Can someone give some examples of $\limsup_{n\to \infty}f_n = \inf_{k\geq 1}\{sup_{i\geq k}f_i(x)\}$?

Question

$$\limsup_{n\to \infty}f_n = \inf_{k\geq 1}\{\sup_{i\geq k}f_i\}$$

Does the sup take the supremum of $\{f_k(x), f_{k+1}(x),...| \text{ all } x\in X\}$ or just at a given $x$?

I know there are some posts explaining this, but can someone provide some simple examples to understand $\limsup_{n\to \infty}f_n$?

For example, $f_n(x)=1/n$, $sup_{n\geq 1}f_n=1$. What is $\limsup_{n\to \infty}f_n$ for this sequence of functions?

Answer 1

Let $(A_n)_{n\in \Bbb N}$ be a bounded sequence of real numbers. That is, the set $\{A_n:n\in \Bbb N\}$ has upper and lower bounds in $\Bbb R.$

For $k\in \Bbb N$ let $B_k=\{A_n:n\geq k\}.$ Since $B_1$ has an upper bound in $\Bbb R$ and $ B_k\subset B_1 $ for every $k\in \Bbb N, $ therefore every $B_k$ has an upper bound in $\Bbb R.$

For $k\in \Bbb N$ let $C_k=\sup B_k. $ Since $B_{k+1}\subset B_k, $ therefore $C_{k+1}\leq C_k$ for every $k.$

Since $B_1$ has a lower bound in $\Bbb R$ and $B_{k+1}\subset B_k\subset B_1, $ therefore $C_k\geq \inf B_1$ for every $k.$

The sequence $(C_k)_{k\in \Bbb N}$ is a decreasing sequence with a lower bound $\inf B_1,$ therefore it has a limit $D.$

Define $\lim \sup_{n\to \infty}A_n=D.$

The notation $\lim\sup_{n\to \infty }A_n$ is an abbreviation for what $D$ actually is, because $$D= \lim_{k\to \infty}C_k=\lim_{k\to \infty}(\sup B_k)=\lim_{k\to \infty} (\sup \{A_n:n\geq k\}).$$

Since $(C_k)_{k\in \Bbb N}$ is a decreasing convergent sequence, therefore $\inf_{k\geq 1}C_k=\lim_{k\to \infty}C_k=D.$

$\bullet \quad $ If $(A_n)_{n\to \infty}$ is a convergent sequence then $\lim_{n\to \infty}A_n=\lim \sup_{n\to \infty}A_n=D.$

$\bullet \quad $ When $(f_n)_{n\in \Bbb N}$ is a sequence of real-valued functions on a common domain $U,$ the notation $f=\lim \sup f_n$ means that $ f(x)=\lim_{k\to \infty} \sup \{f_n(x): n\geq k\}$ for each $x\in U.$

Examples:

Let $A_n=1/n.$ Since $\lim_{n\to \infty}A_n=0,$ therefore $D=0.$

Let $0\leq A_{2n-1}\leq 1$ and $A_{2n}=1+\frac {1}{2n}.$ Then $C_{2k-1}=C_{2k}=1+\frac {1}{2k}$ so $D=\lim_{k\to \infty}C_k=1.$

A sub-sequence of $(A_n)_{n\in \Bbb N}$ is a sequence $(A_{g(n)})_{n\in \Bbb N}$ for some (any) strictly increasing $g:\Bbb N\to \Bbb N.$ Then $D$ is the largest value of a limit of a convergent sub-sequence of $(A_n)_{n\in \Bbb N}.$