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

Question

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.

Answer 1

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 $-1$.

Answer 2

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

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

Answer 3

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.