Is $f(x,y) = x\log(x)+y\log(y)$ a coercive function?

Question

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

A function is coercive if

$\lim\limits_{\|x\| \to \infty} f(x) \to \infty$

and super-coercive if,

$\lim\limits_{\|x\| \to \infty} \dfrac{f(x)}{\|x\|} \to \infty$

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

If I apply the definition, I get,

$\lim\limits_{\|(x,y)\| \to \infty} x\log(x)+y\log(y)$

or

$\lim\limits_{\|(x,y)\| \to \infty} \dfrac{x\log(x)+y\log(y)}{\sqrt{x^2 + y^2}}$

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.

Answer 1

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

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

Answer 2

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

First, presume without loss of generality that $x\geq y\geq0$. Then $\left\|(x,y)\right\|=\sqrt{x^2+y^2}\leq\sqrt{2}x$.

Secondly, we have $g(x)$ is bounded below hence $f(x,y)=g(x)+g(y)\geq g(x)+C$ for some finite constant $C=\min_{x\geq0}g(x)$.

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