Let $C$ be a convex subset of $\mathbb{R}^{n}$ and let $\bar{x} \in C$. Then the normal cone $N_{C}(\bar{x})$ is closed and convex. Here, we're defining the normal cone as follows:

$$N_{C}(\bar{x}) = \{v \in \mathbb{R}^{n} \vert \langle v, x - \bar{x} \rangle \le 0, \forall x \in C \}.$$

Proving convexity is straightforward, as is proving $N_{C}(\bar{x})$ is closed when $C$ is open ($i.e.$ every $x \in C$ is an interior point). However, I'm not sure how to prove that $N_{C}(\bar{x})$ is closed more generally?


Write $N_C(\bar{x}) = \cap_{x \in C} \{v | \langle v, x-\bar{x} \rangle \le 0 \}$.

Hence $N_C(\bar{x})$ is the intersection of closed hyperplanes which is closed (the function $v \mapsto \langle v, x-\bar{x} \rangle$ is continuous).

This approach also shows that $N_C(\bar{x})$ is convex.

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    $\begingroup$ ....and now I see why this proof has been omitted from so many books that I've looked through. Thanks @copper.hat. $\endgroup$ – StevenA Jan 9 '19 at 23:25
  • $\begingroup$ @StevenA: There are many cute tricks, after a while you will add them to your quiver :-). $\endgroup$ – copper.hat Jan 9 '19 at 23:28

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