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:
- $\lambda_3 \le \lambda_1$
$\blacksquare$