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Let $f(x)=\min_{y \in C}\|y-x\|$ where $C$ is a closed set in $\mathbb{R}^n$ and $x,y \in \mathbb{R}^n$.

Show $f(x)$ is a Lipschitz function with constant $L=1$.

My try

I need to show $$ |f(x_2)-f(x_1)| \leq \|x_2-x_1\| \quad \forall x_1,x_2 \in \mathbb{R}^n $$

When $x_1,x_2 \in C$, it is trivial because $f(x_2)=f(x_1)=0$. When one of them is in $C$, let's say $x_1$, then $f(x_1)=0$ and $|f(x_2)-f(x_1)|=\min_{y \in C}\|y-x_2\|\leq \|x_1-x_2\|$ because $x_1 \in C$.

Question

How can I show it when neither $x_1$ nor $x_2$ is in $C$?

Saeed
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  • It is not even well defined. You probably forgot to assume convexity of $C$. – Moishe Kohan Aug 08 '23 at 22:06
  • @Moishe Kohan: It is well-defined. Then projection vector is not unique but the projection error value for a given point $x$ is well-defined. – Saeed Aug 08 '23 at 22:08
  • @MoisheKohan If $C\ne\emptyset$, then $f$ is well-defined: let $y_0\in C$. Then $\inf_{y\in C} \lVert x-y\rVert=\inf_{y\in {z\in C,:, \lVert z-x\rVert\le\lVert y_0-x\rVert}} \lVert y-x\rVert$. Since ${z\in C,:, \lVert z-x\rVert\le\lVert y_0-x\rVert}$ is compact, that $\inf$ is achieved. If it had $\inf$ instead of $\min$, then it would also be defined in all metric spaces, and for all non-empty subsets. –  Aug 08 '23 at 22:12
  • In a more general setting we have: https://math.stackexchange.com/a/8068/470072 – Sudix Aug 08 '23 at 22:20
  • @Sudix: Very good point. How can I tailor that proof for my question. Can you please help me with that? – Saeed Aug 08 '23 at 23:28
  • The minimum exists but need not be unique. That's why the map is not well defined. You have to invoke the axiom of choice to choose the projection. This map will not even be continuous. – Moishe Kohan Aug 09 '23 at 00:33
  • @Moishe Kohan: what do you mean? There might be multiple $y$'s but all of them have the same function value otherwise we get a contradiction. – Saeed Aug 09 '23 at 00:35
  • Oh, I see what you mean – Moishe Kohan Aug 09 '23 at 00:36

1 Answers1

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I am tailoring the general solution posted here to answer your question:

Let $z \in C$ and $x_1, x_2 \in \mathbb{R}^n$ be arbitrary points. Using triangle inequality you can write the following:

$$ \begin{aligned} \|z-x_1\| &\leq \|z-x_2\|+\|x_2-x_1\| \\ \min_{y \in C}\|y-x_1\|\leq \|z-x_1\| &\leq \|z-x_2\|+\|x_2-x_1\| \\ \min_{y \in C}\|y-x_1\| &\leq \min_{y \in C} \|y-x_2\|+\|x_2-x_1\| \end{aligned} $$ where in the second line we know that $\min_{y \in C}\|y-x_1\|\leq \|z-x_1\|$ because $z \in C$, and in the third line we know that $z$ is an arbitrary point so we can let it be a minimizer of $\|y-x_2\|$ over $C$ to achieve $\min_{y \in C} \|y-x_2\|$. Thus we have $$ \min_{y \in C}\|y-x_1\| - \min_{y \in C} \|y-x_2\| \leq \|x_2-x_1\| \tag{1}. $$ Switch $x_1$ and $x_2$ to get the following: $$ \min_{y \in C}\|y-x_2\| - \min_{y \in C} \|y-x_1\| \leq \|x_2-x_1\| \tag{2}. $$ Put (1) and (2) together, then you get $$ \Big| \min_{y \in C}\|y-x_2\| - \min_{y \in C} \|y-x_1\| \Big| \leq \|x_2-x_1\| $$ which means $|f(x_2)-f(x_1)| \leq \|x_2-x_1\|$.