Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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...

share|cite|improve this question
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

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

share|cite|improve this answer
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
Is there a condition I can add to make sure this example will not happen? – James Rohal Feb 4 '12 at 17:32
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

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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