# Equivalence of Definitions of Norm of Linear Transformation/Definition 2 Greater or Equal Definition 1

## Theorem

Let $H, K$ be Hilbert spaces.

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

Let:

$\lambda_1 = \sup \set {\norm {A h}_K: \norm h_H \le 1}$

and

$\lambda_2 = \sup \set {\dfrac {\norm {A h}_K} {\norm h_H}: h \in H, h \ne 0_H}$

Then:

$\lambda_2 \ge \lambda_1$

## Proof

### Lemma

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

$\Box$

By definition of the supremum:

$\forall h \in H, h \ne \mathbf 0_H: \dfrac {\norm {A h}_K} {\norm h_H} \le \lambda_2$

Hence:

$\forall h \in H, h \ne \mathbf 0_H: \norm {A h}_K \le \lambda_2 \norm h_H$

From Lemma:

$\norm{A 0_H}_K = \lambda_2 \norm{0_H}_H$

Hence:

$\forall h \in H: \norm {A h}_K \le \lambda_2 \norm h_H$

In particular:

 $\ds \forall h \in H, \norm h_H \le 1 : \ \$ $\ds \norm {A h}_K$ $\le$ $\ds \lambda_2 \norm h_H$ $\ds$ $\le$ $\ds \lambda_2$

From Continuum Property:

$\lambda_1 = \sup \set {\norm {A h}_K: \norm h_H \le 1}$ exists

By definition of the supremum:

$\lambda_1 \le \lambda_2$

$\blacksquare$