Convergent Sequences form Invariant Subspace of Bounded Sequences wrt Cesàro Summation Operator

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Theorem

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

Let $c$ be the space of convergent sequences.

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


Then $c$ is an invariant subspace of $\ell^\infty$ with respect to $A$.


Proof

Let $\sequence {x_n}_{n \mathop \in \N} \in c$ be a sequence.

By definition, $\sequence {x_n}_{n \mathop \in \N}$ converges.

Let $\ds L = \lim_{n \mathop \to \infty} x_n$ be the limit of $\sequence {x_n}_{n \mathop \in \N}$.

Then:

$\forall \epsilon \in \R_{> 0} : \exists N_1 \in \N : \forall n \in \N : n > N_1 \implies \size {x_n - L} < \epsilon$

Furthermore:

$\exists M \in \R_{> 0} : \forall n \in \N : \size {x_n} < M$

$A \mathbf x$ converges to $L$

\(\ds \size{\frac 1 n \paren {\sum_{i \mathop = 1}^{n} x_i} - L}\) \(=\) \(\ds \frac 1 n \size{\sum_{i \mathop = 1}^{N_1} x_i + \sum_{j \mathop = N_1 + 1}^n x_i - n L}\)
\(\ds \) \(\le\) \(\ds \frac 1 n \paren{\sum_{i \mathop = 1}^{N_1} \size {x_i - L} + \sum_{j \mathop = N_1 + 1}^n \size {x_j - L} }\) Triangle Inequality for Real Numbers
\(\ds \) \(<\) \(\ds \frac 1 n \sum_{i \mathop = 1}^{N_1} \paren {\size {x_i} + \size L} + \frac {n - N_1} n \epsilon\) Triangle Inequality for Real Numbers
\(\ds \) \(\le\) \(\ds \frac {N_1} n \paren{M + \size L} + \paren {1 - \frac {N_1} n} \epsilon\)

Let $\epsilon' \in \R_{> 0 }$.

Let $n \in \N$ be such that:

$\ds \frac {N_1} n \paren{M + \size L} + \paren {1 - \frac {N_1} n} \epsilon < \epsilon'$

Suppose $n > N_1$.

Then:

\(\ds \frac {N_1} n \paren{M + \size L} + \paren {1 - \frac {N_1} n} \epsilon\) \(<\) \(\ds \frac {N_1} n \paren{M + \size L} + \epsilon\)
\(\ds \) \(<\) \(\ds \epsilon'\)

Suppose also that $\ds n > N_1 \frac {M + \size L}{\epsilon' - \epsilon}$.

Then:

\(\ds \frac {N_1} n \paren{M + \size L} + \paren {1 - \frac {N_1} n} \epsilon\) \(<\) \(\ds \frac {N_1} n \paren{M + \size L} + \epsilon\)
\(\ds \) \(<\) \(\ds \paren{\epsilon' - \epsilon} + \epsilon\)
\(\ds \) \(=\) \(\ds \epsilon'\)

Altogether, $n$ has to satisfy $n > \tilde N$ where:

$\ds \tilde N = \max \set {N_1, N_1 \frac {M + \size L}{\epsilon' - \epsilon} }$.

$\epsilon'$ was arbitrary.

Hence:

$\ds \forall \epsilon' \in \R{>0} : \exists \tilde N : \forall n \in \N : n > \tilde N \implies \size {A \mathbf x - L} < \epsilon'$

Therefore, $A \mathbf x \in c$.

$\Box$

Thus, $A c \subseteq c$.

By definition, $c$ is an invariant subspace of $\ell^\infty$ with respect to $A$.

$\blacksquare$

Sources

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