# A question on $\liminf$ and $\limsup$

Let us take a sequence of functions $f_n(x)$. Then, when one writes $\sup_n f_n$, I understand what it means: supremum is equal to upper bound of the functions $f_n(x)$ at every $x$. Infimum is defined similarly. Then when one writes $\lim \sup f_n$, then I understand following: There are convergent subsequences of $f_n$, let us call them as $f_{n_k}$ and their limits as a set $E$. Then, $$\limsup f_n = \sup E$$

First question: Are these definitions right?

Second question: I do not understand the notion of convergent subsequences. What does it mean really? And why they are necessary at the first place, why they are important?

Thanks.

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A good way to think about subsequences is the following: say someone gives you a "random" bounded sequence. What does it "converge" to? Likely nothing, it'll just be a random bunch of numbers. But, because of the Bolzano-Weierstrauss theorem, you can "filter out" this random sequence so that it converges to something by carefully picking some of the entries. You can do this basically because you have an infinite amount of numbers that have 'nowhere else to go' (because they're bounded). –  icurays1 Nov 12 '12 at 18:18
@icurays1: Thanks! –  oeda Nov 12 '12 at 20:51

Your definitions are good, but I usually prefer this definition of $\limsup$: $$\limsup_{n\to\infty} y_n=\lim_{n\to\infty}\sup_{k\ge n} y_k$$ the point being that $\sup_{k\ge n} y_k$ decreases with $n$ (it is a supremum over smaller and smaller sets), so it has a limit, or converges to $-\infty$.

It is a good exercise to show that there always exists a subsequence converging to $\limsup y_n$, and that the limit of any convergent subsequence is at most as large as $\limsup y_n$.

For functions, the $\limsup$ is defined pointwise, just as for limits.

Oh, and why are subsequences needed? A trivial example is $y_n=(-1)^n$. It has no limit, but the superior limit is $1$, and the inferior limit is $-1$.

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Last part of your answer is really clarified the thing. Thanks :) –  oeda Nov 12 '12 at 18:21

Question $(2)$:
I think one thing that gets confusing, when referring to subsequences of a sequence, is the use of indices to denote them, e.g. when using a "double subscript" to denote a subsequence $f_{n_k}(x)$ of $f_n(x)$, or a subsequence $\{a_{n_k}\}$ of $\{a_n\}$.

Perhaps an example of a subsequence would help illustrate what is meant by a subsequence:

Let, e.g., $a_n = 1, \frac{1}{2}, \frac{1}{3}\dots \frac{1}{n}$.

We can define a subsequence $\{a_{n_k}\} \text{ of}\; \{a_n\}$ where $a_{n_k}$ is defined by $a_{2n} = \frac{1}{2}, \frac{1}{4}, \dots \frac{1}{2n}$.

The important thing to note is that a sequence $\{b_n\}$ is a subsequence of $\{a_n\}$ if and only if for each $j\ge 1$ both of the following hold:

$(1)$ $b_j$ is one of the terms of the sequence $a_1, a_2, \dots$ and
$(2)$ the term $b_{j+1}$ appears later than $b_j$ in the sequence $a_1, a_2, \dots$

Note:
It may happen that sequence $\{a_n\}$ does not converge, while some subsequence(s) of $\{a_n\}$ does converge. But a sequence $\{a_n\}$ converges if and only if every subsequence of $\{a_n\}$ converges.

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amWhy, thank you very much. –  oeda Nov 13 '12 at 7:45

1 For any $x$ there are $n_{k(x)}$ such that $$\limsup f_n(x) = \lim f_{n_{k(x)}}(x)$$

2 Maybe $f_n(x)$ can not converges. But, there are subindices $n_k$ such that $f_{n_k}(x)$ converges.

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