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Let $X$ be a convex set and $f:X\mapsto \mathbb R$ be a continously differentiable convex function and $x_0$ be an element of $X$ such that $$\forall x\in X:f(x_0)\le f(x).$$

Let $y_0\in Y\subset X$ ($Y$ is not necessarily connected) satisfy $$\forall y\in Y:\|x_0-y_0\|\le\|x_o-y\|.$$

Is it true that $\forall y\in Y:f(y_0)\le f(y)$ ?

davcha
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  • I may be wrong, I tried to take $X=\mathbb{R}$, $f(x) = \begin{cases} x^2, & \mbox{if } x \leq 0 \ 0.05x, & \mbox{if } x>0 \end{cases}$ and $Y=[-2,-1] \cup [10,20]$. Thus the closest point to $x_0=0$ is $y_0=-1, \ f(y_0)=1$, but $f(10)=0.5$. – Slowpoke Aug 25 '15 at 12:40
  • This implies that $Y$ should be connected, at least. – Slowpoke Aug 25 '15 at 12:42
  • What if $f$ is continously differentiable ? And, yes, $Y$ is not connected in my case. – davcha Aug 25 '15 at 12:45
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    My idea is that one part of $f$ that corresponds to one part of $Y$ grows much slower than other; we can "smooth" my function to make it have as many derivatives as needed, but it won't change the result. Just draw it on the paper and you will see the problem. – Slowpoke Aug 25 '15 at 12:49
  • It seems to work in my case. Even in two dimensions, it suffice that one dimension grows much faster than the other and that a point $y$ exists close to $x_0$ in the direction of the dimension that grows faster. Thanks. – davcha Aug 25 '15 at 12:53
  • @hcl14: Your choice of $f$ is not continuously differentiable. – dohmatob Aug 25 '15 at 18:21
  • @dohmatob The function can be smoothed. – Slowpoke Aug 25 '15 at 18:24
  • Short story: Your proposed $f$ doesn't satisfy the requirements of the question. It's strange you're trying to back it up. What do you mean by "the function can be smoothed" ? – dohmatob Aug 25 '15 at 18:30
  • The requirement continuously differentiable was added after his answer. – davcha Aug 25 '15 at 18:34
  • @dohmatob The question was edited, you may check it by link below the post (you can see that edit incorporated my comment actually). The moment I was commenting there was no such a requirement. By smoothing I mean, for example, removing the problem at 0 by making the function more smooth here and replacing the line with the curve; I was taught that as a common technique to make the function have as much derivatives as needed, in particular to have continuous 1st derivative. – Slowpoke Aug 25 '15 at 18:57
  • @hc14: OK, it didn't occur to me the question was edited to its current form only after your answer. It'd been much better to say so instead of (or alongside) your comment about smoothability, as it appeared you we rather trying to justify why your $f$ was "ok" for the current state of the question. This makes sense now. – dohmatob Aug 25 '15 at 19:18
  • @dohmatob I also haven't noticed that it was edited, so answered you thinking that it was a resumption of the previous discussion. – Slowpoke Aug 25 '15 at 20:04

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Consider the function $$ f(x) = \begin{cases} x^2, & x > 0, \\ 0, & x \le 0 \end{cases} $$ on the real line with $Y = \mathbb R\setminus (-2, 1)$.

The function $f$ is continuously differentiable, convex, and minimal at $x_0 = 0$. However, the projection of $x_0$ onto $Y$ is $y_0 = 1$ and $f(y_0) = 1 > 0 = f(-2)$.

Comments:

  • Connectivity doesn't matter. Consider $g(x,y) = f(x)$ on $\mathbb R^2$ and $Y = \mathbb R^2 \setminus (-2, 1)\times(-2, 2)$.

  • It might help, if $Y$ is convex and compact.

user251257
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