Limit Superior and Limit Inferior

We first extend the notion of convergence by allowing the values $+\infty$ and $$-\infty$.$ We then associate with each tail of a sequence its supremum and infimum. The limiting behavior of these tail bounds leads to the notions of limit superior and limit inferior. Throughout this section, all suprema and infima are understood to be taken in $\overline{\mathbb R}$

Definition 3.4.1. Let $$(x_n)$$ be a sequence in $$\mathbb R$.$ If for every $y\in \mathbb R$ there is $n_0\in \mathbb N^{+}$ such that

\[n\ge n_0\ \Longrightarrow \ x_n\ge y,\]

then we write $x_n\to +\infty$ or $$\lim_{n\to \infty}x_n = +\infty$.$ Similarly, if for every $y\in \mathbb R$ there is $n_0\in \mathbb N^{+}$ such that

\[n\ge n_0 \ \Longrightarrow \ x_n\le y,\]

then we write $x_n\to -\infty$ or $$\lim_{n\to \infty}x_n = -\infty$.$ A number $x\in\overline{\mathbb R}$ is called the extended limit of $$(x_n)$$ if $$x_{n}\to x$,$ and is called an extended subsequential limit of $$(x_n)$$ if $x_{n_k}\to x$ for some subsequence $$(x_{n_k})$.$

Definition 3.4.2. Let $$(x_n)$$ be a sequence in $$\mathbb R$.$ For each $$k$,$ define

\[s_k = \sup {\{x_n : n\ge k\}}\quad \text{and} \quad t_k = \inf {\{x_n : n\ge k\}}.\]

The limit superior and limit inferior of $$(x_n)$$ are defined, respectively, by

\[\limsup_{n\to\infty}{x_n} = \inf_{k\ge 1}{s_k} \quad \text{and} \quad \liminf_{n\to\infty}x_n = \sup_{k\ge 1}{t_k}.\]

The sequence $$(s_k)$$ is monotonically decreasing, while $$(t_k)$$ is monotonically increasing. Thus $$s_k$$ and $$t_k$$ form, respectively, upper and lower envelopes for the successive tails of $$(x_n)$.$ The limit superior and limit inferior record the eventual values of these two envelopes. The tail-based definition also has a subsequential interpretation: the limit superior and limit inferior are precisely the largest and smallest extended limits that can be obtained along subsequences.

Theorem 3.4.3. Let $$(x_n)$$ be a sequence in $$\mathbb R$,$ and let $$A$$ be the set of all extended subsequential limits of $$(x_n)$.$ Then

\[\limsup_{n\to\infty}x_n=\max A\quad \text{and} \quad \liminf_{n\to\infty}x_n=\min A.\]

Proof. Let $s_k = \sup {\{x_n : n\ge k\}}$ and $$L = \limsup_{n\to\infty}x_n$.$ First suppose $$L\in \mathbb R$.$ We show that no extended subsequential limit can be greater than $$L$.$ Suppose $x_{n_j}\to y$ and fix $$k$.$ Since $$n_j\to \infty$,$ there is $$j_0$$ such that $j\ge j_0$ implies $$n_j\ge k$.$ It follows that $$x_{n_j}\le s_k$.$ If $$y\in \mathbb R$,$ then for any $$\varepsilon>0$,$ there is $$j_0'$$ such that $j\ge j_0'$ implies $$|x_{n_j}-y|<\varepsilon$.$ Taking maximum of $$j_0$$ and $$j_0'$,$ eventually $$y-\varepsilon < s_k$.$ Since $$y$$ and $$s_k$$ are the supremum and an upper bound for $\{y-\varepsilon : \varepsilon > 0\}$ respectively, so $$y\le s_k$.$ If $$y = +\infty$,$ then for any $$N>0$,$ there exists $$j_0'$$ such that $j\ge j_0'$ implies $$x_{n_j}\ge N$.$ Taking maximum, eventually $$N\le x_{n_j}\le s_k$,$ which implies $$s_k=+\infty$.$ Hence $$y\le s_k$.$ If $$y=-\infty$,$ then automatically $$y\le s_k$.$ Thus $y\le s_k$ in every case. Since $$k$$ was arbitrary, $$y\le \inf_{k\ge 1} {s_k} = L$.$ Furthermore, $$L$$ is itself a subsequential limit. Indeed, put $$n_0 = 0$.$ Since $$s_k\downarrow L$,$ after $n_{j-1}$ has been chosen, we can choose $k_j>n_{j-1}$ such that $$s_{k_j}<L+\frac1j$.$ Since $s_{k_j}$ is the supremum for $$\{x_n:n\ge k_j\}$,$ there exists $n_j\ge k_j$ such that $$x_{n_j}>s_{k_j}-\frac 1 j$.$ Thus $n_j>n_{j-1}$ and

