Submultiplicity of Operator Norm

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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:

$\forall h \in H: \norm {Ah}_K \le \norm A \norm{h}_H$

Proof

From Operator Norm is Finite:

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

and

$\norm A < \infty$


Let $x \in H \setminus \set{0_H}$

Let $\lambda \in \set {c > 0: \forall h \in H: \norm {A h}_K \le c \norm h_H}$.

Then:

\(\ds \norm {A x}_K\) \(\le\) \(\ds \lambda \norm x_H\)
\(\ds \leadstoandfrom \ \ \) \(\ds \dfrac {\norm{A x}_K} {\norm x_H}\) \(\le\) \(\ds \lambda\)


As $c$ was arbitrary, then:

$\forall \lambda \in \set {c > 0: \forall h \in H: \norm {A h}_K \le c \norm h_H}: \dfrac {\norm{A x}_K} {\norm x_H} \le \lambda$

By the definition of the infimum:

$\dfrac {\norm{A x}_K} {\norm x_H} \le \norm A$

Hence:

$\norm{A x}_K \le \norm A \norm x_H$


Since $x$ was arbitrary:

$\forall h \in H \setminus \set{0_H}: \norm{A h}_K \le \norm A \norm h_H$


Lastly, we have:

\(\ds \norm{A 0_H}_K\) \(=\) \(\ds \norm{0_K}_K\)
\(\ds \) \(=\) \(\ds 0\)
\(\ds \) \(=\) \(\ds \norm A \cdot 0\)
\(\ds \) \(=\) \(\ds \norm A \norm{0_H}\)

It follows that:

$\forall h \in H: \norm {Ah}_K \le \norm A \norm{h}_H$

$\blacksquare$

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