# Equivalence of Definitions of Norm of Linear Transformation/Definition 1 Greater or Equal Definition 3

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## 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

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

Then:

$\lambda_1 \ge \lambda_3$

## Proof

By definition of the supremum:

$\forall h \in H, \norm h_H \le 1 : \norm{A h}_K \le \lambda_1$

In particular:

$\forall h \in H, \norm h_H = 1 : \norm{A h}_K \le \lambda_1$

From Continuum Property:

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

By definition of the supremum:

$\lambda_3 \le \lambda_1$

$\blacksquare$