I asked this and now I am wondering if the previous proof can be used to show the following:
Let $f(x)=\min_{y \in C}\|y-x\|^2$ where $C$ is a closed set in $\mathbb{R}^n$ and $x,y \in \mathbb{R}^n$.
Show $f(x)$ is a continuous function.
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Counter-example: $C=\{0\}$ in $\mathbb R$. In this case $f(x)=x^{2}$ which is not Lipschitz (since $f'(x)=2x$ is not bounded).
Answer for the modified question: $f(x)=[g(x)]^{2}$ where $g(x)=\min_{y \in C} \|y-x\|$. By the link you have provided, $g$ is continuous. Hence, $f$ is also continuous.
geetha290krm
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What if I add the condition of $0 \in C$? – Saeed Aug 09 '23 at 05:05
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I don't understand. Doesn't $0 \in{0}$? BTW, any one point set $C$ will also serve as a counter-example. – geetha290krm Aug 09 '23 at 05:08
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Never mind. I got confused. My goal is to show it is continuous that is why I tried to show it is Lipschitz. I will edit the question. Could you modify your answer. – Saeed Aug 09 '23 at 05:10
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@Sepide I have edited my answer. – geetha290krm Aug 09 '23 at 05:18
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$f(x)\neq (g(x))^2$ because $\min_{y\in C}|y-x|^2 \neq (\min_{y\in C}|y-x|)^2$. – Saeed Aug 09 '23 at 05:35
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1@Sepide If $E$ is a set of non-negative real numbers then $\min {x^{2}: x \in E}=(\min {x: x \in E})^{2}$ – geetha290krm Aug 09 '23 at 05:43
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Can you please prove this result for me or provide me with a source? – Saeed Aug 09 '23 at 05:59
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Let us continue this discussion in chat. – geetha290krm Aug 09 '23 at 06:06