How to determine Coercive functions 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?
 A: 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.
A: 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 $\sqrt{x^2 + y^2} > K$.
A: For (c),
use
$e^x \ge 1+x$,
so
$e^{x^2} \ge 1+x^2$.
