0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.