Supremum Operator Norm as Universal Upper Bound

Theorem
Let $\map {CL} {X, Y}$ be the continuous linear transformation space.

Let $T \in \map {CL} {X, Y}$.

Let $\norm {\, \cdot \,}$ be an operator norm on $\map {CL} {X, Y}$ defined as:


 * $\norm T := \sup \set {\norm {Tx}_Y : x \in X : \norm x_X \le 1}$

Then:


 * $\forall x \in X : \norm {Tx}_Y \le \norm T \norm x_X$

Proof
Let $S = \set {\norm {Tx}_Y : x \in X : \norm x_X \le 1}$.

Suppose $x = \mathbf 0$.

Then:

Suppose $x \ne \mathbf 0$.

Let $\hat x \in X$ such that:


 * $\ds \hat x = \frac x {\norm x_X}$

Then:


 * $\ds \norm {\hat x}_X = \frac {\norm x_X}{\norm x_X} = 1$.

Thus:


 * $\norm {T \hat x}_Y \in S$

and

$\norm {T \hat x}_Y \le \sup S = \norm T$

Furthermore:

Rearranging, we get:


 * $\norm {T x}_Y \le \norm T \norm x_X$