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I am stuck on proofs with subsequences. I do not really have a strategy or starting point with subsequences.

NOTE: subsequential limits are limits of subsequences

Prove: $a_n$ is bounded $\implies \liminf a_n \leq \limsup a_n$

Proof:

Let $a_n$ be a bounded sequence. That is, $\forall_n(a_n \leq A)$.

If $a_n$ converges then $\liminf a_n = \lim a_n = \limsup a_n$ and we are done.

Otherwise $a_n$ has a set of subsequential limits we need to show $\liminf a_n \leq \limsup a_n$:

This is where I am stuck...

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  • $\begingroup$ Actually, \liminf and \limsup is a LaTeX command. $\endgroup$ – Asaf Karagila Oct 13 '12 at 23:58
  • $\begingroup$ I've changed (sequences) to (sequences-and-series). From FAQ about tags: Try to avoid creating new tags. Instead, check if there is some synonym that already has a popular tag. It's not easy to keep balance between too specific tags and not having enough tags, but it is always good to search first and to ask yourself, whether newly created tag is not too specific. (Of course, you can disagree with the removal of the tag you've created, and there is possibility for further discussion, if needed.) $\endgroup$ – Martin Sleziak Oct 14 '12 at 8:35
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    $\begingroup$ The boundedness hypothesis is irrelevant. $\endgroup$ – Did Oct 14 '12 at 8:46
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Let $\{a_n\}$ be a bounded sequence. Then we define the sequences $\{a_n^+\}$ and $\{a_n^-\}$ by

$$a_n^+=\sup\{a_n,a_{n+1}\dots\}$$

$$a_n^-=\inf\{a_n,a_{n+1}\dots\}$$

We (may) then define

$$\lim a_n^+=\limsup a_n$$ $$\lim a_n^-=\liminf a_n$$

Now, you need two things to work this out:

$(1)$ Let $A$ be any bounded nonempty subset of $\Bbb R$. Then

$$\inf A\leq \sup A$$

$(2)$ Let $\{\alpha_n\}$ be a sequence such that $a_n\geq0 $ for each $n\in \Bbb N$. Then $$\lim a_n\geq 0$$

With $(1)$ you should show $$a_n^-\leq a_n^+$$ for each $n\in \Bbb N$. Monotone convergence says both $\{a_n^+\}$ and $\{a_n^-\}$ converge, since they are bounded (above/below) and are monotone (increasing/decreasing)$^{(*)}$. But $$a_n^+- a_n^-\geq 0$$

for each $n\in \Bbb N$, so use $(2)$ to show

$$\lim a_n^+-\lim a_n^-\geq 0$$

that is:

$$\liminf a_n\leq \limsup a_n$$

$(*)$ To prove this, you need to show that if $A\subseteq B$, then $$\sup A\leq \sup B$$ $$\inf A\geq \inf B$$ Then, observe that

$$\{a_{n+1},a_{n+2},\cdots\}\subseteq \{a_n,a_{n+1},a_{n+2},\cdots\}$$

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  • $\begingroup$ In 2) Why must $a_n \geq 0$? $\endgroup$ – CodeKingPlusPlus Oct 14 '12 at 1:50
  • $\begingroup$ It is the hypothesis. "If $a_n\geq0$ then $\lim a_n\geq 0$" $\endgroup$ – Pedro Tamaroff Oct 14 '12 at 1:53
  • $\begingroup$ I don't see how to use this fact... $\endgroup$ – CodeKingPlusPlus Oct 14 '12 at 2:03
  • $\begingroup$ nevermind... hahah $\endgroup$ – CodeKingPlusPlus Oct 14 '12 at 2:07
  • $\begingroup$ @CodeKingPlusPlus Did you get it? $\endgroup$ – Pedro Tamaroff Oct 14 '12 at 2:24
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Hint: Think about what the definitions mean. We have $$\limsup a_n = \lim_n \sup \{ a_k \textrm{ : } k \geq n\}$$ and $$\liminf a_n = \lim_n \inf \{ a_k \textrm{ : } k \geq n\}$$

What can you say about the individual terms $\sup \{a_k \textrm{ : } k \geq n\}$ and $\inf \{a_k \textrm{ : } k \geq n\}$ ?

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  • $\begingroup$ What do you mean by the individual terms $\sup\{a_k: k \geq n\}$ $\endgroup$ – CodeKingPlusPlus Oct 14 '12 at 0:33
  • $\begingroup$ My understanding of $\limsup$ is the supremum of subsequential limits and $\liminf$ is the infimum of subsequential limits. Where subsequential limits are limits of subsequences. $\endgroup$ – CodeKingPlusPlus Oct 14 '12 at 0:34
  • $\begingroup$ For instance $\limsup a_n$ is by definition the limit as $n$ goes to infinity of the sequence $b_n$, where $b_n = \sup \{a_k \textrm{ : } k \geq n\}$. Each $b_n$ is an `individual term' in the sequence. $\endgroup$ – treble Oct 14 '12 at 0:36
  • $\begingroup$ Your definition is equivalent to mine (under the assumption that the sequence is bounded -think about why). But let's use your definition. What you are trying to prove is that the supremum of the subsequential limits must be bigger than (or equal to) the infimum of subsesquential limits. Tell me why that's true :) $\endgroup$ – treble Oct 14 '12 at 0:39
  • $\begingroup$ What is the $n$ in $k \geq n$? Also, I don't understand how $b_n$ is a sequence. If you take the supremum of a set of numbers greater than $k$ Isn't $b_n$ just set to that supremum? $\endgroup$ – CodeKingPlusPlus Oct 14 '12 at 0:47
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To make @Pedro's answer more complete, here's how things work in the unbounded case. Now we're given any sequence $\{a_n\}$, and would like to show $\liminf a_n \leq \limsup a_n$. I use the more general definition: $\liminf a_n := \sup_N a_N^- = \sup_N \inf_{n\geq N} a_n$, and $\limsup a_n := \inf_N a_N^+ = \inf_N \sup_{n\geq N} a_n$. I also use the standard definition of $\sup$ for a subset $E$ of the extended reals $\mathbb{R}^* = \mathbb{R} \cup{ \{\infty, -\infty \}}$ (see definition 6.2.6 in Analysis I by Terrance Tao).

Suppose $\{a_n\}$ does not have an upper bound. Then $\forall n, a_n^+ = \infty$, so $\limsup a_n = \inf \{\infty\} = \inf \{\} = \infty$; the claim immediately follows from the standard arithmetic on $\mathbb{R}^*$.

Similarly, if $\{a_n\}$ does not have a lower bound, we have $\liminf a_n =\sup \{-\infty\} = \sup \{\} = -\infty \leq \limsup a_n$.

We're now left with the case where $\{a_n\}$ is bounded, which is handled by @Pedro's answer.

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