Supremum Operator Norm as Universal Upper Bound

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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:

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

$\blacksquare$


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