# Nonsmooth analysis Prove a set to be convex

• Jun 24th 2013, 06:12 AM
watashi618
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!
• Jun 25th 2013, 01:25 AM
chiro
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?
• Jun 25th 2013, 09:01 PM
watashi618
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.
• Jun 25th 2013, 11:20 PM
chiro
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?
• Jun 26th 2013, 12:01 AM
watashi618
Re: Nonsmooth analysis Prove a set to be convex
I think it is very difficult to do this.