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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 $x1=x2$ $(x1,x1)$?

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  • $\begingroup$ This screams "polar coordinates"; let $x_1=r\cos \phi$ and $x_2=r\sin \phi$. $\endgroup$ – Giuseppe Negro Dec 2 at 14:11
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Let $x=(x_1,x_2)$ such that $||x||=r$ If $x_1=x_2$, then $(x_1-x_2)^2 =0.$ Hence

$$ \min_{||x||=r}[(x_{1}-x_{2})^2]=0.$$

This gives

$$l=0.$$

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