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We have a problem:

For a convex set $C$ and a point $y \notin C$, $x \in C$ is the minimum distance mapping from $y$ to $C$.

For $\forall z \in C$, it's safe to say that 'the shorter the distance between $z$ and $x$, the shorter the distance between $z$ and $y$'？

Thanks a lot!!!

nwu.hly
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    It's definitely unsafe. It's not clear what relationship $x$ and $y$ have between them. If $C$ is an unbounded convex set, then $z$ can be as far as you want from $y$, while $x$ could be as close as you like to $y$. – Theo Bendit Jun 28 '22 at 03:57
  • Thanks for your comments! – nwu.hly Jun 28 '22 at 07:42

1 Answers1

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Let $C= \{ (x,y) | y \ge 0 \}$, $y=(0,-1), x=(0,0)$.

Choose $z = (1,0)$, $z'=(0,1-t)$, where $t \le 1$.

Then $\|z-x\|^2 = 1$, $\|z'-x\|^2 = (1-t)^2$, $\|z-y\|^2 = 2$ and $\|z'-y\|^2 = (2-t)^2$.

If $t \in (0,1)$ then $\|z'-x\| < \|z-x\|$ and if $t \in (2-\sqrt{2}, 2+ \sqrt{2})$ then $\|z-y\| < \|z'-y\|$.

It is easy to check that these intervals intersect.

copper.hat
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