lim sup of two sequences

Question

Let $(a_n)_{n \in\ \mathbb{N}}$ a bounded sequence in $\mathbb{R}$. For $n \in \mathbb{N}$ let

$$v_n=\sup\{a_k; ~k \geq n\},\quad u_n=\inf\{a_k; ~k \geq n\},\quad s_n=\sup\{|a_k-a_l|; ~k,l \geq n\}$$

Show that:

(a) $s_n=v_n-u_n$ for all $n \in\ \mathbb{N}$

(b) $\lim\limits_{n \rightarrow \infty} \sup a_n- \lim\limits_{n \rightarrow \infty} \inf a_n = \inf\{s_n;~ n \in\ \mathbb{N}\}$

Unfortunately, i have no idea how to start...

Answer 1

a) Let n > 1 be given. Then for all i > n and j > n: v(n) >= a(i) and u(n) =< a(j) so: v(n) - u(n) >= a(i) - a(j). But we also have: v(n) >= a(j) and u(n) =< a(i). So we have: v(n) - u(n) >= a(j) - a(i). Thus v(n) - u(n) >= max(a(i) - a(j), a(j) - a(i)) = /a(i) - a(j)/. So v(n) - u(n) >= sup{ /a(k) - a(l)/: k, l > n} = s(n). So: v(n) - u(n) >= s(n). Now fix j with j > n. We have: a(i) - a(j) =< /a(i) - a(j)/ for all i > n. So a(i) - a(j) =< sup{/a(i) - a(j)/: i > n} =< sup{/a(i) - a(j)/: i, j > n} = s(n). So v(n) - a(j) =< s(n) and so -a(j) =< s(n) - v(n). So: inf{-a(j): j > n} =< s(n) - v(n). Bu -inf{a(j): j > n} = inf{-a(j): j > n}. So: -u(n) =< s(n) - v(n). So v(n) - u(n) =< s(n). This means s(n) = v(n) - u(n) for all natural n's.