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?
Best Answer
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.