Norm of Bounded Linear Transformation in terms of Closed Unit Ball

Theorem
Let $\GF \in \set {\R, \C}$.

Let $\struct {X, \norm {\, \cdot \,}_X}$ and $\struct {Y, \norm {\, \cdot \,}_Y}$ be normed vector spaces over $\GF$.

Let $T : X \to Y$ be a linear transformation.

Let $B_X^-$ and $B_Y^-$ be the closed unit balls of $X$ and $Y$ respectively.

Let $M > 0$.

Then $T$ is bounded with $\norm T_{\map B {X, Y} } \le M$ :


 * $T \sqbrk {B_X^-} \subseteq M B_Y^-$

Necessary Condition
Let $T$ be a bounded linear transformation with $\norm T_{\map B {X, Y} } \le M$.

Then for each $x \in X$ we have:
 * $\norm {T x}_Y \le M \norm x_X$

Then if $x \in B_X^-$, we have $\norm x_X \le 1$ and hence:
 * $\norm {T x}_Y \le M$

So we have:
 * $T \sqbrk {B_X^-} \subseteq M B_Y^-$

Sufficient Condition
Suppose that:
 * $T \sqbrk {B_X^-} \subseteq M B_Y^-$

Let $x \in X \setminus \set { {\mathbf 0}_X}$

Then, we have:
 * $\ds \frac x {\norm x_X} \in B_X^-$

and hence:
 * $\ds \norm {\map T {\frac x {\norm x_X} } }_Y \le M$

From, we have:
 * $\ds \norm {T x}_Y \le M \norm x_X$

Clearly this also holds for $x = {\mathbf 0}_X$.

So $T$ is bounded with $\norm T_{\map B {X, Y} } \le M$.