Nonsmooth analysis Prove a set to be convex

Suppose that$\displaystyle $S\subset X$$ is such that $\displaystyle $\forall x \in S$, $\forall \zeta \in N_S^P(x)$$, we have $\displaystyle $$\langle \zeta, x'-x\rangle \leq 0 \ \ \forall x'\in S$$$. Prove that S is convex. Here X is a Hilbert space and $\displaystyle $N_S^P(x)$$ means the proximal normal cone.

It is urgent for me to get the answer. Thank you very much!

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?

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.

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?

Re: Nonsmooth analysis Prove a set to be convex

I think it is very difficult to do this.