Equivalence of Definitions of Norm of Linear Transformation

From ProofWiki
Jump to navigation Jump to search

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

Retrieved from "https://proofwiki.org/w/index.php?title=Equivalence_of_Definitions_of_Norm_of_Linear_Transformation&oldid=592989"