# Fundamental Property of Norm on Bounded Linear Transformation

## Theorem

Let $\HH, \KK$ be Hilbert spaces.

Let $A: \HH \to \KK$ be a bounded linear transformation.

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

$\norm A = \inf \set {c > 0: \forall h \in \HH: \norm {A h}_\KK \le c \norm h_\HH}$

Then:

$\forall h \in \HH: \norm {A h}_\KK \le \norm A \norm h_\HH$

## Proof

$\norm A = \inf \set {c > 0: \forall h \in \HH: \norm {A h}_\KK \le c \norm h_\HH}$ exists

and

$\norm A < \infty$

Let $x \in \HH \setminus \set{0_\HH}$

Let $\lambda \in \set {c > 0: \forall h \in \HH: \norm {A h}_\KK \le c \norm h_\HH}$.

Then:

 $\ds \norm {A x}_K$ $\le$ $\ds \lambda \norm x_\HH$ $\ds \leadstoandfrom \ \$ $\ds \dfrac {\norm {A x}_\KK} {\norm x_\HH}$ $\le$ $\ds \lambda$

As $c$ was arbitrary, then:

$\forall \lambda \in \set {c > 0: \forall h \in \HH: \norm {A h}_\KK \le c \norm h_\HH}: \dfrac {\norm {A x}_\KK} {\norm x_\HH} \le \lambda$

By the definition of the infimum:

$\dfrac {\norm {A x}_\KK} {\norm x_\HH} \le \norm A$

Hence:

$\norm {A x}_\KK \le \norm A \norm x_\HH$

Since $x$ was arbitrary:

$\forall h \in \HH \setminus \set {0_\HH}: \norm {A h}_\KK \le \norm A \norm h_\HH$

Lastly, we have:

 $\ds \norm {A 0_\HH}_\KK$ $=$ $\ds \norm {0_\KK}_\KK$ $\ds$ $=$ $\ds 0$ $\ds$ $=$ $\ds \norm A \cdot 0$ $\ds$ $=$ $\ds \norm A \norm {0_\HH}$

It follows that:

$\forall h \in \HH: \norm {A h}_\KK \le \norm A \norm h_\HH$

$\blacksquare$