Show that $f(x_1,x_2)=2x_1+(x_2-x_1^2)^2+(1-x_1)^2$ is coercive

Question

I am trying to show that the function $$f(x_1,x_2)=2x_1+(x_2-x_1^2)^2+(1-x_1)^2$$ is coercive on $\mathbb{R}^2$.

To show the function is coercive, we require $\|(x_1,x_2)\|\rightarrow+\infty\implies f(x_1,x_2)\rightarrow +\infty.$ We proceed by using polar coordinates. This gives \begin{align} f(r,\theta)&=2r\cos\theta+(r\sin\theta-r^2\cos^2\theta)^2+(1-r\cos\theta)^2 \\ &=r^2+1+r^4\cos^4\theta-2r^3\sin\theta\cos^2\theta\\ &\geq r^2+1-2r^3\sin\theta\cos^2\theta. \\ \end{align} However, I am unsure on how to proceed. I require $r\rightarrow +\infty\implies f(r,\theta)\rightarrow +\infty$, but I am unable to show how $r^2+1-2r^3\sin\theta\cos^2\theta\rightarrow +\infty$.

Answer 1

Note that \begin{align} f(x_1,x_2)&=2x_1+(x_2-x_1^2)^2+(1-x_1)^2\\ &= (x_2-x_1^2)^2+1+x_1^2 \end{align}

Let $M>0$. Let $R = \sqrt{M^2 + (M^2 + M^{1/2})^2}$. We show that $|f(x_1, x_2)|\ge M$ whenever $\|(x_1, x_2)\|\ge R$. We split into two cases:

Case 1: if $|x_1| \ge M$, then $|f(x_1, x_2)| \ge 1+M^2 >M$.

Case 2: If $|x_1|<M$ then $$ |x_2|^2 = \| (x_1, x_2)\|^2 - |x_1|^2 \ge R^2 - M^2 =(M^2 + M^{1/2})^2,$$ which implies $$ |f(x_1, x_2)|> (x_2 - x_1^2)^2\ge (|x_2|-M^2)^2 \ge M.$$

That is, for all $M>0$ there is $R$ so that $|f(x_1, x_2)|\ge M$ whenever $\|x_1, x_2)\|\ge R$. Thus $f$ is coercive.