Cesàro Summation Operator is Continuous Linear Transformation

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\ell^\infty$ be the space of bounded sequences.

Let $A : \ell^\infty \to \ell^\infty$ be the Cesàro summation operator.


Then $A$ is a continuous linear transformation.


Proof

Well-Definedness

Let $\mathbf x = \sequence {x_n}_{n \mathop \in \N} \in \ell^\infty$.

Then:

\(\ds \forall n \in \N: \, \) \(\ds \size {x_n}\) \(\le\) \(\ds \sup_{n \mathop \in \N} \size {x_n}\)
\(\ds \) \(=\) \(\ds \norm {\mathbf x}_\infty\) Definition of Supremum Norm


Thus:

\(\ds \size {\sum_{i \mathop = 1}^n\frac{x_i} n }\) \(\le\) \(\ds \frac 1 n \sum_{i \mathop = 1}^n \size {x_i}\) General Triangle Inequality for Complex Numbers
\(\ds \) \(\le\) \(\ds \frac 1 n \sum_{i \mathop = 1}^n \sup_{n \mathop \in \N} \size {x_i}\)
\(\ds \) \(=\) \(\ds \frac 1 n n \norm {\mathbf x}_\infty\) Definition of Supremum Norm
\(\ds \) \(=\) \(\ds \norm {\mathbf x}_\infty\)


Hence:

$A \mathbf x \in \ell^\infty$


Further research is required in order to fill out the details.
In particular: Clarify connection to the Definition of Well-Defined Mapping
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by finding out more.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Research}} from the code.


$\Box$


Linearity

Let $\mathbf x = \sequence {x_n}_{n \mathop \in \N}, \mathbf y = \sequence {y_n}_{n \mathop \in \N} \in \ell^\infty$.

Let $\lambda \in \C$.

By Space of Bounded Sequences with Pointwise Addition and Pointwise Scalar Multiplication on Ring of Sequences forms Vector Space:

$\mathbf x + \lambda \mathbf y \in \ell^\infty$


Then:

\(\ds \map A {\mathbf x + \lambda \mathbf y}\) \(=\) \(\ds \tuple {x_1 + \lambda y_1, \frac {x_1 + x_2 + \lambda \paren {y_1 + y_2} } 2, \frac {x_1 + x_2 + x_3 + \lambda \paren {y_1 + y_2 + y_3} } 3, \ldots}\) Definition of Cesàro Summation Operator
\(\ds \) \(=\) \(\ds \tuple {x_1, \frac {x_1 + x_2} 2, \frac {x_1 + x_2 + x_3} 3, \ldots} + \tuple {\lambda y_1, \frac {\lambda \paren {y_1 + y_2} } 2, \frac {\lambda \paren {y_1 + y_2 + y_3} } 3, \ldots}\) Space of Bounded Sequences with Pointwise Addition and Pointwise Scalar Multiplication on Ring of Sequences forms Vector Space
\(\ds \) \(=\) \(\ds \tuple {x_1, \frac {x_1 + x_2} 2, \frac {x_1 + x_2 + x_3} 3, \ldots} + \lambda \tuple {y_1, \frac {y_1 + y_2} 2, \frac {y_1 + y_2 + y_3} 3, \ldots}\)
\(\ds \) \(=\) \(\ds \map A {\mathbf x} + \lambda \map A {\mathbf y}\) Definition of Cesàro Summation Operator

By definition, $A$ is a linear transformation.

$\Box$


Continuity

Let $\mathbf x = \sequence {x_n}_{n \mathop \in \N} \in \ell^\infty$.

Then:

\(\ds \norm {A \mathbf x}_\infty\) \(=\) \(\ds \sup_{n \mathop \in \N} \size {\sum_{i \mathop = 1}^n \frac {x_i} n }\) Definition of Cesàro Summation Operator
\(\ds \) \(\le\) \(\ds \norm {\mathbf x}_\infty\)

By Continuity of Linear Transformation between Normed Vector Spaces, $A$ is continuous.

$\Box$


All together:

$A \in \map {CL} {\ell^\infty}$

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Cesàro_Summation_Operator_is_Continuous_Linear_Transformation&oldid=628458"