Prove that $\overline{ \lim} ~u_n\overline{\lim}~v_n\geq \overline{\lim}~u_nv_n$

Question

Problem: Let $\{u_n\}_n$ and $\{v_n\}_n$ be two bounded sequences and $u_n>0, v_n>0$ for all $n\in \mathbb{N}$. Prove that $$\overline{\lim }~u_n\cdot \overline{\lim}~v_n\geq \overline{\lim}~u_n v_n.$$

Progress:
Since $\{u_n\}_n$ and $\{v_n\}_n$ are bounded , there exis $K_1,K_1>0$ such that $|u_n|\leq K_1$ and $|v_n|\leq K_2$ for all $n\in \mathbb{N}$. What can I do next to prove that result?

Does $\bar{\lim}$ just means $\lim_{n\to\infty}$? – Yuta May 01 '19 at 06:30 — Yuta, May 01 '19 at 06:30
$\overline{\lim} u_n$ means uppler limit of ${u_n}_n$@Yuta – MKS May 01 '19 at 06:32 — MKS, May 01 '19 at 06:32
Yes, $\overline{\lim} u_n=\limsup$ @Yuta – MKS May 01 '19 at 06:34 — MKS, May 01 '19 at 06:34

score 0 · Answer 1 · answered May 01 '19 at 06:32

0

For $n \geq m$ we have $u_ny_n \leq (\sup \{u_k: k \geq m\}) (\sup \{v_k: k \geq m\})$. Taking sup over $n$ we get $(\sup \{u_kv_k: k \geq m\}\leq (\sup \{u_k: k \geq m\}) (\sup \{v_k: k \geq m\})$. Now take limit as $m \to \infty$.

answered May 01 '19 at 06:32

Kavi Rama Murthy

311,013

Prove that $\overline{ \lim} ~u_n\overline{\lim}~v_n\geq \overline{\lim}~u_nv_n$

1 Answers1