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

That is, $\norm A$ is submultiplicative.

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:

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:

It follows that:
 * $\forall h \in \HH: \norm {A h}_\KK \le \norm A \norm h_\HH$