Operator Norm is Finite

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Theorem

Let $H, K$ be Hilbert spaces, and let $A: H \to K$ be a bounded linear transformation.

Let $\norm A$ denote the norm of $A$ defined by:

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


Then:

$\norm A < \infty$


Proof

By definition of a bounded linear transformation:

$\exists c \in \R_{> 0}: \forall h \in H: \norm{A h}_K \le c \norm h_H$

Hence:

$\set {\lambda > 0: \forall h \in H: \norm {A h}_K \le \lambda \norm h_H} \ne \O$


By definition:

$\set {\lambda > 0: \forall h \in H: \norm {A h}_K \le \lambda \norm h_H}$ is bounded below.


From Corollary to Continuum Property:

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


We have:

\(\ds \norm A\) \(\le\) \(\ds c\) Definition of Infimum
\(\ds \) \(<\) \(\ds \infty\) As $c \in \R_{> 0}$

The result follows.

$\blacksquare$