Supremum Operator Norm is Norm

Theorem
Let $\map {CL} {X, Y}$ be the continuous linear transformation space.

Let $\norm {\, \cdot \,} : \map {CL} {X, Y} \to \R$ be the supremum operator norm such that:


 * $\forall T \in \map {CL} {X, Y} : \norm T := \sup \set {\norm {Tx}_Y : x \in X : \norm x_X \le 1}$

Then $\norm {\, \cdot \,}$ is a norm.

Positive definiteness
Let $T \in \map {CL} {X, Y}$.

By definition of norm:


 * $\norm {Tx}_Y \ge 0$

Then


 * $\ds \norm T = \sup_{\begin{split} x \mathop \in X \\ \norm x \mathop \le 1 \end{split} } \norm {T x}_Y \ge 0$

Suppose $\norm T = 0$.

Then:

Hence:


 * $\norm {T x}_Y = 0$

By Norm Axiom $N1$:


 * $\forall x \in X : Tx = \mathbf 0_Y$

where $\mathbf 0_Y$ is the zero vector in the normed vector space $Y$.

Therefore:


 * $T = \mathbf 0$.

Positive homogeneity
Let $\alpha \in \set{\R, \C}$.

Let $T \in \map {CL} {X, Y}$.

Then:

Triangle inequality
Take the supremum of both sides with the condition that $\forall x \in X : \norm x_X \le 1$.

Then:


 * $\norm {T + S} \le \norm T + \norm S$