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:
- $\norm {T x}_Y \le \norm T \norm x_X$
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 2.3$: The normed space $\map {CL} {X,Y}$. Operator norm and the normed space $\map {CL} {X, Y}$