ANA1_L9

• $\lim {\sup }_{n\to \infty }{p}_n=\infty ⟺(p_n)\text{ is not bounded above.}$
• A is a subset of B. inf A >= inf B, sup A ⇐ sup B. sequence of suprema form weakly decreasing sequence, and the sequence of infima form a weakly increasing sequence.
  Quick exercise: show that $i_k\le s_{k^{\prime }}$ for all $k,k^{\prime }\in \mathbb{N}$.
  $i_1\le i_2\le i_3\le \cdots \le s_3\le s_2\le s_1$.
• Note that $\lim {\sup }_{n\to \infty }{p}_n=\infty$ means that every $s_i$ is $\infty$.
• proof: liminf ⇐ limsup
  Pick any $s_k$. $i_m\le s_k$ for all $m$. Thus, $s_k$ is an upper bound for the set of all infima. Thus, $\lim {\inf }_{n\to \infty }{p}_n=\sup \{i_m\}\le s_k$. Since $s_k$ was arbitrary, the preceding statement is valid for all $k$. Thus, $\lim {\inf }_{n\to \infty }{p}_n$ is a lower bound for the set of all suprema. Thus, $\lim {\inf }_{n\to \infty }\le \inf \{s_k\}=\lim {\sup }_{n\to \infty }{p}_n$.
• proof that liminf = limsup only when the sequence converges.

Claim: $(p_n)$ converges $⟺$ limsup = liminf

Suppose $(p_n)\to p$.
pf of ⇒
Cases:

• $p\in \mathbb{R}$
• $p\in \{\infty ,-\infty \}$

Case 1:
Let $ϵ>0$.
There exists a $N\in \mathbb{N}$ such that for all $n>N$, $|p_n-p|<ϵ$.
Thus, $s_n$ and and $i_n$ must lie in $(p-ϵ,p+ϵ)$.
Thus, $\inf s_n$ and $\sup i_n$ must also lie in the same interval.
Since $ϵ$ was arbitrary, this proves that $\lim {\sup }_{n\to \infty }{p}_n=p=\lim {\inf }_{n\to \infty }{p}_n$.

pf of ⇐
same cases.

Case 1:
limsup = liminf = p
i.e, sequence of suprema of tails converges to p (from above monotonically)
sequence of infima of tails converges to p (from below monotonically)
We have to show that $(p_n)\to p$.
let $ϵ>0$.
Find $N$ such that for all $n>N$
$0\le s_n-p<ϵ⟹p_n-p<ϵ$
$0\le p-i_n<ϵ⟹p-p_n<ϵ$.
$⟹|p-p_n|<ϵ$.

Proof of theorem 3.19

Reconciling Rudin and Kulkarni