Example from Wikipedia.

$$f(x)=(x_{1}-x_{2})^2$$

Who can it bee shown that the function is not radially unbounded.

$$l=lim_{x \to \infty}min_{||x||=r}[(x_{1}-x_{2})^2]$$

by using $$min_{||x||=r}$$ as a step before taking the limit or done step by step.

who do you read $$min||x||=r[f(x)]$$ is it possible to check first $$(x1,0)$$, $$(0,x2)$$ and at the end $$x1=x2$$ $$(x1,x1)$$?

• This screams "polar coordinates"; let $x_1=r\cos \phi$ and $x_2=r\sin \phi$. – Giuseppe Negro Dec 2 at 14:11

Let $$x=(x_1,x_2)$$ such that $$||x||=r$$ If $$x_1=x_2$$, then $$(x_1-x_2)^2 =0.$$ Hence
$$\min_{||x||=r}[(x_{1}-x_{2})^2]=0.$$
$$l=0.$$