I would like to prove that the following functions $f :\mathbb{R}^2\to\mathbb{R}$ have a global minimum:

  1. $f_1(x,y)=x^4+2x^2y+y^2-4x^2-8x-8y=(x^2+y)^2-4(x^2+2x+2y)$
  2. $f_2(x,y)=(x-2y)^4+64xy$

I've found the gradient and the hessian of both functions, along with their local minima. I need to prove that those local minima are also global minimums.

  • $f_1$ has strict local minimum at $f_1(1,3)=-20$
  • $f_2$ has strict local minima at $f_2(1,-1/2)=-16$ and $f_2(-1,1/2)=-16$

I think that what I need to do is to show that those minimum points are also their lower bounds, but I couldn't prove that. Another idea is to show that they (maybe) coercive functions.

Please advise.

Thank you.


Answer to first question:

Write $t = x^2+y$, then $y= t-x^2$ so we have $$f(x,y)=(x^2+y)^2-4(x^2+2x+2y)$$

$$ = t^2 -4(x^2+2x+2t-2x^2)$$

$$ =t^2-8t +\color{red}{16}+ 4x^2-8x +\color{red}{4} -20 $$

$$ = (t-4)^2+4(x-1)^2-20$$

$$\geq -20$$

It achieves minimum iff $x=1$ and $t=4$ so $y=3$.


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