= Solution
A <linear operator> $T:X\to Y$ is a \b[<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 \b[<Schauder theorem for compact operators>] says that for a <bounded linear operator> between <Banach spaces>,
$$
\boxed{T:X\to Y\text{ compact}\quad\Longleftrightarrow\quad T^*:Y^*\to X^*\text{ compact},}
$$
where the <dual map of bounded linear operator> is $T^*f=f\circ T$. 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>.
Back to article page
