The ℓ1 Sequences Normed Linear Space

# The ℓ1 Sequences Normed Linear Space

 Definition: The $\ell^1$ Sequence Space is defined to be the set $\displaystyle{\ell^1 = \left \{ (a_i)_{i=1}^{\infty} : \sum_{i=1}^{\infty} |a_i| < \infty \right \}}$ with the norm $\| \cdot \|_1 : \ell_1 \to [0, \infty)$ defined for all $(a_i)_{i=1}^{\infty} \in \ell^1$ by $\displaystyle{\| (a_i)_{i=1}^{\infty} \|_1 = \sum_{i=1}^{\infty} |a_i|}$.

So $\ell^1$ consists of all sequences of real (or complex) numbers whose sum is absolutely convergent, and the norm of the sequence is defined to be the absolutely convergent sum. For example, the sequence $\left ( \frac{1}{2^i} \right ) \in \ell^1$ since:

(1)
\begin{align} \quad \sum_{i=1}^{\infty} \left | \frac{1}{2^i} \right | = \sum_{i=1}^{\infty} \frac{1}{2^i} = 1 < \infty \end{align}
 Proposition 1: $(\ell^1, \| \cdot \|_1)$ is a normed linear space.
• Proof: The set of all infinite sequences is a linear space and $\ell^1$ is a subset of that space, so to show that $\ell^1$ 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 $(a_i), (b_i) \in \ell^1$. Consider the sequence $(a_i) + (b_i) = (a_i + b_i)$. Then:
(2)
\begin{align} \quad \sum_{i=1}^{\infty} |a_i + b_i| \leq \sum_{i=1}^{\infty} [|a_i| + |b_i|] \overset{(*)}= \sum_{i=1}^{\infty} |a_i| + \sum_{i=1}^{\infty} |b_i| < \infty \end{align}
• The equality at $(*)$ is not true in general, but since the series $\sum_{i=1}^{\infty} |a_i|$ and $\sum_{i=1}^{\infty} |b_i|$ are known to converge, it is true. Hence [[$((a_i) + (b_i)) \in \ell^1 • Let$\alpha \in \mathbb{R}$and let$(a_i) \in \ell^1$. Consider the sequence$\alpha (a_i) = (\alpha a_i). Then: (3) \begin{align} \quad \sum_{i=1}^{\infty} |\alpha a_i| = \sum_{i=1}^{\infty} |\alpha| |a_i| = |\alpha| \sum_{i=1}^{\infty} |a_i| < \infty \end{align} • Hence\alpha (a_i) \in \ell^1$. • Lastly, the zero sequence$(0) = (0, 0, ... )$is clearly such that$\sum_{i=1}^{\infty} 0 = 0 < \infty$, so$(0) \in \ell^1$. Hence$\ell^1$is a linear space. • All that remains to show is that$\| \cdot \|_1$is a norm on$\ell^1$. • Showing that$\| (a_i) \|_1 = 0$if and only if$(a_i) = (0)$: Suppose that$\| (a_i) \|_1 = 0$. Then$\sum_{i=1}^{\infty} |a_i| = 0$. A sum of nonnegative terms equals zero if and only if each term in the sum is zero. Hence$(a_i) = (0)$. Conversely, suppose that$(a_i) = (0)$. Then$\| (a_i) \|_1 = \| (0) \|_1 = \sum_{i=1}^{\infty} |0| = 0$. • Showing that$\| \alpha (a_i) \|_1 = |\alpha| \| (a_i) \|_1$: Let$\alpha \in \mathbb{R}$and let$(a_i) \in \ell^1. Then: (4) \begin{align} \quad \| \alpha (a_i) \|_1 = \| (\alpha a_i) \|_1 = \sum_{i=1}^{\infty} |\alpha a_i| = \sum_{i=1}^{\infty} |\alpha||a_i| = |\alpha| \sum_{i=1}^{\infty} |a_i| = |\alpha| \| (a_i) \|_1 \end{align} • Showing that\| (a_i) + (b_i) \|_1 \leq \| (a_i) \|_1 + \| (b_i) \|_1$: Let$(a_i), (b_i) \in \ell^1. Then: (5) \begin{align} \quad \| (a_i) + (b_i) \|_1 = \| (a_i + b_i) \|_1 = \sum_{i=1}^{\infty} |a_i + b_i| \leq \sum_{i=1}^{\infty} [|a_i| + |b_i|] = \sum_{i=1}^{\infty} |a_i| + \sum_{i=1}^{\infty} |b_i| = \| (a_i) \|_1 + \| (b_i) \|_1 \end{align} • Therefore(\ell^1, \| \cdot \|_1)$is a normed linear space.$\blacksquare\$