approximation property as a hypothesis; without it, nothing here produces finite-rank operators close to $T$ on the compact set $C$. In particular no counterexample for a target $Y$ failing the approximation property is asserted, and the converse implication — that approximability of every compact operator into $Y$ forces the approximation property of $Y$ — is a separate statement that is neither proved nor used here.