Let $f:\mathbb{R}\to\mathbb{R}\cup\{-\infty, +\infty\}$ be a function. My definition of coervit is that $f$ is coercive if $$ (D)\qquad \lim_{|x|\to +\infty} f(x) = +\infty.$$
I am trying to justify some equivalent definitions I found surfing on the web. For example:
a) $f(x)\ge |x|$;
It is clear that, for $|x|\to +\infty$, it follows that $f(x)\to +\infty$, too, and then $a)\implies (D)$. However I am not able to understand how to prove (if it holds!) that $(D)\implies a)$.
b) $\displaystyle\lim_{|x|\to +\infty} \frac{f(x)}{|x|} =\alpha>0$. This one is of my particular interest, but I am not able to prove the equivalence with $(D)$. Intuitively I justify it saying that "infinities with same order", something like $\displaystyle\lim_{|x|\to +\infty} \frac{\alpha x}{x}$, but I am not so sure.
Could someone please help me?
Thank you in advance!
The definitions are not equivalent for example $f(x) $ = $\log(|x|)$ and $-1$ at $0$ satisfies $D$ but does not satisfy $a)$ since $|x|>f(x)$ for all $x$ .$f(x)$ also does not satisfy $b)$ since $\log(|x|)/ |x| $ goes to $0$ as $|x|$ goes to infinity