# Submultiplicity of Operator Norm

## Theorem

Let $H, K$ be Hilbert spaces, and let $A: H \to K$ be a bounded linear transformation.

Let $\norm A$ denote the norm of $A$ defined by:

$\norm A = \inf \set {c > 0: \forall h \in H: \norm {A h}_K \le c \norm h_H}$

Then:

$\forall h \in H: \norm {Ah}_K \le \norm A \norm{h}_H$

## Proof

$\norm A = \inf \set {c > 0: \forall h \in H: \norm {A h}_K \le c \norm h_H}$ exists

and

$\norm A < \infty$

Let $x \in H \setminus \set{0_H}$

Let $\lambda \in \set {c > 0: \forall h \in H: \norm {A h}_K \le c \norm h_H}$.

Then:

 $\ds \norm {A x}_K$ $\le$ $\ds \lambda \norm x_H$ $\ds \leadstoandfrom \ \$ $\ds \dfrac {\norm{A x}_K} {\norm x_H}$ $\le$ $\ds \lambda$

As $c$ was arbitrary, then:

$\forall \lambda \in \set {c > 0: \forall h \in H: \norm {A h}_K \le c \norm h_H}: \dfrac {\norm{A x}_K} {\norm x_H} \le \lambda$

By the definition of the infimum:

$\dfrac {\norm{A x}_K} {\norm x_H} \le \norm A$

Hence:

$\norm{A x}_K \le \norm A \norm x_H$

Since $x$ was arbitrary:

$\forall h \in H \setminus \set{0_H}: \norm{A h}_K \le \norm A \norm h_H$

Lastly, we have:

 $\ds \norm{A 0_H}_K$ $=$ $\ds \norm{0_K}_K$ $\ds$ $=$ $\ds 0$ $\ds$ $=$ $\ds \norm A \cdot 0$ $\ds$ $=$ $\ds \norm A \norm{0_H}$

It follows that:

$\forall h \in H: \norm {Ah}_K \le \norm A \norm{h}_H$

$\blacksquare$