# 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)

• 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. – AlexR Sep 8 '15 at 17:20

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 $f$.
• I don't understand the answer for 2) A theorem in my book states that if x* is a critical point then it is a strict global maximizer if Hf(x) is negative definite on $R^n$. So for a global min/max shouldn't Hessian be positive/negative definite for all $x \in R^2$. – sam_rox Sep 8 '15 at 17:47
• @sam_rox Your book says if, not if and only if. So if the Hessian is $\prec 0$ everywhere (and the function is $C^2$), any critical point is a global maximum. The other way around carries no weight - if the Hessian is not $\prec 0$ everywhere (and the function is $C^2$), we don't know anything about the global maximum. – AlexR Sep 8 '15 at 18:01
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.
• The "evident" part of the 1D gaussian is just as evident as is the 2D case in my opinion. In other words, when practicing these topics, none of the two should be deemed "evident" ($\pm0$). – AlexR Sep 8 '15 at 19:34