Space of Zero-Limit Sequences with Supremum Norm forms Banach Space
Theorem
Let $c_0$ be the space of zero-limit sequences.
Let $\norm {\, \cdot \,}_\infty$ be the supremum norm.
Then $\struct {c_0, \norm {\, \cdot \,}_\infty}$ is a Banach space.
Proof
Let $\sequence {a_n}_{n \mathop \in \N}$ be a Cauchy sequence in $\struct {c_0, \norm {\, \cdot \,}_\infty}$.
Let $\struct {\ell^\infty, \norm {\, \cdot \,}_\infty}$ be the normed vector space of bounded sequences.
By Space of Zero-Limit Sequences with Supremum Norm forms Normed Vector Space, $\struct {c_0, \norm {\, \cdot \,}_\infty}$ is a subspace of $\struct {\ell^\infty, \norm {\, \cdot \,}_\infty}$.
Hence, $\sequence {a_n}_{n \mathop \in \N}$ is also a Cauchy sequence in $\struct {\ell^\infty, \norm {\, \cdot \,}_\infty}$.
By Space of Bounded Sequences with Supremum Norm forms Banach Space, $\sequence {a_n}_{n \mathop \in \N}$ converges to $a \in \ell^\infty$.
Denote $a_n = \sequence {a_n^{\paren m}}_{m \mathop \in \N}$ and $a = \sequence {a^{\paren m}}_{m \mathop \in \N}$.
By definition of convergent sequences:
- $\forall \epsilon \in \R_{> 0} : \exists N \in \N : \forall n \in \N : n > N \implies \norm {a_n - a}_\infty < \epsilon$
Then:
\(\ds \forall m \in \N: \, \) | \(\ds \size {a_n^{\paren m} - a^{\paren m} }\) | \(\le\) | \(\ds \max_{m \mathop \in \N} \size {a_n^{\paren m} - a^{\paren m} }\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \norm {a_n - a}_\infty\) | Definition of Supremum Norm | |||||||||||
\(\ds \) | \(<\) | \(\ds \epsilon\) |
Since $a_n \in c_0$:
- $\forall \epsilon \in \R_{\mathop > 0} : \exists M \in \R_{\mathop > 0} : \forall m \in \N : m > M \implies \size {a_n^{\paren m}} < \epsilon$
For all $m > M$ we also have that:
\(\ds \size {a^{\paren m} }\) | \(\le\) | \(\ds \size {a^{\paren m} - a_n^{\paren m} + a_n^{\paren m} }\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \size {a^{\paren m} - a_n^{\paren m} } + \size {a_n^{\paren m} }\) | Norm Axiom $\text N 3$: Triangle Inequality | |||||||||||
\(\ds \) | \(<\) | \(\ds \epsilon + \epsilon\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 \epsilon\) |
In other words:
- $\forall \epsilon' \in \R_{>0} : \exists M' \in \R_{>0} : \forall m \in \N : m > M' \implies \size {a^{\paren m} } < \epsilon'$
where $\epsilon' = 2\epsilon$ and $M' = M$.
By definition of zero-limit sequences, $a \in c_0$.
Therefore, in $\struct {c_0, \norm {\, \cdot \,}_\infty}$ a Cauchy sequence is also convergent in $\struct {c_0, \norm {\, \cdot \,}_\infty}$.
By definition, $\struct {c_0, \norm {\, \cdot \,}_\infty}$ is a Banach space.
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces