# 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.

## Proof

### 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:

 $\ds \norm {T x}_Y$ $\le$ $\ds \norm T \norm x_X$ Supremum Operator Norm as Universal Upper Bound $\ds$ $=$ $\ds 0 \cdot \norm x_X$ $\ds$ $=$ $\ds 0$

Hence:

$\norm {T x}_Y = 0$
$\forall x \in X : Tx = \mathbf 0_Y$

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

Therefore:

$T = \mathbf 0$.

$\Box$

### Positive homogeneity

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

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

Then:

 $\ds \norm {\alpha \cdot T}$ $=$ $\ds \sup_{\begin{split} x \mathop \in X \\ \norm x \mathop \le 1 \end{split} } \norm {\paren{\alpha \cdot T}x}_Y$ Definition of Supremum Operator Norm $\ds$ $=$ $\ds \sup_{\begin{split} x \mathop \in X \\ \norm x \mathop \le 1 \end{split} } \norm {\alpha \cdot \paren {Tx} }_Y$ Definition of Pointwise Scalar Multiplication of Mappings $\ds$ $=$ $\ds \sup_{\begin{split} x \mathop \in X \\ \norm x \mathop \le 1 \end{split} } \size \alpha \norm {\paren {Tx} }_Y$ Definition of Norm Axiom $(N2)$ : Positive Homogeneity $\ds$ $=$ $\ds \size \alpha \sup_{\begin{split} x \mathop \in X \\ \norm x \mathop \le 1 \end{split} } \norm {\paren {Tx} }_Y$ $\ds$ $=$ $\ds \size \alpha \norm T$ Definition of Supremum Operator Norm

$\Box$

### Triangle inequality

 $\ds \norm {\map {\paren{T + S} } x}_Y$ $=$ $\ds \norm {Tx + Sx}_Y$ Definition of Pointwise Addition of Mappings $\ds$ $\le$ $\ds \norm {Tx}_Y + \norm {Sx}_Y$ Definition of Norm Axiom $(N3)$ : Triangle Inequality $\ds$ $\le$ $\ds \paren {\norm T + \norm S} \norm x_X$ Supremum Operator Norm as Universal Upper Bound

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

Then:

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

$\Box$

$\blacksquare$