Relation between coercivity and boundness from below

Question

Let $f:\mathbb{R}\to\mathbb{R}$ be a function such that $$f(x)\ge \| x\|.$$

It is clear that, if $\|x\|\to +\infty$, it follows that $f(x)\to +\infty$, so $f$ is coercive. However, the inequality $f(x)\ge \| x\|$ also states that $f$ is bounded from below.

My question is: it is true that all coercive funtions are bounded from below (I don't think so..) or it a special case?

I am sorry for my possibly dumb question, but I am a bit in trouble right now.

Thank you in advance!

Answer 1

If $f$ is additionally lower semicontinuous, it is also bounded from below. Indeed, there is some $\delta \ge 0$ such that $$ f(x) \ge 0 \qquad\forall \|x\| \ge \delta.$$ Moreover, on the compact set $[-\delta,\delta]$, the lower semicontinuous function $f$ attains its infimum. Therefore, there is $x_0 \in [-\delta,\delta]$ with $$ f(x) \ge f(x_0) \qquad\forall x_0\in [-\delta,\delta].$$ This shows that $f$ is bounded from below by $\min\{0,f(x_0)\}$.

Answer 2

Since you know that $f(x)>|x|$ you know that $f(x)>-1$ so it’s bounded from below , In general coercive function need not be bounded below for example you can let your function be $\log(|x|)$ and $0$ for $x=0$ it’s coercive but not bounded below