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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