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If $S$ is a closed and convex subset of $\mathbb{R}^n$, then there is a unique nearest point in $S$ for any $x \in \mathbb{R}^n$, i.e., there exists a unique $y' \in S$ such that $\|x-y'\| = \inf_{y \in S} \|x-y\|$.

Then define $f : \mathbb{R}^n \to S$ by $f(x)=$ nearest point to $x$ in $S$. That is, via the equality $ \inf \|x - y\| = \|x - y'\|$, we take $f(x)=y'$.

How do you prove $f$ is continuous?

Mark
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    A closed convex subset of $\mathbb R$ should be a closed interval. –  Jun 07 '16 at 19:39
  • @Epsilon Hence compact. I thought about applying the extreme value theorem, but didn't go very far with that. – Mark Jun 07 '16 at 19:41
  • Note that $f$ is the identity on $S$ and outside $S$ it is constant with value the left resp. right endpoint of your interval. This is clearly a continuous function. –  Jun 07 '16 at 19:43
  • @user264885: $S$ need not be compact. But the point is, there are only a few options for the form $S$ can take, so you should be able to write $f$ down explicitly. – carmichael561 Jun 07 '16 at 19:44
  • @carmichael561 And if I wanted to generalize this to $\mathbb{R}^n$, then what do I do? I'm looking for a proof that works without explicitly writing down $f$. – Mark Jun 07 '16 at 19:46
  • This might be too much sophistication, but switch to using the minkowski functional on your convex set, which induces a new (but equivilent) metric on $\mathbf{R}^n$. This functional identifies a homeomorphism from $\mathbf{R}^n$ and itself, which maps your convex set to the unit ball. Perhaps in this new `coordinate system' you have a much nicer representation for your function $f$. – Jacob Denson Jun 07 '16 at 20:19
  • Can $S$ take any other form then a closed interval ? –  Jun 07 '16 at 20:23
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    Perhaps edit the question on this post to address the more general question of $\mathbf{R}^n$, since the one dimensional question is trivial. – Jacob Denson Jun 07 '16 at 20:25

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I'm not sure this is the simplest solution, but this is what I got.

Remember that $\|x-f(x)\| \leq \|x-y\|$ for all $x\in R^n$, $y\in S$.

Consider a convergent sequence $x_n\to x_0$. We want to prove that $f(x_n) \to f(x_0)$. A convergent sequence is bounded, hence: so is $(f(x_n))_n$, because $$ \|f(x_n)\| \leq \|x_n\| + \|f(x_n) - x_n\| \leq \|x_n\| + \|x_n - f(x_0)\| \leq 2 \max_n \|x_n\| + \|f(x_0)\|. $$ Then it has an accumulation point, which lies in $S$, as this set is closed: $$ \exists y\in S \ \exists (n_k): \ f(x_{n_k}) \xrightarrow{k\to\infty} y. $$ Consider the distance $$ \|x_{n_k} - f(x_{n_k})\|. $$ On the one hand, it tends to $\|x_0 - y\|$, as $k\to\infty$. But on the other hand, it can be shown (see at the end) that the function of distance to $S$, ie. $\|x-f(x)\|$, is continuous. So the limit is also equal $$ \|x_0 - y\| = \lim_{k\to\infty} \|x_{n_k} - f(x_{n_k})\| = \|x_0 - f(x_0)\|. $$ This means that $y=f(x_0)$, as the nearest point is unique.

We've chosen any accumulation point, so it follows that $(f(x_n))_n$ has no other accumulation points but $f(x_0)$. Does it mean that this is the limit? Yes, because the sequence is bounded! If this wasn't the limit, then there would be a subsequence staying away from $f(x_0)$; however, the subsequence is still bounded, so there is an accumulation point, but we already know that this has to be $f(x_0)$ - contradiction.

Finally, why is $\|x - f(x)\|$ continuous? Let's denote for brevity $$ d_S(x) \mathrel{:=} \|x-f(x)\|. $$ This is a real-valued function and continuity can be proved by preimages of open intervals: $$ d_S^{-1}((a,b)) = d_S^{-1}((a,\infty)) \cap d_S^{-1}((-\infty,b)). $$ To check that $d_S^{-1}((a,\infty))$ is open, choose some element $x_0$. If $\|x-x_0\| < d_S(x_0) - a$, then $d_S(x) > a$, because otherwise: $$ d_S(x_0) \leq \|x_0 - f(x)\| \leq \|x_0 - x\| + \|x-f(x)\| < d_S(x_0) - a + a = d_S(x_0). $$ Thus, $d_S^{-1}((a,\infty))$ is open. It goes similarly for $d_S^{-1}((-\infty,b))$ (even easier).

eudes
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  • I wanted to add that I don't use convexity anywhere, only uniqueness of the nearest point, but realized that this might be the same. And indeed, for closed sets, it is!: http://math.stackexchange.com/questions/274810/is-a-closed-set-with-the-unique-nearest-point-property-convex – eudes Jun 08 '16 at 00:37