Equivalence of Definitions of Norm of Linear Transformation/Definition 3 Greater or Equal Definition 4
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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:
- $\lambda_4 \le \lambda_3$
$\blacksquare$