Linear function can be coercive?

Question

we know that $\,f:\mathbb R^n \to \mathbb R\,$ is coercive if for any $\, x\in \,\mathbb R^n\,$ $%there exists a constant \,c\in \mathbb R\, such that$ $ %\big\langle \,f\left(x\right),\,x\,\big\rangle \ge c\, \left\| \, x \, \right\|^2 %= c \,\big\langle \,x,\,x\,\big\rangle \left\| \, x \, \right\| \to \infty \implies \left\lvert \,f\left(x\right)\,\right\rvert \to +\infty.$

let $\\f(x)=c^Tx$, this function satisfies above condition still it is not coercive, why ?

Answer 1

Suppose $n>1.$ For any $c\in\mathbb R^n$, there is a non-zero $u\in\mathbb R^n$ such that $c^T u = 0$. Now let $x_n = nu$, and observe that $\|x_n\|\to\infty$ as $n\to\infty$, but yet $f(x_n)=0$ for all $n$.

Answer 2

Your definition of coercive is wrong.

$$\,f:\mathbb R^n \to \mathbb R\, \text{ is coercive if } f(x) \to \infty \text{ as }\left\| \, x \, \right\| \to \infty.$$