Supremum Operator Norm of Cesàro Summation Operator
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Theorem
Let $A : \ell^\infty \to \ell^\infty$ be the Cesàro summation operator.
Let $\norm {\, \cdot \,}$ be the supremum operator norm.
Then $\norm A = 1$.
Proof
By Cesàro Summation Operator is Continuous Linear Transformation:
- $\norm {A \mathbf x}_\infty \le \norm {\mathbf x}_\infty$
where $\norm {\, \cdot \,}_\infty$ denotes the supremum norm.
By Universal Upper Bound greater than Supremum Operator Norm:
- $\norm A \le 1$
Let $\mathbf 1 := \tuple {1, 1, \ldots} \in \ell^\infty$
Then:
\(\ds \norm A\) | \(=\) | \(\ds \norm A \cdot 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm A \norm {\mathbf 1}_\infty\) | Definition of Supremum Norm | |||||||||||
\(\ds \) | \(\ge\) | \(\ds \norm {A \mathbf 1}_\infty\) | Supremum Operator Norm as Universal Upper Bound | |||||||||||
\(\ds \) | \(=\) | \(\ds \norm {\tuple {1, 1, \ldots} }_\infty\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1\) |
Altogether, $\norm A = 1$.
$\blacksquare$
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 $\map {CL} {X, Y}$