The ℓp Sequences Normed Linear Space

# The ℓp Sequences Normed Linear Space

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

So $\ell^p$ consists of all sequences of real (or complex) numbers whose sum of is $p$-absolutely convergent, and the norm of the sequence is defined to be the $p^{\mathrm{th}}$ roots of the $p$-absolutely convergent sum.

 Proposition 1: $(\ell^p, \| \cdot \|_p)$ is a normed linear space.
• Partial Proof: Again, since $\ell^p$ is a subset of the set of all infinite sequences (which is a linear space), all we need to show is that $\ell^p$ is closed under addition, closed under scalar multiplication, and contains the zero sequence $(0)$ to show it is a linear space.
• Let $(a_i), (b_i) \in \ell^p$. Consider the sequence $(a_i) + (b_i) = (a_i + b_i)$. Then:
(1)
\begin{align} \quad \sum_{i=1}^{\infty} |a_i + b_i|^p \leq \sum_{i=1}^{\infty} |2 \max \{ a_i, b_i \}|^p = \sum_{i=1}^{\infty} 2^p |\max \{ a_i^p, b_i^p \}| \leq \sum_{i=1}^{\infty}2^p [|a_i|^p + |b_i|^p] = 2^p \sum_{i=1}^{\infty} |a_i|^p + 2^p \sum_{i=1}^{\infty} |b_i|^p < \infty \end{align}
• So $[(a_i) + (b_i)] \in \ell^p$.
• Now let $\alpha \in \mathbb{R}$ and let $(a_i) \in \ell^p$. Consider the sequence $\alpha (a_i) = (\alpha a_i)$. Then:
(2)
\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}
• So $\alpha (a_i) \in \ell^p$.
• Lastly, $(0) \in \ell^p$ since $\sum_{i=1}^{\infty} |0|^p = 0 < \infty$. So $\ell^p$ is a linear space.
• All that remains to show is that $\| \cdot \|_p$ is a norm on $\ell^p$.
• Showing that $\| (a_i) \|_p = 0$ if and only if $(a_i) = (0)$: Suppose that $\| (a_i) \|_p = 0$. Then $\left ( \sum_{i=1}^{\infty} |a_i|^p \right )^{1/p} = 0$ which implies that $\sum_{i=1}^{\infty} |a_i|^p = 0$ so $a_i = 0$ for each $i \in \mathbb{N}$. Hence $(a_i) = (0)$. Conversely, suppose that $(a_i) = (0)$. Then $\| (a_i) \|_p = \| (0) \|_p = \left ( \sum_{i=1}^{\infty} |0|^p \right )^{1/p} = 0$.
• Showing that $\| \alpha (a_i) \|_p = | \alpha | \| (a_i) \|_p$: Let $\alpha \in \mathbb{R}$ and let $(a_i) \in \ell^p$. Then:
(3)
\begin{align} \quad \| \alpha (a_i) \|_p = \| (\alpha a_i) \|_p = \left ( \sum_{i=1}^{\infty} |\alpha a_i|^p \right )^{1/p} = \left ( \sum_{i=1}^{\infty} |\alpha|^p |a_i|^p \right )^{1/p} = \left ( |\alpha|^p \sum_{i=1}^{\infty} |a_i|^p \right )^{1/p} = |\alpha| \left ( \sum_{i=1}^{\infty} |a_i|^p \right )^{1/p} = |\alpha| \| (a_i) \|_p \end{align}
• Showing that $\| (a_i) + (b_i) \|_p \leq \| (a_i) \|_p + \| (b_i) \|_p$: Unfortunately, showing the triangle inequality holds for $\| \cdot \|_p$ requires some additional preparation, and is known as Minkowski's Inequality for $\ell^p$. We will prove this later. $\blacksquare$