Is Proving (ii) Implies (i) in Limit Theorems Trivial?

[mathstatnoob]

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!

 

[spamiam]

Welcome to PF, mathstatnoob!

I'm not sure, but I think breaking the proof into two cases may help, one if $x_n$ is bounded and the other if it is unbounded.

What does $1/b_n$ converge to? Consequently what do $\frac{x_n}{b_n}$ and $\frac{\max_{k \leq n}x_k}{b_n}$ converge to in the first case?

In the second case, I think you can use the unboundedness of $x_n$ to show (i) and (ii) are equivalent.

Hope this helps!

 

[mathstatnoob]

Welcome to PF, mathstatnoob!

I'm not sure, but I think breaking the proof into two cases may help, one if $x_n$ is bounded and the other if it is unbounded.

What does $1/b_n$ converge to? Consequently what do $\frac{x_n}{b_n}$ and $\frac{\max_{k \leq n}x_k}{b_n}$ converge to in the first case?

In the second case, I think you can use the unboundedness of $x_n$ to show (i) and (ii) are equivalent.

Hope this helps!

Thanks for your input, spamiam.

So in the case when $x_n$ is bounded, both $\frac{x_n}{b_n}$ and $\frac{\max_{k \leq n}x_k}{b_n}$ approach 0 as $n$ approaches infinity. I showed this by employing the squeeze theorem. Let me think about the case when $x_n$ is unbounded a bit more (I just wanted to quickly acknowledge your suggestion).