limit infimum as $x\to \infty$ If $$\liminf_{x\to \infty}f(x)>M,$$ for some $M>0$, where $f$ is continuous function on $\mathbb R$.
Does this imply that: there exist $x_{o}$ such that for all $x\geq x_{o}$ we have $f(x)>M$? If so, how I can prove it?
I also have another question, just to be sure: If $f(x)\leq g(x)$ for all $x\in \mathbb R$, both are continuous on $\mathbb R$, then
 $\liminf_{x\to\infty}f(x)\leq \liminf_{x\to\infty} g(x)$
 A: *

*Let $L = \liminf_{x\to\infty}f(x) = \sup_{N>0}\inf_{x\ge N}f(x)$.  Then we have $L>M$, so $M$ isn't a supremum of $\left\{\inf_{x\ge N}f(x)\mid N>0\right\}$.  Therefore, $\exists x_0>0$ such that $\inf_{x\ge x_0}f(x)>M$.  Hence, $f(x) > M \;\forall x \ge x_0$.

*The answer is yes.  We're given
$$f(x)\leq g(x) \;\forall x \in \Bbb R.\tag{1}$$
Let $N>0$.  We first take infimum on the LHS over all $x\ge N$.
$$\inf_{x\ge N}f(x) \le g(x) \;\forall x \ge N\tag{2}$$
Then $\inf_{x\ge N}f(x)$ is a lower bound of $\left\{g(x)\mid x \ge N\right\}$, while $\inf_{x\ge N}g(x)$ is the greatest lower bound of $\left\{g(x)\mid x \ge N\right\}$.  Therefore, one has
$$\inf_{x\ge N}f(x) \le \inf_{x\ge N}g(x).\tag{3}$$
From (1) to (3), we realize that we can take infimum on both sides.  Similarly, we can take supremum over $N>0$ on both sides.
$$\sup_{N>0}\inf_{x\ge N}f(x) \le \sup_{N>0}\inf_{x\ge N}g(x)\tag{4}$$
In other words, we have the statement in the OP.
$$\liminf_{x\to\infty}f(x) \le \liminf_{x\to\infty}g(x)$$

