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Consider a function $g$ with the following properties.

  • It is smooth.
  • $g > 0$.
  • $g \to 0$ at infinity.
  • It has at least two critical points.
  • There are finitely many critical points.
  • Each critical point is isolated.

Thanks to the answer below, I am going to add one additional restriction on $g$.

  • $g$ is a rational function.

I am adding yet another condition after seeing an edit below.

  • Each critical point of $g$ is non-degenerate; that is, if $x$ is a critical point then $\det g''(x) \neq 0$.

In the example below, the critical point that is not a saddle has a zero eigenvalue and hence the determinant is zero.

Notice at least one of the critical points has to be a local max.

The question is: does $g$ have a saddle point?

In particular, for $g \colon \mathbb{R}^n \to \mathbb{R}$, does $g$ have a critical point of index $n-1$?

If there is a reference you can point me to that would be terrific. I believe a variant of the Mountain Pass Theorem may work...

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Where does your function $g$ live? When you write "in particular, for $g:{\mathbb R}^n\to{\mathbb R}$", which other environments are you envisaging? –  Christian Blatter Feb 4 '12 at 18:52
    
I am just interested in real functions in $n$ independent variables. –  James Rohal Feb 9 '12 at 21:22
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2 Answers

Consider functions of the form $$ g(x,y,a)=a e^{-((x-1)^2+y^2)}+e^{-((x+1)^2+y^2)} $$ where $a\geq 1$. For suitable value of $a$ you can get exactly to critical points. One of them will point of maximum, another just a critical point. Necessary condition for $a$ is $$ \frac{\partial g}{\partial x}(x_0,0,a)=0 $$ $$ \frac{\partial g}{\partial x}(x,0,a)\geq0\quad\text{ for all } x\text{ in the neighborhood of }x_0 $$ Here is a graph of such a function. Approximately $a\approx 3$. enter image description here

If we make an additional requirement that functions are rational the answer is still no. Indeed consider function of the form $$ g(x,y,a)=\frac{a}{(x-1)^2+y^2+1}+\frac{1}{(x+1)^2+y^2+1} $$ where $a\geq 1$. For the appropriate value of $a$ you still get one point of maximum, one critical point and no saddle points. This value is approximately equal to $a\approx 2.39$ enter image description here

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Terrific! That answers my original question. I'm going to make an edit and add one more restriction to $g$ to eliminate this case. –  James Rohal Feb 4 '12 at 2:55
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Is there a condition I can add to make sure this example will not happen? –  James Rohal Feb 4 '12 at 17:32
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I don't think so. If can make an effective condition, we can describe all saddle points. –  Norbert Feb 4 '12 at 17:56
    
I realized that if I add one more condition then this example will not hold. I made the change above. –  James Rohal Feb 9 '12 at 21:17
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This doesn't satisfy all the conditions of the question, but it's a little too long for a comment. Maybe you'll find it interesting.

Here is an example of a rational function that has two isolated local maxima and no saddle point: $$g(x,y) = \frac1{(x-1)^2+\big(y-\frac1x\big)^2+1} + \frac1{(x+1)^2+\big(y-\frac1x\big)^2+1}$$ This is what it looks like. It is just the sum of two "rational bumps", $1/\big((x-1)^2+y^2+1\big)$ and $1/\big((x+1)^2+y^2+1\big)$, composed with a transformation $(x,y) \mapsto \big(x,y-\frac1x\big)$ that sends the saddle point $(0,0)$ to infinity. I first saw this, or something very much like it, on the home page of a math professor who is also an active user on this site, but I can't remember who it was now.

Unfortunately, every point on the line $x = 0$ is also a critical point, which violates a couple of your criteria.

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You can vanish this critical point by summing $g$ with function $s=\arctan(x)\omega(y)$, where $\omega(y)$ is the usual bump function from the theory of distributions. I think you can improve ypur post up to acceptable answer. –  Norbert Feb 9 '12 at 22:58
    
@Norbert: I don't follow. The bump function has compact support, say $[-1,1]$. That still leaves all points of the form $(0,y)$ with $|y| > 1$ as critical points. –  Rahul Feb 9 '12 at 23:11
    
$s$ function vanish critical points on the line $x=0$ and doesn't chhange anything outside domain $\mathbb{R}\times [-1,1]$ –  Norbert Feb 10 '12 at 9:48
    
@Norbert, perhaps you mean $s(x,y) = \omega(x) \tan^{-1}(y)$ so that the support of $s$ includes the entire line $x = 0$? In either case, $\tan^{-1}$ does not vanish at infinity, so that would violate another of James's criteria. –  Rahul Feb 10 '12 at 9:54
    
The support of your $s$ function is $[-1,1]\times R$. As for vanishing at infinity, we can conseder $s(x,y)=x e^{-x}\omega(y)$ –  Norbert Feb 10 '12 at 9:58
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