I am reading a real analysis textbook, and need help clarifying/unpacking the definition of $\limsup$ and $\liminf$.

Given a sequence $\{a_n\}$ of real numbers,$$\limsup_{n \to \infty} a_n = \inf_n \sup_{m \ge n} a_m, \quad \liminf_{n \to \infty} a_n = \sup_n \inf_{m \ge n} a_m.$$ We use analogous definitions when we take a limit along the real numbers. For example,$$\limsup_{y \to x} f(y) = \inf_{\delta > 0} \sup_{|y - x| < \delta} f(y).$$

Any help would be well-appreciated. Thanks!

  • 7
    $\begingroup$ What exactly do you need clarified? $\endgroup$ Aug 16 '16 at 4:56
  • 1
    $\begingroup$ Note that $n \mapsto \sup_{m \ge n} a_m$ is a non increasing sequence. $\endgroup$
    – copper.hat
    Aug 16 '16 at 4:57
  • $\begingroup$ One alternative (and equivalent) definition of $\limsup_{t\to\infty}f(t)$ is that it is the largest possible limiting value of $f(t_k)$ over all subsequences $\{t_k\}$ for which $t_k\to\infty$ and $\lim_{k\to\infty} f(t_k)$ exists (possibly being $\infty$ or $-\infty$). For $\liminf$, it is similar but "smallest possible." For example, $\cos(t)$ has no limit, but $\limsup_{t\to\infty} \cos(t) = 1$ and $\liminf_{t\to\infty} \cos(t)=-1$ since we can find subsquences with limits that achieve these, and no limiting value of $\cos(t)$ over any subsequence can be larger than 1 or smaller than -1. $\endgroup$
    – Michael
    Aug 17 '16 at 0:17
  • $\begingroup$ You can show these "largest possible" and "smallest possible" limiting values always exist (possibly being $\infty$ or $-\infty$) and are achievable over some particular subsequences. $\endgroup$
    – Michael
    Aug 17 '16 at 0:19

$\limsup, \liminf$ definitions in most textbooks are cryptic (like the one in your question) in the sense that they are written very concisely using symbols and are normally not accompanied with some detailed explanation (note that examples are not substitute for explanation, rather they complement it). The best treatment (in my opinion) of these concepts of $\limsup, \liminf$ is in Hardy's classic textbook A Course of Pure Mathematics. I present the same below.

Let $a_{n}$ be a sequence and consider any number $K$. We have three mutually exclusive and exhaustive possibilities:

  1. There is a positive integer $m$ such that $a_{n} \leq K$ for all $n \geq m$ i.e. $a_{n}$ does not exceed $K$ for all large values of $n$.
  2. There is a positive integer $m$ such that $a_{n} \geq K$ for all $n \geq m$ i.e. $K$ does not exceed $a_{n}$ for all large values of $n$.
  3. Given any positive integer $m$ there are two integers $n_{1}, n_{2}$ both greater than $m$ such that $a_{n_{1}} > K$ and $a_{n_{2}} < K$ i.e. there are infinitely many values of $a_{n}$ greater than $K$ and also infinitely many values of $a_{n} < K$.

If $a_{n} = (-1)^{n}\left(1 + \dfrac{1}{n}\right)$ and $K = 1.5$ then the first possibility occurs. If $K = -1.1$ then second possibility occurs and if $K = 0$ then the third possibility occurs. You should convince yourself that these numbers fit the possibilities described above.

Hardy calls the numbers of type $K$ in the first possibility as superior numbers (with respect to sequence $a_{n}$). And those numbers $K$ which fit the second possibility are called inferior numbers. The numbers $K$ fitting third possibility are called intermediate numbers.

It is easily seen via these definitions that

  1. If $K$ is superior then any number greater than $K$ is also superior.
  2. If $K$ is inferior then any number less than $K$ is also inferior.
  3. Any intermediate number is less than any superior number and greater than any inferior number.

