# Submultiplicity of Operator Norm

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## 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

From Operator Norm is Finite:

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

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

$\blacksquare$

## Sources

- 1990: John B. Conway:
*A Course in Functional Analysis*... (previous) ... (next) $\S II.1$