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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.

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Consider the blue region in this image... –  Rahul Mar 11 '12 at 6:09
    
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. –  Rahul Mar 11 '12 at 7:16
    
@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. –  user782220 Mar 11 '12 at 9:00
    
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? –  Rahul Mar 11 '12 at 9:21
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...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? –  Rahul Mar 11 '12 at 10:07

2 Answers 2

up vote 1 down vote accepted

Consider the set $$A:=[-1,3]\times[-3,3]\ \setminus\ \{(x,y)\ |\ 0<|y|<x\leq3\}$$ (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 $f(c)=0$, $f(c')=2$.

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@Rahul Narain: There was a typo in the definition of the isosceles triangle. Thank you. –  Christian Blatter Mar 11 '12 at 13:19

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.

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A square is not a non-convex set. –  user782220 Mar 11 '12 at 9:01
    
@user782220, I believe WimC means the boundary of the square, not its interior. –  Rahul Mar 11 '12 at 10:23
    
@user782220 As Rahul Narain commented: The boundary of the square. (Otherwise $f$ would clearly only have a single local minimum.) –  WimC Mar 11 '12 at 12:50

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