The ℓ∞ Sequences Normed Linear Space

# The ℓ∞ Sequences Normed Linear Space

 Definition: The $\ell^{\infty}$ Sequence Space is defined to be the set $\displaystyle{\ell^{\infty} = \left \{ (a_i)_{i=1}^{\infty} : (a_i)_{i=1}^{\infty} \: \mathrm{is \: bounded} \right \}}$ with the norm $\| \cdot \|_{\infty} : \ell_{\infty} \to [0, \infty)$ defined for all $(a_i)_{i=1}^{\infty} \in \ell^1$ by $\displaystyle{\| (a_i)_{i=1}^{\infty} \|_1 = \sup_{i \geq 1} \{ |a_i| \}}$.

So $\ell^{\infty}$ consists of all sequences of real (or complex) numbers that are bounded, and the norm of the sequence is defined to be the supremum of the absolute value of the terms in the sum.

 Proposition 1: $(\ell^{\infty}, \| \cdot \|_{\infty})$ is a normed linear space.
• Proof: Again the set of all infinite sequences is a linear space and $\ell^{\infty}$ is a subset of that space, so to show that $\ell^{\infty}$ is a linear space we only need to show that it is closed under addition, closed under scalar multiplication, and contains the zero sequence.
• Let $(a_i), (b_i) \in \ell^{\infty}$. Consider the sequence $(a_i) + (b_i) = (a_i + b_i)$. Since $(a_i) \in \ell^{\infty}$ there exists an $M > 0$ such that $|a_i| \leq M$ for all $i \in \mathbb{N}$. Since $(b_i) \in \ell^{\infty}$ there exists an $N > 0$ such that $|b_i| \leq N$ for all $i \in \mathbb{N}$. Hence, for all $i \in \mathbb{N}$:
(1)
\begin{align} \quad |a_i + b_i| \leq |a_i| + |b_i| \leq M + N \end{align}
• So $(a_i + b_i)$ is bounded. Hence $[(a_i) + (b_i)] \in \ell^{\infty}$.
• Let $\alpha \in \mathbb{R}$ and let $(a_i) \in \ell^{\infty}$. Consider the sequence $\alpha (a_i) = (\alpha a_i)$. Since $(a_i) \in \ell^{\infty}$ there exists an $M > 0$ such that $|a_i| \leq M$ for all $i \in \mathbb{N}$. Hence for all $i \in \mathbb{N}$:
(2)
\begin{align} \quad |\alpha a_i| = |\alpha||a_i| \leq |\alpha|M \end{align}
• So $(\alpha a_i)$ is bounded. Hence $\alpha (a_i) \in \ell^{\infty}$.
• Lastly, we clearly have that the zero sequence $(0)$ is bounded, so $(0) \in \ell^{\infty}$.
• All that remains to show is that $\| \cdot \|_{\infty}$ is a norm on $\ell^{\infty}$.
• Showing that $\| (a_i) \|_{\infty} = 0$ if and only if $(a_i) = (0)$: Suppose that $\| (a_i) \|_{\infty} = 0$. Then $\sup_{i \geq 1} \{ |a_i| \} = 0$. So $|a_i| \leq 0$ for each $i \in \mathbb{N}$. Hence $a_i = 0$ for each $i \in \mathbb{N}$ implying that $(a_i) = (0)$. 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 $(a_i) \in \ell^{\infty}$. Then:
(3)
\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 $(a_i), (b_i) \in \ell^{\infty}$. Then:
(4)
\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 $(\ell^{\infty}, \| \cdot \|_{\infty})$ is a normed linear space. $\blacksquare$