Investigating the Limit Superior

Some analysis courses introduce the idea of the limit superior, denoted $\displaystyle \limsup_{n \to \infty}$ , and its dual cousin the limit inferior, denoted $\displaystyle \liminf_{n \to \infty}$ . Why should we care?

For the rest of the problems, let $\{x_n\}$ be a bounded $\mathbb R$ -sequence satisfying $|x_n| \leq M$ for any $n \in \mathbb N$ .

Problem 1. Let $\mathcal X$ denote the subset of real numbers $x$ where there exists a subsequence $\{x_{n_k}\}$ of $\{x_n\}$ such that $x_{n_k} \to x$ . Then $\sup \mathcal X$ and $\inf \mathcal X$ exist.

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Solution. Since $\{x_n\}$ is bounded, by the Bolzano-Weierstrass theorem, $\{x_n\}$ contains a convergent subsequence with limit $L \in \mathcal X$ , so that $\mathcal X$ is nonempty.

Fix $x \in \mathcal X$ , and suppose $x_{n_k} \to x$ . Then

$\displaystyle x = \lim_{k \to \infty} x_{n_k} \in [-M, M].$

Thus, $\mathcal X \subseteq [-M, M]$ and $\mathcal X$ is bounded. By the completeness of $\mathbb R$ , $\sup \mathcal X$ and $\inf \mathcal X$ exist.

Problem 2. Prove that the sequence $\displaystyle \{x_n^{\sup}\}$ defined by $\displaystyle x_n^{\sup} := \sup_{k \geq n} x_k$ converges. Note that $\displaystyle \{x_n^{\sup}\}$ may not be a subsequence of $\{x_{n_k}\}$ .

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Solution. By definition, $x_n^{\sup} \geq -M$ is non-increasing. By the monotone convergence theorem, $x_n^{\sup}$ converges.

Problem 3. Prove that $\displaystyle \lim_{n \to \infty} x_n^{\sup} = \sup \mathcal X$ .

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Solution. Denote $\displaystyle L_1 := \lim_{n \to \infty} x_n^{\sup}$ and $L_2 := \sup \mathcal X$ .

We aim to prove $L_1 \leq L_2$ and $L_2 \leq L_1$ .

We first prove $L_1 \leq L_2$ . Define $n_0 := 1$ and define $n_k$ for $k \in \mathbb N_0$ inductively as follows. Suppose we know $n_k$ . Then there exists $M_{k+1} \in \mathbb N$ such that for $j > \max\{n_k, M_{k+1}\}$ ,

$\displaystyle |x_j^{\sup} - L_1| < \frac 1{2(k+1)}.$

In particular, the estimate holds for $j = \max\{n_k, M_{k+1}\}+1 > n_k$ . By the definition of $x_j^{\sup}$ , find $n_{k+1} \geq j > n_k$ such that

$\displaystyle x_{n_{k+1}} > x_j^{\sup} - \frac{1}{2(k+1)}.$

Combining the estimates yields

$\displaystyle |x_{n_{k+1}} - L_1| < \frac 1{2(k+1)} \cdot 2 = \frac 1{k+1}.$

Now it is clear that we have constructed a subsequence $\{x_{n_k}\}$ such that $x_{n_k} \to L_1$ , and

$\displaystyle L_1 = \lim_{k \to \infty} x_{n_k} \leq L_2,$

as required.

We next prove $L_2 \leq L_1$ . For any $x \in \mathcal X$ , let $\{x_{n_k}\}$ be a subsequence of $\{x_n\}$ such that $x_{n_k} \to x$ . Then

$\displaystyle x = \lim_{k \to \infty} x_{n_k} \leq \lim_{k \to \infty} \left( \sup_{j \geq n_k} x_j \right) = \lim_{k \to \infty} x_{n_k}^{\sup} = L_1.$

Since $x\in\mathcal X$ was arbitrary, $L_1$ is an upper bound for $\mathcal X$ , so that $L_2 \leq L_1$ , being the least upper bound for $\mathcal X$ .

On the left-hand side, we have the limit of the supremum of the tail sequence. On the right-hand side, we have the supremum of the limits of subsequences. Problem 3 therefore motivates the notation $\displaystyle \limsup_{n \to \infty}$ , since

$\displaystyle \limsup_{n \to \infty}x_n := \lim_{n \to \infty} x_n^{\sup} \equiv \sup \mathcal X.$

Problem 4. Prove that the sequence $\displaystyle \{x_n^{\inf}\}$ defined by $\displaystyle x_n^{\inf} := \inf_{k \geq n} x_k$ converges to $\inf \mathcal X$ .

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Solution. We observe that

$\displaystyle x_n^{\inf} := \inf_{k \geq n} x_k = - \sup_{k \geq n} (-x_k) \to -\sup(-\mathcal X) = \inf\mathcal X.$

It is obvious that convergent sequences are bounded, and it is not always true that bounded sequences converge. But, for any bounded sequence $\{x_n\}$ , $\displaystyle \limsup_{n \to \infty} x_n$ , $\displaystyle \liminf_{n \to \infty} x_n$ always exist. We can use their existence to formulate an equivalent criterion (i.e. a characterisation) for bounded sequences to converge.

Problem 5. Prove that a bounded sequence $\{x_n\}$ converges if and only if $\displaystyle \liminf_{n \to \infty} x_n = \limsup_{n \to \infty} x_n$ .

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Solution. The right-hand side is equivalent to $\mathcal X = \{x\}$ for some $x \in \mathbb R$ , which yields the direction $(\Rightarrow)$ trivially.

For the direction $(\Leftarrow)$ , we observe that $x_n^{\inf} \leq x_n \leq x_n^{\sup}$ . Since $x_n^{\inf} \to x$ and $x_n^{\sup} \to x$ , $x_n \to x$ by the squeeze theorem.