# 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 the supremum operator norm on $\map {CL} {X, Y}$.

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:

 $\ds \norm {T x}_Y$ $=$ $\ds \norm {T \mathbf 0}_Y$ $\ds$ $=$ $\ds \norm {\mathbf 0}_Y$ $\ds$ $=$ $\ds 0$ $\ds$ $=$ $\ds \norm T 0$ $\ds$ $=$ $\ds \norm T \norm {\mathbf 0}_X$ $\ds$ $=$ $\ds \norm T \norm x_X$

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:

 $\ds \frac {\norm {Tx}_Y }{\norm x_X}$ $=$ $\ds \norm {\frac {Tx}{\norm x_X} }_Y$ Definition of Norm on Vector Space $\ds$ $=$ $\ds \norm {\map T {\frac x {\norm x_X} } }_Y$ Definition of Linear Transformation on Vector Space $\ds$ $=$ $\ds \norm {T \hat x}_Y$ $\ds$ $\le$ $\ds \norm T$

Rearranging, we get:

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

$\blacksquare$