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Convexe and serie

  1. Feb 19, 2016 #1
    Hello every one, let be V an open convex a a normed vectorial space E.
    Let be ##(a_{n}) \in \mathbb{R}^{n}## with ##\sum_{i \in \mathbb{N}} a_{i} = 1##.
    Let be ##(v_{n}) \in V^{\mathbb{N}}## as ##\sum_{i \in \mathbb{N}} a_{i}v_{i}## exists.
    Does necessarly ##\sum_{i \in \mathbb{N}} a_{i}v_{i} \in V## please?

    Thank you in advance and have a nice afternoon:oldbiggrin:.
     
  2. jcsd
  3. Feb 19, 2016 #2
    I suppose a positive.
    In fact for my need I can suppose ##a_{n} = \frac{1}{2^{n+1}}##.
     
  4. Feb 19, 2016 #3

    mathman

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    Your original statement needs proofreading. I assume you meant V an open convex set (but I'm not sure). Also [itex]V^N[/itex]? In any case, the sum is not necessarily in V, if it exists, but in the closure of V. The sum existence depends on whether on not E is complete.
     
  5. Feb 20, 2016 #4

    Samy_A

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    Here is a theorem about open convex sets that might help (based on theorem 3.4 in Rudin's Functional Analysis).

    Suppose ##X## is a topological vector space over ##\mathbb R##, ##A## and ##B## disjoint, nonempty, convex subsets of ##X## with ##A## open, then there exists a continuous linear function ##f: X \to \mathbb R## and a real number ##\gamma## such that ##f(a) \lt \gamma \leq f(b)## for all ##a \in A,\ b\in B##.

    This should help answer the question.

    For reference, the theorem from the book:
    rudin34.jpg
     
    Last edited: Feb 20, 2016
  6. Feb 20, 2016 #5
    No I suppose my sum exist . And I clearly wroght V is an open convex.
     
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