Determine whether $f$ is coercive or not.

Question

I have difficulty in determine whether those functions are coercive or not. This is part of an exercise in Amir Beck's book "Introduction to Nonlinear Optimization".

a) $f(x_1,x_2)=2x_1^2-8x_1x_2+x_2^2$.

b) $f(x_1,x_2)=4x_1^2+2x_1x_2+2x^2_2$.

C) $f(x_1,x_2)=x_1^3+x_2^3+x_3^3$.

I'm trying to estimate $f(x,y)$ to the norm $\|x\|$.

For the definition of coercice function, the following is known

Let $f : \mathbb{R}^n \to \mathbb{R}$ is continuous function. The function f is called coercive if $\lim\limits_{\|x\| \to \infty} f(x) = \infty $.

Any help would be highly appreciated.

Answer 1

So coercice function are functions that go to infinity regardless of the direction. Therefore $(a)$ is not coercice function because along the line $(x_1,x_2)=(t,t(4+\sqrt{14}))$ is constantly zero. Although $||(x_1,x_2)||\to\infty$ as $t\to \infty$. Next $(b)$ function is coercice. Its graph is a paraboloid which grows in every direction. In point $(c)$ you can see that it is enough to consider function along the line $(x_1,x_2,x_3)=(t,-t,0)$.

Answer 2

For $a)$ and $b)$,you can use the Lemma2.42 in Amir Beck's book "Introduction to Nonlinear Optimization",Lemma 2.42 is

Let $f(\mathbf{x})=\mathbf{x}^{\top}A\mathbf{x}+2\mathbf{b}^{\top}\mathbf{x}+c$,where $A \in \mathbb{R}^{n \times n}$ is symmetric,$\mathbf{b} \in \mathbb{R}^n$,and $c \in \mathbb{R}$.Then $f$ is coercive if and only if $A \succ0$

For $a)$,$A=\begin{bmatrix}2 & -4\\-4 & 1\end{bmatrix}$,${\rm trace}(A)=2 > 0,{\rm det}(A)=-14 < 0$,$A$ is indefinite,therefore (a) is not coercive function

For $b)$,$A=\begin{bmatrix}4 & 1\\1 & 2\end{bmatrix}$,${\rm trace}(A)=6 > 0,{\rm det}(A)=7> 0$,$A \succ$ ,therefore (b) is coercive function