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

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.

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 by some simple calculus.

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.

• $g(y)$ can be negative for small $y,$ so the lower bound may not hold. – dezdichado Dec 3 '18 at 3:45
• Thanks, need bounded below instead – obscurans Dec 3 '18 at 3:50