[proofplan] We use the definition that a linear operator is compact precisely when it maps every bounded subset of its domain to a relatively compact subset of its codomain. One implication is immediate because the unit ball is bounded. For the converse, every bounded subset of $X$ lies inside a scalar multiple of $B_X$, and scalar multiplication by a fixed scalar is a homeomorphism of $Y$, so relative compactness of $T(B_X)$ transfers to relative compactness of the image of any bounded set. [/proofplan]

[step:Use compactness of $T$ on the bounded unit ball] Assume first that $T$ is compact. By definition, for every bounded subset $A \subset X$, the image $T(A)$ is relatively compact in $Y$. The set $B_X$ is bounded because $\|x\|_X \leq 1$ for every $x \in B_X$. Applying compactness of $T$ to $A := B_X$ gives that $T(B_X)$ is relatively compact in $Y$. [/step]

[step:Scale an arbitrary bounded set into the unit ball]Assume conversely that $T(B_X)$ is relatively compact in $Y$. Let $A \subset X$ be an arbitrary bounded subset. By boundedness of $A$, there exists a constant $M > 0$ such that \begin{align*} \|x\|_X \leq M \end{align*} for every $x \in A$. Hence $M^{-1}x \in B_X$ for every $x \in A$, and therefore \begin{align*} A \subset M B_X := \{Mz : z \in B_X\}. \end{align*} By linearity of $T$, \begin{align*} T(A) \subset T(MB_X) = M T(B_X). \end{align*}[/step]

[guided]We want to prove that $T$ is compact, so we must start with an arbitrary bounded subset $A \subset X$ and prove that $T(A)$ is relatively compact in $Y$. Since $A$ is bounded, there is a number $M > 0$ such that \begin{align*} \|x\|_X \leq M \end{align*} for every $x \in A$. This lets us compare $A$ with the unit ball: for each $x \in A$, the vector $M^{-1}x$ satisfies \begin{align*} \|M^{-1}x\|_X = M^{-1}\|x\|_X \leq 1, \end{align*} so $M^{-1}x \in B_X$. Equivalently, $x \in M B_X$, and therefore \begin{align*} A \subset M B_X. \end{align*} Applying the [linear map](/page/Linear%20Map) $T: X \to Y$ to this inclusion gives \begin{align*} T(A) \subset T(MB_X). \end{align*} Because $T$ is linear, $T(Mz) = M T(z)$ for every $z \in B_X$, so \begin{align*} T(MB_X) = M T(B_X). \end{align*} Thus \begin{align*} T(A) \subset M T(B_X). \end{align*} This is the key reduction: the image of every bounded set is contained in a scalar multiple of the image of the unit ball.[/guided]

[step:Transfer relative compactness through scalar multiplication] Define the scalar multiplication map \begin{align*} S_M: Y &\to Y \\ y &\mapsto My. \end{align*} The map $S_M$ is a homeomorphism, with inverse $S_{M^{-1}}: Y \to Y$ given by $y \mapsto M^{-1}y$. Since $T(B_X)$ is relatively compact, $\overline{T(B_X)}$ is compact in $Y$. Therefore \begin{align*} S_M(\overline{T(B_X)}) \end{align*} is compact in $Y$ as the continuous image of a compact set. Since $S_M(T(B_X)) = M T(B_X)$, we have \begin{align*} M T(B_X) \subset S_M(\overline{T(B_X)}). \end{align*} The set $S_M(\overline{T(B_X)})$ is compact, hence closed in the normed space $Y$. Therefore \begin{align*} \overline{M T(B_X)} \subset S_M(\overline{T(B_X)}), \end{align*} so $\overline{M T(B_X)}$ is compact as a closed subset of a compact set. Since $T(A) \subset M T(B_X)$, it follows that \begin{align*} \overline{T(A)} \subset \overline{M T(B_X)}. \end{align*} The set $\overline{T(A)}$ is closed in $Y$, and $\overline{M T(B_X)}$ is compact. Hence $\overline{T(A)}$ is compact, so $T(A)$ is relatively compact in $Y$. Because $A \subset X$ was an arbitrary bounded subset, $T$ maps every bounded subset of $X$ to a relatively compact subset of $Y$. Thus $T$ is compact. [/step]