Existence of Saddle Point 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...
 A: 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 $a\approx 3$.

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 $a\approx 2.39$

A: 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.
