$ℓ 𝑝$ space

For $𝑝 \in [1, \infty]$ , the $ℓ 𝑝$ space is a Banach space defined as the set of all sequences $𝑥 ∙$ in $ℂ$ with finite $𝑝$ -norm given by

‖ 𝑥 ∙ ‖ 𝑝 = (\infty \sum 𝑛 = 1 | 𝑥 𝑛 | 𝑝) 1 / 𝑝 < \infty

where in the case of $𝑝 = \infty$ we get the supremum

‖ 𝑥 ∙ ‖ \infty = s u p {𝑥 𝑛} \infty 𝑛 = 1

Thus $ℓ 𝑝$ is equivalent to the Lebesgue space $𝐿 𝑝 (ℕ, 𝜇) = L 𝑝 (ℕ, 𝜇)$ ¹ where $𝜇$ is the counting measure. More generally one defines $ℓ 𝑝 (𝑆) = 𝐿 𝑝 (𝑆, 𝜇)$ with the counting measure for any set $𝑆$ .

Proof of Banach space

This all follows from the general case of a Lebesgue space, however we will explicitly show completeness for $𝑝 \in [1, \infty)$ .

Let $(𝑥 (𝑛) ∙) \infty 𝑛 = 1$ be a Cauchy sequence in $ℓ 𝑝$ , i.e. for every $𝜖 > 0$ there exists some $𝑁 \in ℕ$ such that for all $𝑛, 𝑚 \geq 𝑁$ we have
$\infty \sum 𝑖 = 1 | 𝑥 (𝑛) 𝑖 - 𝑥 (𝑚) 𝑖 | 𝑝 < 𝜖 𝑝$
Now for any fixed $𝑗 \in ℕ$
$| 𝑥 (𝑛) 𝑗 - 𝑥 (𝑚) 𝑗 | 𝑝 \leq \infty \sum 𝑖 = 1 | 𝑥 (𝑛) 𝑖 - 𝑥 (𝑚) 𝑖 | 𝑝 < 𝜖 𝑝$
so $𝑥 (∙) 𝑗$ is a Cauchy sequence in $ℝ$ convergent to some $𝑥 𝑗$ . It remains to show that $𝑥 ∙ \in ℓ 𝑝$ and $(𝑥 (𝑛) ∙) \infty 𝑛 = 1 \to 𝑥 ∙$ . For any fixed $𝑘 \in ℕ$
$𝑘 \sum 𝑖 = 1 | 𝑥 (𝑚) 𝑖 - 𝑥 (𝑛) 𝑖 | 𝑝 \leq \infty \sum 𝑖 = 1 | 𝑥 (𝑚) 𝑖 - 𝑥 (𝑛) 𝑖 | < 𝜖 𝑝$
whence
$𝑘 \sum 𝑖 = 1 | 𝑥 (𝑚) 𝑖 - 𝑥 𝑖 | 𝑝 < 𝜖 𝑝$
and by the triangle inequality
$(𝑘 \sum 𝑖 = 1 | 𝑥 𝑖 | 𝑝) 1 / 𝑝 \leq (𝑘 \sum 𝑖 = 1 | 𝑥 (𝑚) 𝑖 - 𝑥 𝑖 | 𝑝) 1 / 𝑝 + (𝑘 \sum 𝑖 = 1 | 𝑥 (𝑚) 𝑖 | 𝑝) 1 / 𝑝 < 𝜖 + (𝑘 \sum 𝑖 = 1 | 𝑥 (𝑚) 𝑖 | 𝑝) 1 / 𝑝$
hence
$‖ 𝑥 ∙ ‖ 𝑝 \leq 𝜖 + ‖ 𝑥 (𝑚) ∙ ‖ 𝑝 < \infty$
so $𝑥 ∙ \in ℓ 𝑝$ . Furthermore
$‖ 𝑥 (𝑚) ∙ - 𝑥 ‖ 𝑝 𝑝 = \infty \sum 𝑖 = 1 | 𝑥 (𝑚) 𝑖 - 𝑥 𝑗 | 𝑝 < 𝜖 𝑝$
so indeed $(𝑥 (𝑛) ∙) \infty 𝑛 = 1 \to 𝑥 ∙$ .

tidy | en | SemBr

There is no need to take a normed quotient here, $‖ \cdot ‖ 𝑝$ is already a full norm due to properties of the counting measure. ↩

$ℓ 𝑝$ space

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ℓ𝑝 space

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$ℓ 𝑝$ space