A useful lemma: Let \(\{x_n\}\) be a real sequence with \(\liminf_{n\to\infty}x_n\), \(\limsup_{n\to\infty}x_n\in\mathbb R\). Show that for any \(\epsilon>0\) there exists some \(N\in\mathbb N\) so that \[ \left(\liminf_{n\to\infty}x_n\right)-\epsilon<x_n< \left(\limsup_{n\to\infty}x_n\right)+\epsilon \qquad\text{for all }n\ge N. \] What does this result imply if \(\limsup_{n\to\infty}x_n<L\)?
Solution
1. Upper bound: eventually \(x_n < \limsup x_n + \epsilon\).Proof: let \(L = \limsup x_n = \inf_N \sup_{n \ge N} x_n\) (finite by assumption).For any \(\epsilon > 0\), \(L + \epsilon\) is not a lower bound of the tail suprema… more precisely, \(L = \inf_N S_N\) with \(S_N = \sup_{n\ge N}x_n\), so there is \(N_1\) with \(S_{N_1} < L + \epsilon\).Then for all \(n \ge N_1\), \(x_n \le S_{N_1} < L + \epsilon\).
2. Lower bound: eventually \(x_n > \liminf x_n - \epsilon\).Proof: let \(\ell = \liminf x_n = \sup_N \inf_{n\ge N} x_n\).For any \(\epsilon > 0\), \(\ell - \epsilon\) is not an upper bound of the tail infima, so there is \(N_2\) with \(\inf_{n\ge N_2} x_n > \ell - \epsilon\); hence for all \(n \ge N_2\), \(x_n > \ell - \epsilon\).
3. Combine.Proof: with \(N = \max(N_1, N_2)\), for all \(n \ge N\): \[\ell - \epsilon < x_n < L + \epsilon.\]
4. Consequence if \(\limsup x_n < L\).Proof: taking \(\epsilon = L - \limsup x_n > 0\) in 1 gives \(x_n < \limsup x_n + \epsilon = L\) for all \(n \ge N\).Hence the sequence is eventually strictly bounded above by \(L\): \(x_n < L\) for all large \(n\).(Similarly, \(\liminf x_n > \ell\) would force \(x_n > \ell\) eventually.)
5. Q.E.D.