Finite-rank operator

Remarks

Equivalently, there are $n \ge 0$, bounded linear functionals $f_1, \ldots, f_n$ on $X$ and vectors $y_1, \ldots, y_n \in Y$ with $$T x = \sum_{k=1}^{n} f_k(x) \, y_k \quad \text{for all } x \in X.$$ Example: for a nonzero bounded linear functional $f$ on $X$ and a nonzero $y \in Y$, the map $x \mapsto f(x) y$ has image $\mathbb{K} y$, so it has rank $1$. Non-example: the identity map of $\ell^2$ is bounded, but its image $\ell^2$ is infinite-dimensional, so it is not of finite rank. For linear maps between finite-dimensional spaces, the rank is the usual rank of a linear map. A bounded finite-rank operator is compact, because bounded subsets of a finite-dimensional normed space are relatively compact. On Hilbert spaces, every compact operator is a norm limit of finite-rank operators. On general Banach spaces this can fail; this is the question of the approximation property.