A continuous function $f(x)$ that is defined on $R^n$ is called coercive if $\lim\limits_{\Vert x \Vert \rightarrow \infty} f(x)=+ \infty$.
I am finding it difficult to understand how the norm of these functions are computed in order to show that they are coercive.
$a) f(x,y)=x^2+y^2
\\b)f(x,y)=x^4+y^4-3xy\\c)f(x,y,z)=e^{x^2}+e^{y^2}+e^{z^2}$
To show that they are coercive I have to show that as norm goes to infinity the function too should go to infinity right?
$\endgroup$ 33 Answers
$\begingroup$To prove that the function is coercive, we need to show that its value goes to $\infty$, as the norm becomes $\infty$.
1)$$ f(x,y)=x^2+y^2= \infty \\ as \left \| x \right \|\rightarrow \infty $$i.e. $||x||=\sqrt(x^2+y^2)$
Hence , $f(x)$ is coercive.
2)$$ f(x,y)=x^4+y^4- 3xy \\ \because ((x+y)^2-(x^2+y^2))=3xy (\frac{2}{3}) \\f(x,y)=x^4+y^4-(\frac{3}{2})( (x+y)^2)-(x^2+y^2)) \\ \leq x^4+y^4 + (\frac{3}{2})(x^2+y^2)\\ \leq (x^2+y^2)^2 + (\frac{3}{2})(x^2+y^2) \\ \therefore f(x,y)=\infty \\ as \left \| x \right \|\rightarrow \infty $$i.e. $||x||=\sqrt(x^2+y^2)$
Hence , $f(x)$ is coercive.
3)$$ f(x,y,z)=e^{x^{2}} + e^{y^{2}}+ e^{z^{2}} \\ \approx (1+x^{2})+(1+y^{2})+(1+z^{2}) = \infty $$$$\\ as \left \| x \right \|\rightarrow \infty $$i.e. $||x||=\sqrt(x^2+y^2+z^2)$
Hence , $f(x)$ is coercive.
$\endgroup$ $\begingroup$Consider the first function $f(x,y) = x^2 + y^2$. This function can be written in terms of vectors as $f(\mathbf{x}) = \|\mathbf{x}\|^2$. Now you can see that $f(\mathbf{x}) \to \infty$ as $\|\mathbf{x}\| \to \infty$.
Here is a hint for the second function. Use the inequality $-\frac{3}{2}(x^2 + y^2) \leq -3xy$ to derive a lower bound for $f(x,y)$. Show that for any $M > 0$ there exists a number $K > 0$ such that $f(x,y) > M$ whenever $\sqrt{x^2 + y^2} > K$.
$\endgroup$ 6 $\begingroup$For (c), use $e^x \ge 1+x$, so $e^{x^2} \ge 1+x^2$.
$\endgroup$
