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Let $C \subset R^n$ with $C \neq \emptyset$ be a closed convex set. Consider some $x \in \mathbb{R}^n$ satisfying $x \notin C$. Prove that there exists some $y \in \mathbb{R}^n$ in the boundary of $C$ which implies the existence of a nonempty hyperplane defined by $\{z \in \mathbb{R}^n : (x-y)^T(z-y) = 0\}$.

Geometrically, I want to show that there exists some $y$ on the boundary of $C$ such that the line passing through $y$ and $x$ is orthogonal to the hyperplane tangent to $C$ at $y$.

David Smith
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  • your geometric description says that there exists $ y \in \partial C$ such that for any $z\in \mathbb{R}^n$, $(y-x)^T z = 0 \implies y+z \not \in C$. because $C$ is convex, its projections and intersections with (hyper)planes are still convex, so I guess you should use that to make a proof by induction on $n$. – reuns Feb 06 '16 at 05:26
  • your statement in the first paragraph is somehow incomplete. What you have written after 'satisfying' does not really impose anything on $y$. Did you intend to claim that the set you've written down there is related to $C$ (e.g. that $C$ is on one side of this set)? – Thomas Feb 06 '16 at 05:46
  • I've edited my question to address your question. – David Smith Feb 06 '16 at 05:52

1 Answers1

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Just a hint, not an answer:

If the boundary of $C$ is of class $C^2$ then this is rather easy. If $c:(-\varepsilon, \varepsilon) \rightarrow \partial C$ is a smooth curve which attains a minimum at $t=0$, then $$\frac{d}{dt}|c(t)-x|^2|_{t=0}=0= \langle x-c(0), c^\prime(0)\rangle $$ In other words, in such a point every tangent to $\partial C$ is orthogonal to $x-c(0)$

You now have to (i) show such points (minimizing the distance to given $x$) always exist and (ii) figure out what to do if $\partial C$ is not differentiable.

Thomas
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