Let $f:(a,b)\to \mathbb{R}$ be a function and define upper and lower limits of function: $$\lim\limits_{x\to x_0} \sup f(x):=\inf_{r>0}\sup \{f(x): 0<|x-x_0|<r,x\in (a,b)\}$$ and $$\lim\limits_{x\to x_0} \inf f(x):=\sup_{r>0}\inf \{f(x): 0<|x-x_0|<r,x\in (a,b)\}.$$
How to show that $\lim\limits_{x\to x_0} \sup f(x)\geq \lim\limits_{x\to x_0} \inf f(x)$ by definition? I have tried but I stucked.
Would be thankful for your help!
Best Answer
Let $L(r)=\inf_{x \in B(x_0,r)\setminus \{x_0\}} f(x), U(r)=\sup_{x \in B(x_0,r)\setminus \{x_0\}} f(x)$.
Note that $L$ is non increasing and $U$ is non decreasing and $L(r) \le U(r)$.
The following coarse schematic might help with the following inequalities:
Suppose $0 < r \le r'$, then $L(r') \le L(r) \le U(r)$. Suppose $0 < r \ge r'$, then $L(r') \le U(r') \le U(r)$.
In particular, we always have $L(r') \le U(r)$.
In particular, $\lim_{r' \to 0} L(r') = \sup_{r'> 0} L(r') \le U(r)$ and so $\sup_{r'> 0} L(r') \le \inf_{r>0} U(r) = \lim_{r \to 0} U(r)$.