Finding global maximizers and minimizers I want to  find if global maximum or minimum exists in $ f(x,y)=e-^{(x^2+y^2)}$    
I found that (0,0) is the only critical point.
In the Hessian matrix $H_{(f)}(0,0)$
was negative definite and so (0,0) is a local maximizer.
There are few things I need to clarify .
1)As $(0,0)$ is the only critical point and it is a local maximum does it indicate that it is a global maximum as well, because it is the only critical point.  
2)Using principal minor method in Hessian matrix $\partial^2f \over \partial^2x$=$2(2x^2-1) *e-^{(x^2+y^2)}$ which is the first minor. And this is $<0$ for (0,0) but for (2,0) it is $>0$.
For a global min/max shouldn't Hessian be positive/negative definite for all $x \in R^2$.
So in this case to have a global maximum  isn't it necessary that the first minor be negative for all $x \in R^2$  
3) $\lim\limits_{x,y \rightarrow +\infty}f(x,y) =0$. Also $\lim\limits_{x,y \rightarrow -\infty}f(x,y) =0$.  From this how to determine if the function has a global maximum or global minimum at (0,0)
 A: 1) Certainly needs more than that. Consider $|1-x^2|$. It has one critical point (at $0$, a local maximum) and no global maximum with global minima at $\pm1$, but these are no critical points since the derivative doesn't exist there.
2) Is wrong and your function is a counter-example. $0$ is in fact a global maximum of your function.
3) You need more than that. Precisely, you must send $x$ and $y$ to $\pm\infty$ independently, as in
$$\lim_{\|x\|_2 \to \infty} f(x_1, x_2)$$
Your particular function is very well behaved since
$$f(x_1, x_2) = e^{-\|x\|_2^2}$$
So the above limit is indeed $0$. To prove that you have a global maximum, observe that $g(r) := e^{-r^2}$ satisfies $g'(x) < 0$ for $x > 0$ and $g(x) = 1$. Thus by some Calculus theorem on differential inequalities, $g$ satisfies
$$g(x) \le 1 \quad \forall\ x > 0$$
and likewise you can show $g(x)\le 1 \forall\ x < 0$. You have now established that $0$ is a global maximum of $g$. Finally find $f(x) = g(\|x\|_2) \le 1$ to establis that $(0,0)$ is a global maximum of $f$.
A: While it is always nice to practice multi-variable extremal analysis, with partial derivatives, gradients, Hessians, etc, sometimes a much simpler argument will do. Since the function $e^{-(x^2+y^2)}$ depends only on the length of the vector $(x,y)$, we can look at the function $e^{-r^2}$. We can justify this formally by passing to polar coordinates: $x=r\cos\theta$ and $y=r\sin\theta$. Now it is evident that $e^{-r^2}$ has a global maximum at $r=0$ and only there, and that it has no minimum, not global nor local. So $f$ has a global maximum at the origin $x=y=0$ and only there, and has no minimum anywhere.
