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:

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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?

  • $\begingroup$ Fix $x$ and take the sequence $n \mapsto f_n(x)$ and take the $\liminf$ of that. That is, $\liminf_n f_n(x) = \lim_{n \to \infty} \inf_{k \ge n} f_k(x)$. I don't understand what you mean by the last function in the sequence. For example, take $f_n(x) = x^n$ on $[0,1]$. Then $\liminf f_n(x) = 1_{\{1\}}(x)$. $\endgroup$
    – copper.hat
    Commented Feb 2, 2015 at 16:05
  • $\begingroup$ I meant that, while for $\inf f_{n}(x)$ we examine all possible functions $f_{n}(x)$ of the sequence and take their infimum in each point $x$, in $\lim_\limits{n \to \infty} \inf_{k \ge n} f_k(x)$ this would be analogous to extracting all functions $f_{k<n}$ of the analysis, so that $\lim_\limits{n \to \infty} f_n(x)$ would leave us with only one function, the infimum being the function itself. In, $f_n(x) = x^n$ on [0,1], I would interpret the $\liminf f_n(x)$ as $\lim_\limits{n \to \infty} x^n$ on [0,1], so that $\liminf f_n(x) = 0$ on [0,1) and $\liminf f_n(x) = 1$ on [1,1] $\endgroup$
    – why
    Commented Feb 2, 2015 at 17:19
  • $\begingroup$ My example was bad in that it satisfied $f_{n+1} \le f_n$. Another example would be $f_n(x) = \sin (nx)$. With this example, $\liminf_n f_n$ is a little more complicated and no $f_n$ matches the $\liminf$. $\endgroup$
    – copper.hat
    Commented Feb 2, 2015 at 17:23

1 Answer 1


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


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 $f_n(x)$, $n=1,2,..$.


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

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

  • 2
    $\begingroup$ Thanks a lot. I couldn't find any explanation such as yours before. Could you give me a hint of where to find one in a book or lecture notes? $\endgroup$
    – why
    Commented Feb 2, 2015 at 17:24

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