The p(C) spaces are complete

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Statement

Let p[1,+]¯R be given and consider p(C) then we claim[1]:

  1. For pR1 that p(C):={(xn)nNC | n=1|xn|p<+}
    is a complete metric space with respect to the metric induced by the norm: (xn)nNp:=(n=1|xn|p)1p
  2. For p=+[Note 1] that (C):={(xn)nNC | SupnN(|xn|)<+}
    is a complete metric space with respect to the metric induced by the norm: (xn)nN:=SupnN(|xn|)

Proof of claims

Grade: A*
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Notes

  1. Jump up Herein just as only one is in the relevant range of p

References

  1. Jump up Functional Analysis - Volume 1: A gentle introduction - Dzung Minh Ha