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In the given Lemma.

Suppose $x$ is a point outside a non-empty closed convex set $C$ . Then there exist a unique point $C(x)$, on the boundary of $C$ , which is closer to $x$ than any other point in $C$ .

What I am not understanding that why the closest point falls on boundary only why not within C. For example if there is disk (which I believe is closed convex set) on some plane in R3 and there is point outside that plane then there could be some point within disk that is more closer to this point than the points on the boundary.

I know my knowledge is limited and this question may seems very silly :) The only way this lemma makes sense to me is that if the point and the closed convex set $C$ both lie on to the same plane. Is this assumption is mandatory in understanding Hyperplane separation theorems.

Arun Chauhan
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Assume $C(x)$ is in the interior of $C$. This means there exists some ball $B(C(x),\epsilon)\subset C$ such that the closure of the ball is also a subset of $C$ (since $C$ is closed, this is obvious).

Now draw a line from $C(x)$ to $x$. Call this line $L$. Now take the point where $L$ intersects with the edge of $B(C(x),\epsilon)$. Call this ppoint $C'(x)$.

Now, clearly, the distance from $C'(x)$ to $x$ is smaller than the distance between $C(x)$ and $x$ (in fact, it's smaller by $\epsilon$). Also, clearly, $C'(x)\in C$, which means $C(x)$ is not closer to $x$ than any other point in $C$.

5xum
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  • Thanks for answer. But where I am wrong in my second paragraph (What I am no...) of question asked? – Arun Chauhan Jul 11 '17 at 11:40
  • @ArunChauhan My answer proves exactly why $C(x)$ does not fall in the interiour of $C$. – 5xum Jul 11 '17 at 11:41
  • @ArunChauhan Also, if you have a two-dimensional disk in $\mathbb R^3$, then the interior of the disk is empty. – 5xum Jul 11 '17 at 11:42
  • Thank you so much for the answer. I am new to this subject :) thats why these silly questions creates trouble for me. Hope I can contact you further for these dump questions :) Anyway thanks for this answer. – Arun Chauhan Jul 11 '17 at 11:45
  • @ArunChauhan You are welcome. If this answer is what you needed, you can also accept it. – 5xum Jul 11 '17 at 11:49