Supremum Operator Norm on Continuous Linear Transformation Space is Submultiplicative
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Theorem
Let $\struct {X, \norm {\, \cdot \,}_X}$ and $\struct {Y, \norm {\, \cdot \,}_Y}$, $\struct {Z, \norm {\, \cdot \,}_Z}$ be normed vector spaces.
Let $A : Y \to Z$ and $B : X \to Y$ be continuous linear transformations.
Let $\norm {\, \cdot \,}$ be the supremum operator norm.
Let $\circ$ denote the composition.
Then $\norm {\, \cdot \,}$ is submultiplicative:
- $\norm {A \circ B} \le \norm A \cdot \norm B$
Proof
\(\ds \norm {A \circ B}\) | \(=\) | \(\ds \sup_{x \mathop \in X \mathop : \norm x_X \mathop \le 1} \norm {\map {\paren {A \circ B} } x}_Z\) | Definition of Supremum Operator Norm | |||||||||||
\(\ds \) | \(\le\) | \(\ds \sup_{x \mathop \in X \mathop : \norm x_X \mathop \le 1} \norm A \cdot \norm {\map B x}_Y\) | Supremum Operator Norm as Universal Upper Bound | |||||||||||
\(\ds \) | \(=\) | \(\ds \norm A \cdot \sup_{x \mathop \in X \mathop : \norm x_X \mathop \le 1} \norm {\map B x}_Y\) | $\norm A$ is independent of $x$ | |||||||||||
\(\ds \) | \(\le\) | \(\ds \norm A \cdot \sup_{x \mathop \in X \mathop : \norm x_X \mathop \le 1} \norm B \cdot \norm x_X\) | Supremum Operator Norm as Universal Upper Bound | |||||||||||
\(\ds \) | \(=\) | \(\ds \norm A \cdot \norm B \cdot \sup_{x \mathop \in X \mathop : \norm x_X \mathop \le 1} \norm x_X\) | $\norm B$ is independent of $x$ | |||||||||||
\(\ds \) | \(\le\) | \(\ds \norm A \cdot \norm B \cdot 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm A \cdot \norm B\) |
$\blacksquare$
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