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
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 2.3$: The normed space $\map {CL} {X,Y}$. Operator norm and the normed space $\map {CL} {X, Y}$