Upper and Lower Limits

Definition

Let $\{s_n\}$ be a sequence of real numbers with the following property:

For every real number $M$ there is an integer $N$ such that $n\ge N$ implies $s_n\ge M$. We then write

$$s_n\to +\infty$$

and likewise, for every real $M$ there is an integer $N$ such that $n\ge N$ implies $s_n\le M$, we write

$$s_n\to -\infty$$

Let $E$ be the set of numbers $x$ such that $s_{n_k}\to x$ for some subsequence $\{s_{n_k}\}$. This set $E$ contains all subsequential limits, and maybe $+\infty ,-\infty$. Then,

$$s^{\ast }=\sup E{s}_{\ast }=\inf E$$

These are the upper and lower limits of $\{s_n\}$, with the notation

$$\underset{n\to \infty }{\mathrm{limsup}}{s}_n=s^{\ast }\underset{n\to \infty }{\mathrm{liminf}}{s}_n=s_{\ast }$$

Lemma (Closed)

Let $\{p_n\}$ be a sequence in $X$. Let $\mathcal{A}$ be the set of subsequence limits of $\{p_n\}$. Then $\mathcal{A}$ is closed.

Proof: Let $q\in {\mathcal{A}}^{\prime }$. The goal is to show that $q\in \mathcal{A}$. That is, there exists subsequence $\{p_{n_i}\}$ such that $p_{n_i}\to q$. We split on the number of occurrences of $q$: infinitely, or finitely.

Case 1: Infinite Suppose there exist indices $n_1<n_2<\cdots$ such that $p_{n_i}=q$ for $i\in \mathbb{N}$. Then certainly $p_{n_i}=q\to q$ so $q\in \mathcal{A}$.

Case 2: Finite We have the case where the sequence stops hitting $q$. Then $\exists N\in \mathbb{N},\forall n\ge N,p_n\ne q$. We will find some $n_1<n_2<\cdots$ that are greater than $N$ such that

$$d(p_{n_i},q)<\frac{2}{i}$$

Consider $r=1$ (where $i=1$) and $N_r(q)$. Since $q\in {\mathcal{A}}^{\prime }$, then $\exists x_1\in \mathcal{A}$ such that $x_1\in N_1(q)$ where $x_1=q$. Now,

$$\begin{array}{rl}{x}_1\in \mathcal{A}& ⟹\exists p_{n_{i_j}}{\to }_{j\to \infty }{x}_1\\ & ⟹\exists n_1>N\text{ such that }{p}_{n_1}\in N_1(x_1)\\ & ⟹d(p_{n_1},q)\le d(p_{n_1},x_1)+d(x_1,q)\le 1+1=2\end{array}$$

Now that we have our first index $n_1$ and thus point, we can repeat for smaller radii. Suppose

$$\exists n_1<n_2<\cdots <n_i$$

such that $\exists x_i\in \mathcal{A},x_i\ne q$ with $x_i\in N_{1/i}(q)$ and $d(p_{n_i},q)<2/i$. Now we can show the inductive step:

Consider $N_{1/i+1}(q)$. Then $\exists x_{i+1}\in \mathcal{A}$ such that $x_{i+1}\ne q,x_{i+1}\in N_{1/i+1}(q)$ since $x_{i+1}\in \mathcal{A}$, so $\exists n_{i+1}>n_i$ such that $p_{n_{i+1}}\in N_{1/(i+1)}(q)$. This implies that

$$d(p_{n_i},q)\le d(p_{n_i},x_{i+1})+d(x_{i+1},q)\le \frac{1}{i+1}+\frac{1}{i+1}=\frac{2}{i+1}$$

Thus we may find

$$n_1<n_2<\cdots$$

such that $d(p_{n_i},q)<\frac{2}{i}$. Let $\epsilon >0$. By Archimedean Property, $\exists I\in \mathbb{N}$ such that

$$\frac{2}{\epsilon }<I⟺\frac{2}{I}<\epsilon$$

so $\forall i\ge I,d(p_{n_i},q)<2/i\le 2/I<\epsilon$. Thus $p_{n_i}\to q$ so $q\in \mathcal{A}$.

Lemma (Checkability)

Let $\{p_n\}$ be a sequence in $\mathbb{R}$. Define $p^{\ast }$and $p_{\ast }$ as before.

1. $p^{\ast },p_{\ast }\in \mathcal{A}$. They are limits of subsequence of $p_n$
2. (Checkable criterion for upper/lower limits): If $x>p^{\ast }$, then $\exists N\in \mathbb{N}$ such that $p_n<x$, $\forall n\ge N$. Similarly, $y<p_{\ast }$, then $\exists N\in \mathbb{N}$ such that $y<p_n,\forall n\ge N,y$.

Proof: We will only consider the case $p^{\ast },p_{\ast }\in \mathbb{R}$. (i.e, it is a finite real, and $-\infty ,\infty$ from Extended Real Numbers).

Case 1: Suppose so, then $\mathcal{A}$ is bounded. From closedness, we have that $\sup (\mathcal{A}),\inf (\mathcal{A})\in \overline{\mathcal{A}}$, so $\overline{\mathcal{A}}=\mathcal{A}$ Case 2: We prove by contradiction. Say $x>p^{\ast }$. Then WTS $\exists N\in \mathbb{N},\forall n\ge N,p_n<x$. Then the contradiction is that $\forall N\in \mathbb{N},\exists n\ge N,p_n\ge x$.

• when $N=1,\exists n_1\ge 1,p_{n_1}\ge x$
• when $N=n_1+1$, $\exists n_2>n_1,p_{n_2}\ge x$ and so on. Indeed, we have

$$n<n_2<n_3<\cdots$$

such that $p_{n_i}\ge x$. This implies $\exists \alpha \ge x$ which is a subsequence limit of $p_{n_i}$. So $\exists$ a subsequence of $\{p_n\}$ which has $\alpha$ as a limit. Indeed, we have 2 cases: 3. $\{p_{n_i}\}$ diverges to $\infty$ and this implies $p^{\ast }=\infty$ which is a contradiction. 4. $\{p_{n_i}\}$ is bounded above, say by $b$. Then $\forall i\in \mathbb{N}$, $x\le p_{n_i}\le b$ implies that by BW there is a convergent subsequence. But then the subsequence is this one, which is a contradiction. Thus $\alpha \ge x>p^{\ast }=\sup (\mathcal{A})$.

It is a Supremum after all. So the proof strategy is similar.

Lemma

Let $(p_n),(q_n)$ be a sequence. Assume $\exists N\in \mathbb{N}$, $p_n\le q_n,\forall n\ge N$. Then

$$\begin{array}{rl}\overline{\lim }{p}_n& \le \overline{\lim }{q}_n\\ \underline{\lim }{p}_n& \le \underline{\lim }{q}_n\end{array}$$

Proof: From homework.