Equivalence of Definitions of Norm of Linear Transformation
Theorem
Let $H, K$ be Hilbert spaces.
Let $A: H \to K$ be a bounded linear transformation.
The following definitions of the concept of Norm on Bounded Linear Transformation are equivalent:
Definition 1
The norm of $A$ is the real number defined and denoted as:
- $\norm A = \sup \set {\norm {A x}_Y : \norm x_X \le 1}$
Definition 2
The norm of $A$ is the real number defined and denoted as:
- $\norm A = \sup \set {\dfrac {\norm {A x}_Y} {\norm x_X}: x \in X, x \ne \mathbf 0_X}$
This supremum is to be taken in $\closedint 0 \infty$ so that $\sup \O = 0$.
Definition 3
The norm of $A$ is the real number defined and denoted as:
- $\norm A = \sup \set {\norm {A x}_X : \norm x_X = 1}$
This supremum is to be taken in $\closedint 0 \infty$ so that $\sup \O = 0$.
Definition 4
The norm of $A$ is the real number defined and denoted as:
- $\norm A = \inf \set {c > 0: \forall x \in X: \norm {A x}_Y \le c \norm x_X}$
Proof
Let:
\(\ds \lambda_1\) | \(=\) | \(\ds \sup \set {\norm {A h}_K: \norm h_H \le 1}\) | ||||||||||||
\(\ds \lambda_2\) | \(=\) | \(\ds \sup \set {\dfrac {\norm {A h}_K} {\norm h_H}: h \in H, h \ne 0_H}\) | ||||||||||||
\(\ds \lambda_3\) | \(=\) | \(\ds \sup \set {\norm {A h}_K: \norm h_H = 1}\) | ||||||||||||
\(\ds \lambda_4\) | \(=\) | \(\ds \inf \set {c > 0: \forall h \in H: \norm {A h}_K \le c \norm h_H}\) |
From Norm on Bounded Linear Transformation is Finite:
- $\lambda_4 < \infty$
We will show that:
- $\lambda_4 \ge \lambda_2 \ge \lambda_1 \ge \lambda_3 \ge \lambda_4$
Lemma
- $\forall \lambda > 0 : \norm{A 0_H}_K = \lambda \norm{0_H}_H$
$\Box$
Inequality: $\lambda_4 \ge \lambda_2$
From Fundamental Property of Norm on Bounded Linear Transformation:
- $\forall h \in \HH : \norm{A h}_\KK \le \lambda_4 \norm h_\HH$
Hence:
- $\forall h \in H, h \ne 0_\HH : \dfrac {\norm{A h}_\KK} {\norm h_\HH} \le \lambda_4$
From Continuum Property:
- $\lambda_2 = \sup \set {\dfrac {\norm {A h}_\KK} {\norm h_\HH}: h \in \HH, h \ne 0_\HH}$ exists
By definition of the supremum:
- $\lambda_2 \le \lambda_4$
$\Box$
Inequality: $\lambda_2 \ge \lambda_1$
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:
- $\lambda_1 \le \lambda_2$
$\Box$
Inequality: $\lambda_1 \ge \lambda_3$
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:
- $\lambda_3 \le \lambda_1$
$\Box$
Inequality: $\lambda_3 \ge \lambda_4$
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:
- $\lambda_4 \le \lambda_3$
$\Box$
It follows that the definitions are all equivalent.
$\blacksquare$
Sources
- 1990: John B. Conway: A Course in Functional Analysis (2nd ed.) ... (previous) ... (next) $\S \text {II}.1$