The ℓp(C) spaces are complete
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[hide]Statement
Let p∈[1,+∞]⊆¯R be given and consider ℓp(C) then we claim[1]:
- For p∈R≥1 that ℓp(C):={(xn)n∈N⊆C | ∞∑n=1|xn|p<+∞}is a complete metric space with respect to the metric induced by the norm: ∥(xn)n∈N∥p:=(∞∑n=1|xn|p)1p
- For p=+∞[Note 1] that ℓ∞(C):={(xn)n∈N⊆C | Supn∈N(|xn|)<+∞}is a complete metric space with respect to the metric induced by the norm: ∥(xn)n∈N∥∞:=Supn∈N(|xn|)
Proof of claims
Grade: A*
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Notes
- Jump up ↑ Herein just ∞ as only one is in the relevant range of p
References