when I was catching up with Evans PDE and I realise I do not have a good understanding about limsup and liminf.

I found the following exercise in a real analysis book that motivates the definition of limsup.But I have no idea how to answer.

Let $(x_n)$ be a bounded sequence. For each $k\in\mathbb{N}$, let

$$y_k=\sup_{n\ge k} x_n=\sup\{x_k,x_{k+1},x_{k+2},\cdots\}.$$

  1. Show that the sequence $(y_k)$ is decreasing and bounded below.
  2. Conclude that $(y_k)$ converges.

[Part 1) was solved in previous posts, but I have not found an answer for part 2)]


marked as duplicate by user99914, 6005, user147263, Mark Viola, Claude Leibovici Nov 13 '15 at 5:53

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ Can you at least go from (1) to (2)? $\endgroup$ – Joey Zou Nov 2 '15 at 4:21

Hint: The second one follows from the first. The intuition behind $\limsup$ ($\liminf$ as well) is this: take $\sup_{n \geq 1}x_{n}$; delete $x_{1}$ and take $\sup_{n \geq 2}x_{n}$; delete $x_{2}$ and take $\sup_{n \geq 3}x_{n}$; repeat this process indefinitely. Clearly, we have $n_{1} > n_{2}$ only if $\{ x_{n} \mid n \geq n_{1}\} \subset \{ x_{n} \mid n \geq n_{2} \}$ and only if $\sup_{n \geq n_{1}}x_{n} \leq \sup_{n \geq n_{2}}x_{n}$; hence the sequence $(\sup_{n \geq N}x_{n})_{N \geq 1}$ is decreasing.

Can you continue from here?

  • $\begingroup$ is it a reference for this, it is a bit too abstract to understand $\endgroup$ – math101 Nov 2 '15 at 4:32

Let $S_k := \{x_k, x_{k+1}, \ldots \}$. Then by definition, $y_k = \sup S_k$. Notice that $S_{k} \supseteq S_{k+1}$. Therefore $\sup S_k \geq \sup S_{k+1}$, i.e. $y_k \geq y_{k+1}$ for all $k$. If this doesn't click for you right away, you need to spend more time thinking about it on your own .

The sequence $\{y_k\}$ is bounded below because, in particular, $y_k \geq \inf \{x_1, x_2, \ldots\}$ for every $k$, and that infimum is finite since $\{x_k\}$ is a bounded sequence.

Part 2 of your question is a standard fact which I would encourage you to spend more time thinking about on your own.

  • $\begingroup$ I got stuck from your first sentence? Could you please explain (or find another way to phase what you meant) By the way, your hint looks plausible for me to make a start and I will need to spend some time once I understand every sentence you wrote. $\endgroup$ – math101 Nov 2 '15 at 4:32
  • $\begingroup$ I'll edit my answer to make it more precise $\endgroup$ – Alex G. Nov 2 '15 at 4:33
  • $\begingroup$ okay take your time, i need to find a quiet place to read too. $\endgroup$ – math101 Nov 2 '15 at 4:33
  • $\begingroup$ ok. reading it line by line. The first sentence make sense, as the least upper bound of a bigger set should be no smaller than that of a smaller set. Hence $y_k\ge y_{k+1}$. $\endgroup$ – math101 Nov 2 '15 at 5:37
  • $\begingroup$ I don't quite follow your reasoning about why $\{y_k\}$ is bounded below, do you mean $y_k\ge\inf\{x_1,x_2,\cdots,x_{k-1}\}$ for every $k$? $\endgroup$ – math101 Nov 2 '15 at 5:58

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