The c0 Sequences Normed Linear Space

Definition: The $c_0$ Sequence Space is the set $\displaystyle{c_0 = \left \{ (a_i)_{i=1}^{\infty} : \lim_{i \to \infty} a_i = 0 \right \}}$ with the norm $\| \cdot \|_{\infty}$ defined for all $(a_i) \in c_0$ by $\displaystyle{c_0 = \sup_{i \geq 1} |a_i|}$.

So $c_0$ consists of all sequences $(a_i)$ which converge to $0$, and the $\infty$-norm of $(a_i)$ is the supremum of all such $(a_i)$.

For example, the sequence $\left ( i^{-1}\right ) \in c_0$ since:

(1)

\begin{align} \quad \lim_{i \to \infty} \frac{1}{i} = 0 \end{align}

We have that $\| (i^{-1}) \|_{\infty} = 1$.

Proposition 1: $(c_0, \| \cdot \|_{\infty})$ is a normed linear space.

• Proof: Again, since $c_0$ is a subset of the set of all infinite sequences (which is a linear space), all we need to show is that $c_0$ is closed under addition, closed under scalar multiplication, and contains the zero sequence $(0)$ to show that it is a linear space.

• Let $(a_i), (b_i) \in c_0$. Then $\lim_{i \to \infty} a_i = 0$ and $\lim_{i \to \infty} b_i = 0$. So:

(2)

\begin{align} \quad \lim_{i \to \infty} (a_i + b_i) = \lim_{i \to \infty} a_i + \lim_{i \to \infty} b_i = 0 + 0 = 0 \end{align}

• Thus $[(a_i)+ (b_i)] \in c_0$.

• Let $\alpha \in \mathbb{R}$ and let $(a_i) \in c_0$. Then $\lim_{i \to \infty} a_i = 0$. So:

(3)

\begin{align} \quad \lim_{i \to \infty} \alpha a_i = \alpha \lim_{i \to \infty} a_i = \alpha \cdot 0 = 0 \end{align}

• Thus $\alpha a_i \in c_0$.

• Lastly, $(0) \in c_0$ since $\lim_{i \to \infty} 0 = 0$. Hence $c_0$ is a linear space.

• All that remains to show is that $\| \cdot \|_{\infty}$ is a norm on $c_0$.

• Showing that $\| (a_i) \|_{\infty} = 0$ if and only if $(a_i) = (0)$: Suppose that $\| (a_i) \|_{\infty} = 0$. Then $\sup_{i \geq 1} \{ |a_i| \} = 0$ which implies that $|a_i| \leq 0$ for each $i \in \mathbb{N}$. But then $a_i = 0$ for each $i \in \mathbb{N}$. Thus $(a_i) = (0)$. Conversely, suppose that $(a_i) = (0)$. Then $\| (a_i) \|_{\infty} \| (0) \|_{\infty} = \sup_{i \geq 1} \{ |0| \} = 0$.

• Showing that $\| \alpha (a_i) \|_{\infty} = |\alpha| \| a_i \|_{\infty}$: Let $\alpha \in \mathbb{R}$ and let $(a_i) \in c_0$. Then:

(4)

\begin{align} \quad \| \alpha (a_i) \|_{\infty} = \| (\alpha a_i) \|_{\infty} = \sup_{i \geq 1} \{ | \alpha a_i | \} = \sup_{i \geq 1} \{ |\alpha| |a_i| \} = |\alpha| \sup_{i \geq 1} \{ |a_i| \} = |\alpha| \| (a_i) \|_{\infty} \end{align}

• Showing that $\| (a_i) + (b_i) \|_{\infty} \leq \| (a_i) \|_{\infty} + \| (b_i) \|_{\infty}$: Let $(a_i), (b_i) \in c_0$. Then:

(5)

\begin{align} \quad \| (a_i) + (b_i) \|_{\infty} = \| (a_i + b_i) \|_{\infty} = \sup_{i \geq 1} \{ |a_i + b_i| \} \leq \sup_{i \geq 1} \{ |a_i| + |b_i| \} = \sup_{i \geq 1} \{ |a_i| \} + \sup_{i \geq 1} \{ |b_i| \} = \| (a_i) \|_{\infty} + \| (b_i) \|_{\infty} \end{align}

• Therefore $\| \cdot \|_{\infty}$ is a norm on $c_0$ and so $(c_0, \| \cdot \|_{\infty})$ is a normed linear space. $\blacksquare$