- #1
mathstatnoob
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Homework Statement
The lemma below is from Sample-Path Analysis of Queueing Systems by El-Taha and Stidham Jr.
Lemma 2.10
Let \{x_n, n \geq 1\} be a sequence of non-negative real numbers and \{b_n, n \geq 1\} a non-decreasing sequence of real numbers such that b_n \rightarrow \infty as n \rightarrow \infty. Then the following are equivalent:
(i)b_n^{-1}x_n \rightarrow c as n \rightarrow \infty;
(ii)b_n^{-1}\max_{k \leq n}x_k \rightarrow c as n \rightarrow \infty.
The authors claim that proving statement (ii) implies statement (i) is trivial and leave no proof (they do go on to prove (i) implies (ii)). However, "trivial" is a subjective term. I'm trying to provide a proof that (ii) implies (i), but I get stuck.
2. Attempt at a solution
Let \epsilon > 0. It follows from the hypothesis that there exists an N_0 such that if n > N_0, then c-\epsilon < b_n^{-1}\max_{k \leq n}x_k < c+\epsilon.
Since b_n \rightarrow \infty as n \rightarrow \infty,there exists an N_1 such that if n > N_1, then b_n > 0 which implies b_n^{-1} > 0 which then implies b_n^{-1}x_n \leq b_n^{-1}\max_{k \leq n}x_k.
So let N=\max\{N_0,\, N_1\}. If n > N, then b_n^{-1}\max_{k \leq n}x_k < c+\epsilon. Hence b_n^{-1}x_n < c+\epsilon. Now either (a) c-\epsilon < b_n^{-1}x_n, or (b) b_n^{-1}x_n \leq c-\epsilon. If case (a) is true, then c-\epsilon < b_n^{-1}x_n < c+\epsilon whenever n > N. Regarding case (b)...
...and this is where I get stuck. My guess is case (b) leads to some contradiction, but I'm not sure how to obtain one. Any suggestions or specific references to look at? Or if there's a simpler way to show that (ii) implies (i), I'm all ears. Thanks!
