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Math Help - Nonsmooth analysis Prove a set to be convex

  1. #1
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    Nonsmooth analysis Prove a set to be convex

    Suppose that  $S\subset X$ is such that  $\forall x \in S$, $\forall \zeta \in N_S^P(x)$, we have  $$\langle \zeta, x'-x\rangle \leq 0 \ \ \forall x'\in S$$. Prove that S is convex. Here X is a Hilbert space and $N_S^P(x)$ means the proximal normal cone.
    It is urgent for me to get the answer. Thank you very much!
    Last edited by watashi618; June 24th 2013 at 07:23 AM.
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  2. #2
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    Re: Nonsmooth analysis Prove a set to be convex

    Hey watashi618.

    Are there results that show if an inner product space is convex, then any subspace is also convex as well?
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    Re: Nonsmooth analysis Prove a set to be convex

    Any subspace is convex, but here S is just a subset of X, not a subspace of X.
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    Re: Nonsmooth analysis Prove a set to be convex

    Can you make it into the appropriate inner product space and prove the corresponding set with its inner product is a sub-inner product space?
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    Re: Nonsmooth analysis Prove a set to be convex

    I think it is very difficult to do this.
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