Equivalence of Definitions of Norm of Linear Transformation/Definition 2 Greater or Equal Definition 1
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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
- $\lambda_2 = \sup \set {\dfrac {\norm {A h}_K} {\norm h_H}: h \in H, h \ne 0_H}$
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:
- $\lambda_1 \le \lambda_2$
$\blacksquare$