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, $\norm A = 1$.

$\blacksquare$


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