$\liminf$ of a sequence of functions

Question

I know that, given $f(x)$, $\inf f(x)$ is its greatest lower bound. I also know that $\lim \inf f(x)$ is its greatest lower bound when $x \longrightarrow \infty$. For a sequence of functions $f_{n}$, I also know that $\inf f_{n}(x)$ is a function which gives me the $ \inf f_{n}(x)\ \forall\ x \in \mathbb{R}$, so it´s a function (right?) like the one on the right in the image below:

I´d like to know what does $\liminf f_{n}(x)$ mean. I thought it should be the function which returns the infimum of the sequence of functions as $n \longrightarrow \infty$, that is, $ \lim\limits_{n \rightarrow \infty} \inf f_{n}(x)\ \forall\ x \in \mathbb{R}$, but that would just be the "last" function of the sequence, what sounds kind of strange. Can someone help?

Answer 1

$\liminf_n a_n $ gives you the infinimum among the limits of all convergent subsequences of $a_n$.

Example:

1) $a_n=1/n$ then $\liminf_n a_n=0$.

2) $a_n=(-1)^{n}$ then $\liminf_n a_n=-1$. The convergent subsequence $a_{2n+1}$ has this limit.

In the case $\liminf_n f_n(x)$ it gives you a function such that for each $x$ it returns the infimum among the limits of convergent subsequences of the sequence $f_n(x)$, $n=1,2,..$.

Example:

1) $f_n(x)=x/n$ then $\liminf_n f_n(x)=0$

2) $f_n(x)=(-1)^nx^n$ then $$\liminf_n f_n(x)=\begin{cases}\infty&x<-1&\text{The whole sequence goes to }\infty.\\-\infty&x>1&n=2k+1\text{ is a subsequence with this limit}\\0& x\in(-1,1)&\text{The whole sequence goes to }0\\-1&x=1&n=2k+1\text{ is a subsequence with this limit}\\1&x=-1&\text{The whole sequence goes to }1\end{cases}$$