Equivalence of Definitions of Norm of Linear Transformation

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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 $A$, denoted $\norm A$, is the real number defined by:

$\norm A = \sup \set {\norm {A h}_K: \norm h_H \le 1}$


Definition 2

The norm of $A$, denoted $\norm A$, is the real number defined by:

$\norm A = \sup \set {\dfrac {\norm {A h}_K} {\norm h_H}: h \in H, h \ne \mathbf 0_H}$


Definition 3

The norm of $A$, denoted $\norm A$, is the real number defined by:

$\norm A = \sup \set {\norm {A h}_K: \norm h_H = 1}$


Definition 4

The norm of $A$, denoted $\norm A$, is the real number defined by:

$\norm A = \inf \set {c > 0: \forall h \in H: \norm {A h}_K \le c \norm h_H}$


Proof

Let:

$\lambda_1 = \sup \set {\norm {A h}_K: \norm h_H \le 1}$
$\lambda_2 = \sup \set {\dfrac {\norm {A h}_K} {\norm h_H}: h \in H, h \ne 0_H}$
$\lambda_3 = \sup \set {\norm {A h}_K: \norm h_H = 1}$
$\lambda_4 = \inf \set {c > 0: \forall h \in H: \norm {A h}_K \le c \norm h_H}$


From Operator Norm 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 Submultiplicity of Operator Norm:

$\forall h \in H : \norm{A h}_K \le \lambda_4 \norm h_H$

Hence:

$\forall h \in H, h \ne 0_H : \dfrac {\norm{A h}_K} {\norm h_H} \le \lambda_4$

From Continuum Property:

$\lambda_2 = \sup \set {\dfrac {\norm {A h}_K} {\norm h_H}: h \in H, h \ne 0_H}$ 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: $\lambda_1 \ge \lambda_4$

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.

$\blacksquare$

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