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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?

Rodrigo de Azevedo
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clarkson
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3 Answers3

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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.

RITI
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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 $\sqrt{x^2 + y^2} > K$.

K. Miller
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  • What I don't understand is what is this $||x||$ – sam_rox Sep 08 '15 at 17:14
  • The function $|\cdot|$ is called a norm. In this case you are using the Euclidean norm or $2$-norm. For any vector $\mathbf{x} = (x_1,\ldots,x_n) \in \mathbb{R}^n$ the $2$-norm of $\mathbf{x}$ is defined by $|\mathbf{x}| = (x_1^2 + \cdots + x_n^2)^{1/2}$. In two variables $x$ and $y$ you have $|(x,y)| = \sqrt{x^2 + y^2}$. – K. Miller Sep 08 '15 at 17:20
  • @K.Miller SO even for $f(x,y)=x^4+y^4-3xy$ as there are only two variables is the norm = $|(x,y)| = \sqrt{x^2 + y^2}$. For $f(x,y,z)=e^{x^2}+e^{y^2}+e^{z^2}$ as there are three variables is the norm = $\sqrt{x^2+y^2+z^2}$ – clarkson Sep 08 '15 at 17:30
  • No. The norm is just a function that maps vectors to a positive real number. You need to rearrange $f(x,y)$ to show that as $x^2 + y^2$ grows large so to does $f(x,y)$. – K. Miller Sep 08 '15 at 17:35
  • @K.Miller I understand that I have to show function grows as norm grows. But my problem is finding out what is norm. Is the norm for $x^4+y^4-3xy$ is $\sqrt{x^8+y^8}$ – clarkson Sep 08 '15 at 17:53
  • You are confusing the norm of a vector and the function $f$. The norm $|\mathbf{x}|$ provides a notion of size for a vector $\mathbf{x}$. The norm $|\mathbf{x}|$ is always defined in the same way. You need to find a way relating the function $f$ and $|\mathbf{x}|$, so that you can make determinations about the size of $f$ based on the size of $\mathbf{x}$. – K. Miller Sep 08 '15 at 18:55
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For (c), use $e^x \ge 1+x$, so $e^{x^2} \ge 1+x^2$.

marty cohen
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