Supremum Operator Norm is Norm

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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:

$\norm {T x}_Y = 0$

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:

$T = \mathbf 0$.

$\Box$

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:

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

$\Box$

$\blacksquare$


Sources

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