Fundamental Property of Norm on Bounded Linear Transformation

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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

From Norm on Bounded Linear Transformation is Finite:

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

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

$\blacksquare$


Sources


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