Suppose that $f(x):R^p \rightarrow R$ is a convex function with global minimum, say 0.

Let $C=(x: f(x)=0)$, i.e. the set of the global minimum. Suppose that there exist at least one point $y$ such that $f(y) = 0$,

It is easy to see that $C$ is convex subset.

Let $a_{\lambda}$ such that $f(a_{\lambda})$ approach 0 as $\lambda$ approach 0 and let $a_{\lambda}^1$ be the closest point to $a_{\lambda}$ in $C$.

Prove that $|a_{\lambda}^1-a_{\lambda}|$ converge to 0, i.e. $a_{\lambda}$ approaches $C$ as $\lambda$ approach 0.

  • $\begingroup$ What about $f(x)=e^x$? $\endgroup$ – daw Feb 7 '15 at 20:53
  • $\begingroup$ the global minimum is attinable, for $e^x$ the global minimum is 0 but for $x=-\infty$, $\endgroup$ – nir Feb 7 '15 at 20:56
  • $\begingroup$ I saw the following: math.stackexchange.com/questions/345865/… $\endgroup$ – nir Feb 7 '15 at 22:10
  • $\begingroup$ Maybe, I need to add another restriction. If I assume that C is compact then it is probably true by argument like the link above $\endgroup$ – nir Feb 7 '15 at 22:18

The answer is NO:

Consider in $X = \mathbb{R}^2$ the function $f(x,y) = x+\sqrt{x^2+y^2}$.
Then the set of minimizers $C=\mathbb{R}_-\times\{0\}$.
Set $\mathbf{x}_n=(-n,1)$.
Then $f(\mathbf{x}_n)\to 0$ but the distance from $\mathbf{x}_n$ to $C$ is $1$.

This is Example 11.24 in Bauschke-Combettes Convex Analysis and Monotone Operator Theory, second edition. That section contains other (some even crazier) examples.

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