1
$\begingroup$

Can someone give an example of a convex function $f$ on a path-connected compact nonconvex set where some point $c$ is a local minimum with $\nabla f(c)=0$ but not a global minimum. Thus showing that the set being not convex makes finding global minimums harder.

  • 1
    Consider the blue region in [this image](https://en.wikipedia.org/wiki/File:Lappo.svg)...2012-03-11
  • 0
    Regarding your edit, if $f$ is convex and $\nabla f(c)=0$, then $c$ has to be the global minimum. But $\nabla f(c)=0$ is not necessary for $c$ to be a local or global minimum when the domain is compact: often, $\nabla f$ is zero nowhere, and the local and global minima lie on the boundary of the domain, as my example shows.2012-03-11
  • 0
    @RahulNarain: if $f$ is convex and $\nabla f(c)=0$, then $c$ has to be the global minimum if the set on which $f$ is defined on is convex. But if the set is not convex then I don't see why $c$ would be the global minimum.2012-03-11
  • 0
    I was implicitly assuming $f$ to be defined on a convex superset of your nonconvex set, in which case the result is immediate. If not, how do you define a convex function on a nonconvex set?2012-03-11
  • 0
    $f(t x_1+ (1-t)x_2) \le t f(x_1) + (1-t)f(x_2)$ if the line connecting $x_1$ and $x_2$ lies in the nonconvex set.2012-03-11
  • 1
    ...So, by your definition, *any* function on the circle $x^2+y^2 = 1$ is convex? I wouldn't call that a very useful definition. Is it part of some standard reference, or did you make it up?2012-03-11
  • 0
    Well presumably the focus would be on set with interior having some meat to it.2012-03-11

2 Answers 2

2

Consider the set $$A:=[-1,3]\times[-3,3]\ \setminus\ \{(x,y)\ |\ 0<|y| (a rectangle minus an isosceles triangle) and define $$f(x,y):=\cases{(x-2)^2 & $(x\leq 0 \ \vee\ y<0)$ \cr 2(x-1)^2+2 & $(x>0\ \wedge \ y>0)$\cr}\ .$$ This $f$ is $C^1$ on $A$, convex, and has local minima at $c:=(2,-{5\over2})$, $c':=(1,2)$ with $f(c)=0$, $f(c')=2$.

  • 0
    @Rahul Narain: There was a typo in the definition of the isosceles triangle. Thank you.2012-03-11
2

For example the function $f(x,y) = x^2+y^2$ restricted to the square with vertices $(-1,-2)$, $(5,-2)$, $(5,4)$ and $(-1,4)$ has four local minima but only one global minimum.

  • 0
    A square is not a non-convex set.2012-03-11
  • 0
    @user782220, I believe WimC means the boundary of the square, not its interior.2012-03-11
  • 0
    @user782220 As Rahul Narain commented: The boundary of the square. (Otherwise $f$ would clearly only have a single local minimum.)2012-03-11