The next step is to consider the set of all numbers which are superior with respect to sequence $a_{n}$ i.e. $$A = \{K\mid K\text{ is superior with respect to }a_{n}\}$$ If the sequence $a_{n}$ is bounded then the set $A$ above is bounded below and hence there is a greatest lower bound for $A$. We define $$\limsup a_{n} = \inf A = \inf\, \{K\mid K\text{ is superior with respect to }a_{n}\}$$ so a $\limsup$ is almost like the smallest number superior to sequence $a_{n}$. For the example sequence we have $\limsup a_{n} = 1$. Similarly $$\liminf a_{n} = \sup\,\{K\mid K \text{ is inferior with respect to }a_{n}\}$$ so that $\liminf a_{n}$ is almost the greatest number inferior to $a_{n}$. For the example given above $\liminf a_{n} = -1$.

What do we gather from the above discussion? Well we have the following two obvious properties for $M = \limsup a_{n}$:

  1. Any number greater than $M$ is superior so if $\epsilon > 0$ then $a_{n} \leq M + \epsilon$ for all large values of $n$. (Why? Because $M$ is the infimum of all superior numbers and by definition we have a superior number as close to $M$ as possible and thus we have a superior $K$ with $M \leq K < M + \epsilon$. Now $M + \epsilon$ is greater than a superior number $K$ and hence $M + \epsilon$ is also superior. And therefore $a_{n}$ does not exceed $M + \epsilon$ for all large $n$.)
  2. Any number less than $M$ is not superior so if $\epsilon > 0$ then $a_{n} > M - \epsilon$ for infinitely many values of $n$. (Why? $M - \epsilon$ is less than $M$ and hence it is not superior. By definition it is either inferior or intermediate. In both cases we have $a_{n}$ exceeding it infinitely many times (see definition of intermediate/inferior numbers))

We have similar properties for $m = \liminf a_{n}$ which you can formulate yourself. Note that in practical applications of the concept of $\limsup$ and $\liminf$ we need the above properties and not their actual definitions.

  • $\begingroup$ Nice written (+1) $\endgroup$
    – Mark Viola
    Feb 9 '17 at 6:35
  • $\begingroup$ very nice explanation. Even though I have a decent understanding of the concepts already, this still illuminated some fine points for me $\endgroup$
    – peek-a-boo
    Jun 4 '20 at 17:48
  • $\begingroup$ @peek-a-boo: thanks man! This is one of primary reasons I love Hardy's book mentioned in this answer. $\endgroup$ Jun 5 '20 at 2:26

Perhaps an example would be illuminating? Let's figure out $\limsup(a_n)$ for the following sequence:

$$a_n = \begin{cases} 1 + 1/n &\mbox{if } n \equiv 0 \\ -1 - 1/n& \mbox{if } n \equiv 1 \end{cases} \pmod{2}$$

For a given $n$, the quantity $\displaystyle \sup_{m \geq n}a_m$ is simply the largest term in the sequence after the first $n$ terms are thrown away. For any $n$, it is easy to see in the sequence above that this value will be the first even-indexed term that was not thrown away. As we let $n$ go to infinity, you can check that the even-indexed terms limit to $1$, or in other words $\displaystyle \sup_{m \geq n} a_m$ limits to $1$, or in other words $\limsup(a_n) = 1$.

Put another way, if $\limsup(a_n) = x$, this means that $x$ is the smallest number such that, for any given $\delta > 0$, only a finite number of terms in the sequence are larger than $x + \delta$.

Likewise, $\liminf(a_n) = x$ means that $x$ is the largest number such that, for any given $\delta > 0$, only a finite number of terms in the sequence are smaller than $x - \delta$. You can check that, in the example sequence above, $\liminf(a_n) = -1$.

Finally, notice that if $\lim(a_n)$ exists, then it will agree with $\liminf(a_n)$ and $\limsup(a_n)$.

I don't know if this graphic will be helpful to you, but I have it in mind pretty much every time I think about these concepts (from Wikipedia): enter image description here


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