Compact linear operator

Let $X$ and $Y$ be normed spaces over the same scalar field $\mathbb{K}$, read in the real case from A norm on a real vector space, the induced metric, and the dictionary with the metric axioms and in the complex case from Real and complex scalar conventions for normed spaces. A linear map $T:X\to Y$ (Linear map between vector spaces over the same field) is a compact operator when the image of every bounded subset of $X$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space) has compact closure in $Y$ (Open cover, subcover, compact metric space, and compact subset of a metric space); explicitly, for every bounded $A\subseteq X$ the closure $\overline{T(A)}$ in $Y$ is a compact subset of $Y$. The set of compact operators $X\to Y$ is written $\mathcal{K}(X,Y)$.

The closed unit ball suffices. Put ${\bar{B}}_X:=\{x\in X:\mathrm{\parallel }x\mathrm{\parallel }\le 1\}$ (Open ball, closed ball and sphere in a metric space). Then $T$ is compact if and only if $\overline{T({\stackrel{‾}{B}}_X)}$ is a compact subset of $Y$.

Indeed, if $T$ is compact then ${\bar{B}}_X$ is bounded, because ${\bar{B}}_X\subseteq B(0,2)$ while $B(0,2)$ is bounded and a subset of a bounded set is bounded, so $T({\bar{B}}_X)$ has compact closure. Conversely assume $\overline{T({\stackrel{‾}{B}}_X)}$ compact and let $A\subseteq X$ be bounded. If $A=\mathrm{\varnothing }$ then $T(A)=\mathrm{\varnothing }$, whose closure is empty and hence compact. Otherwise $A\subseteq B(x_0,r)$ for some $x_0\in X$ and real $r>0$, so every $x\in A$ satisfies $\mathrm{\parallel }x\mathrm{\parallel }\le \mathrm{\parallel }{x}_0\mathrm{\parallel }+r=:R$; if $R>0$ then $A\subseteq R{\bar{B}}_X$ and if $R=0$ then $A\subseteq \{0\}$, so in either case $T(A)\subseteq R\overline{T({\stackrel{‾}{B}}_X)}$. Scalar multiplication by $R$ is continuous (Vector addition and scalar multiplication are continuous in a normed space), so $R\overline{T({\stackrel{‾}{B}}_X)}$ is a compact subset of $Y$ (A continuous image of a compact space is compact; a continuous real-valued map on a nonempty compact space attains a maximum and a minimum; and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism), hence closed (A compact subset of a metric space is closed and bounded); therefore $\overline{T(A)}\subseteq R\overline{T({\stackrel{‾}{B}}_X)}$ is a closed subset of a compact set, hence compact, and $T$ is compact.

A compact operator is bounded. If $T$ is compact then the compact set $\overline{T({\stackrel{‾}{B}}_X)}$ is bounded (A compact subset of a metric space is closed and bounded), so there is a real $C\ge 0$ with $\mathrm{\parallel }Tx\mathrm{\parallel }\le C$ for every $x\in {\bar{B}}_X$ and hence $\mathrm{\parallel }Tx\mathrm{\parallel }\le C\mathrm{\parallel }x\mathrm{\parallel }$ for every $x\in X$; thus $T$ is a bounded linear operator (A bounded linear operator between normed spaces). This is a consequence of compactness, not a hypothesis of the definition.