I have the following question.
Consider the function $$f(x,y) = 8xy +\dfrac14(x-y)^{4}$$
(a) find all the local minimum points of $f$ on $\mathbb R^2~$.
(b) Does $~f~$ have any global minimum point on $\mathbb R^2~$?
Justify your answer rigorously.
For part (a), I obtained two local minimum, which are attained at $~(1,-1)~$ and $~(-1,1)~$.
However, I have no idea how to show that there is no global minimum point on $\mathbb R^2$.
Is it related to the connect of strictly convex of $f$ over $\mathbb R^2$?
If so, how can I proof there is/ is no global minimum point on $\mathbb R^2$?
Use the variable changes
$$x=u+v, \>\>\> \>\>y = u-v$$
to reexpress the function as,
$$f(u,v) = 8(u^2-v^2)+4v^4 = 8u^2 + 4(v^2-1)^2 -4 \ge -4$$
Thus, -4 is the absolute minimum value that $f(x,y)$ takes, hence the global minima.
The above global minima are at $u=0$ and $v^2=1$, which in the original $x$ and $y$ coordinates are at the points (1,-1) and (-1,1). So, for this function, the local minima are also the global ones.