Is this function coercive

Question

I am trying to figure out whether $f(x_1,x_2)=(x_2-2x_1)^4+64x_1x_2$ is coercive. I.e.: as $\Vert x\Vert \to \infty$, does $f\to \infty$? $x\in \mathbb{R}^2$

My instinct is yes, but I've been unable to show it explicitly. Any advice on how to proceed?

Thank you!

Answer 1

For this type of problem, the blow-up resolution of singularity method generally works - and here it is very simple as we will see.

Let $A$ the domain where $|x_1| \ge |x_2|$ and B the domain where $|x_2| \ge |x_1|$; it is enough to prove that $f \to \infty$ when $||x|| \to \infty$ in $A$ and $B$ separately.

For $A$ we use the transformation $\frac{x_2}{x_1}=u, \frac{1}{x_1}=v$ so $|u| \le 1, |v| \to 0, x_1=\frac{1}{v}, x_2=\frac{u}{v}$ when $||x|| \to \infty$ $f(x_1,x_2)$ becomes $f_A(u,v)=\frac{(u-2)^4+64uv^2}{v^4}$.

But $1 \le (u-2)^4 \le 81$ clearly implies that $f_A(u,v) \to \infty$ as $|u| \le 1, v \to 0$ since the numerator is dominated by $(u-2)^4$ when $v \to 0$ and that stays away from $0$

For $B$ we use the transformation $\frac{x_1}{x_2}=u, \frac{1}{x_2}=v$ so $|u| \le 1, |v| \to 0, x_2=\frac{1}{v}, x_1=\frac{u}{v}$ when $||x|| \to \infty$ and $f$ becomes $f_B(u,v)=\frac{(1-2u)^4+64uv^2}{v^4}$; here we have a little more work since $1-2u$ can get small, but then $u$ has to be around $.5$ and then $64u/v^2 \to \infty$.

Formally, one can split $B$ into the parts where $.25 <u<.75$ and the rest of $|u| \le 1$ and call those $B_1,B_2$ and show that the limit goes to infinity on $B_1, B_2$.

In $B_2$ as in the $A$ case $(1-2u)^4$ stays away from zero and dominates the numerator, while in $B_1$, $64uv^2> 16v^2, (1-2u)^4 \ge 0$ so $f_B(u,v) > \frac{16}{v^2} \to \infty, v \to 0$ so we are done.

Answer 2

Using polar coordinates representation, let $x_1=r\cos(\theta)$ and $x_2=r\sin(\theta)$. Then, we have $$f(x_1,x_2)=(x_2-2x_1)^4+64x_1x_2=r^4(\sin(\theta)-2\cos(\theta))^4+64r^2\cos(\theta)\sin(\theta)=:g(r,\theta).$$

Since $0\le (\sin(\theta)-2\cos(\theta))^4\le 3^4$ and $-1\le\cos(\theta)\sin(\theta)\le 1 $, we see that $ \lim_{r\to\infty} g(r,\theta)= \infty.$ Equivalently, $\lim_{\|x\|\to\infty} f(x_1,x_2)= \infty.$