How can we say a function has a global minimum without calculating any critical points or major calculations? I want to know about the right theorem and how do we reduce the theorem to a problem like this.
\begin{align*}f(x,y) = x^{2n} + y^{2n} -nx^2 + 2nxy – ny^2\end{align*}
on $\mathbb{R}^2$ and $n$ be an integer greater than or equal to 2
Theorem: If $f$ is continuous on the region $R$ (the function should also be continuous at the boundary), then the global maximum and global minimum exist, either in the interior or on the boundary.
I know that there is a global minimum and the function is coercive then there is no global maximum. Could you please give me a motivation to answer this question "Is there a maximum/minimum value of $f$? Give the reason"
Best Answer
Denote $f_z = \partial f /\partial z$.
To find convexity of $f$, note that $$ \vec{\nabla} f^T = \begin{bmatrix} f_x \\ f_y \end{bmatrix} = \begin{bmatrix} 2nx^{2n-1}-2nx + 2ny \\ 2ny^{2n-1}-2ny + 2nx \end{bmatrix} = 2n \begin{bmatrix} x^{2n-1}-x + y \\ y^{2n-1}-y + x \end{bmatrix} $$ and the Hessian is given by $$ H = \begin{bmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{bmatrix} = 2n \begin{bmatrix} (2n-1) x^{2n-2}-1 & 1 \\ 1 & (2n-1) y^{2n-2}-1 \end{bmatrix} $$ To prove $f$ is convex, it suffices to show $H$ is positive semidefinite, i.e. $u^THu \ge 0$ for any vector $u \in \mathbb{R}^2$. If you let $\vec{u} = (v,w)$, we get $$ \begin{split} \vec{u}^TH\vec{u} &= 2n \left(v^2 (2n-1) x^{2n-2} + w^2 (2n-1) y^{2n-2}\right) \\ &= 2n(2n-1) \left(v^2 x^{2n-2} + w^2 y^{2n-2}\right) \\ &\ge \vec{0}, \end{split} $$ since it only has square terms, with equality iff $v=w=0$.
Thus, $f$ is convex everywhere.