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From Peressini, Sullivan, Uhl, the mathematics of nonlinear programming,

A function is coercive if

$limlimits_{|x| to infty} f(x) to infty$

and super-coercive if,

$limlimits_{|x| to infty} dfrac{f(x)}{|x|} to infty$

I want to know if $f(x,y) = xlog(x)+ylog(y)$ is a (super) coercive function

If I apply the definition, I get,

$limlimits_{|(x,y)| to infty} xlog(x)+ylog(y)$

or

$limlimits_{|(x,y)| to infty} dfrac{xlog(x)+ylog(y)}{sqrt{x^2 + y^2}}$

But there is no obvious dependence on $|(x,y)|$ for $f(x,y)$. Is there any way I can argue that this function is or isn't coercive.

Let $x = rcos t, y = rsin t$ with $0leq tleq frac pi 2.$ Then:
$$dfrac{f(x,y)}{r} = dfrac{rcos tlog(rcos t)+rsin tlog(rsin t)}{r}=$$
$$=log(r)(cos t+sin t)+cos tlogcos t+sin tlogsin t.$$
Can you take it from here?

You can prove that:
$$g(t) = cos tlogcos t+sin tlogsin t$$ is bounded below when $0<t<pi/2$ by some simple calculus.

Since $f(x,y)=g(x)+g(y)$ where $g(x)=xlog(x)$ you can bound it with a one-dimensional function.

First, presume without loss of generality that $xgeq ygeq0$. Then $left|(x,y)right|=sqrt{x^2+y^2}leqsqrt{2}x$.

Secondly, we have $g(x)$ is bounded below hence $f(x,y)=g(x)+g(y)geq g(x)+C$ for some finite constant $C=min_{xgeq0}g(x)$.

Now we can get
$$lim_{stackrel{left|(x,y)right|rightarrowinfty}{xgeq y}}frac{f(x,y)}{left|(x,y)right|}geqlim_{xrightarrowinfty}frac{g(x)+C}{sqrt{2}x}$$
and you can simply prove that $g(x)$ (hence $g(x)/sqrt{2}$) is (super-)coercive in one dimension.