# Equivalence of Definitions of Norm of Linear Transformation/Definition 3 Greater or Equal Definition 4

## Theorem

Let $H, K$ be Hilbert spaces.

Let $A: H \to K$ be a bounded linear transformation.

Let:

$\lambda_3 = \sup \set {\norm {A h}_K: \norm h_H = 1}$

and

$\lambda_4 = \inf \set {c > 0: \forall h \in H: \norm {A h}_K \le c \norm h_H}$

Let:

$\lambda_3 \ge \lambda_4$

## Proof

### Lemma

$\forall \lambda > 0 : \norm{A 0_H}_K = \lambda \norm{0_H}_H$

$\Box$

Let $h \in H: h \ne 0_h$.

We have:

 $\ds \norm{\dfrac 1 {\norm h_H} h }_H$ $=$ $\ds \dfrac {\norm h_H}{\norm h_H}$ Norm Axiom $\text N 2$: Multiplicativity $\ds$ $=$ $\ds 1$

and

 $\ds \dfrac {\norm{A h}_K} {\norm h_H}$ $=$ $\ds \norm{\dfrac 1 {\norm h_H} A h}_K$ Norm Axiom $\text N 2$: Multiplicativity $\ds$ $=$ $\ds \norm{A \paren{ \dfrac 1 {\norm h_H} h } }_K$ Linear Transformation Maps Zero Vector to Zero Vector $\ds$ $\le$ $\ds \lambda_3$ Definition of Supremum of Set

Hence:

$\forall h \in H: h \ne 0_h: \norm{A h}_K \le \lambda_3 \norm h_H$

From Lemma:

$\norm{A 0_H}_K = \lambda_3 \norm {0_H}_H$

Hence:

$\forall h \in H: \norm{A h}_K \le \lambda_3 \norm h_H$

That is,

$\lambda_3 \in \set {c > 0: \forall h \in H: \norm {A h}_K \le c \norm h_H}$

By definition of the infimum:

$\lambda_4 \le \lambda_3$

$\blacksquare$