We know that a function defined on a closed bounded domain must have a maximum and a minimum, But $f(x,y) = xe^{−2x^2−y^2}$ is defined on the domain $R^2$, which is not closed bounded, so how to show that it actually has a global max and a global min?
I am trying to prove that outside a disk with distance d, the absolute value of the function value can be as small as possible. That is, for all $\epsilon>0$, exist $d>0$ such that $||(x,y)-(0,0)||>d$ implies $||f(x,y)||<\epsilon$ But how may I prove that? Am I on the right track?
Take $(x,y) \in R^2$ such that $r:=||(x,y)|| > 0$. Then we have $2x^2+y^2 \ge x^2+y^2=r^2$, thus $e^{-2x^2-y^2} \le e^{-r^2}$. This gives
$$|f(x,y)| \le \frac{r}{e^{r^2}}$$