Fundamental Property of Norm on Bounded Linear Transformation

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 Leigh.Samphier/Sandbox/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:

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:

It follows that:
 * $\forall h \in H: \norm {Ah}_K \le \norm A \norm{h}_H$