\[L-\frac1j\le s_{k_j}-\frac 1 j <x_{n_j}\le s_{k_j}<L+\frac1j.\]

Therefore, we obtain $$x_{n_j}\to L$.$ Hence $$L$$ is an upper bound for $$A$$ and $$L\in A$,$ we conclude that $$L = \max A$.$ We now consider the infinite cases. If $$L=+\infty$,$ then $\{x_n : n\ge k\}$ is unbounded above for every $$k$.$ Choose $$n_1$$ such that $$x_{n_1}>1$.$ Having chosen $$n_1,\ldots,n_{j-1}$,$ we can choose $n_j>n_{j-1}$ so that $$x_{n_j}>j$.$ It follows that $$x_{n_j}\to+\infty$,$ which implies $$\max A = +\infty$.$ Hence $$L=\max A$.$ If $$L=-\infty$,$ then $$s_k\to-\infty$.$ Given $$y\in \mathbb R$,$ we have $$s_k<y$$ eventually. It follows that $$x_k\le s_k < y$.$ Thus eventually $$x_k<y$,$ giving $$x_k\to -\infty$.$ Therefore $A=\{-\infty\}$ and $$\max A=-\infty$.$ Hence $$L=\max A$.$ In the case of the limit inferior, the proof is analogous. $\square$

Example 3.4.4. Let $$(x_n)$$ be a sequence in $\mathbb{R}$ such that $$\{x_n: n\ge 1\} = \mathbb{Q}$.$ Let $y\in \mathbb R$ be given. Suppose $n_1<\cdots<n_{k-1}$ have been chosen. Since $\{x_1,\ldots,x_{n_{k-1}}\}$ is finite and the interval

\[\left(y-\frac 1 k, y+\frac 1 k\right)\]

has infinitely many rationals, there is a rational $$q$$ which is not in $\{x_1,\ldots,x_{n_{k-1}}\}$ but belongs to the interval. Since $$q$$ appears in $$(x_n)$$ at least once, there exists $n_k>n_{k-1}$ such that $$x_{n_k} = q$.$ In this way, we obtain a subsequence $(x_{n_k})$ that converges to $$y$.$ Since $$y$$ was arbitrary, every real number is a subsequential limit. Let $$A$$ be the set of all extended subsequential limits of $$(x_n)$.$ Since $$\sup A\ge\sup {\mathbb R}$,$ by Theorem 3.4.3,

\[\limsup_{n\to \infty}{x_n} = \sup A = +\infty.\]

Similarly,

\[\liminf_{n\to \infty}{x_n} = \inf A = -\infty.\]

While Theorem 3.4.3 describes the values attained along suitable subsequences, the tail formulation gives useful bounds for the entire sequence.

Theorem 3.4.5. Let $$(x_n)$$ be a sequence in $\mathbb R$ and let $$y\in \mathbb R$.$

If $y>\limsup_{n\to\infty} x_n$ then eventually $$x_n<y$.$
If $y<\liminf_{n\to \infty} x_n$ then eventually $$x_n>y$.$

Proof. (1) Let $$S$$ be the set of every supremum $$s_k$$ for $$\{x_n:n\ge k\}$.$ Since $$y>\inf S$,$ $$y$$ is not a lower bound for $$S$.$ So there exists $$k_0$$ such that $$s_{k_0}<y$.$ Then $$y$$ is an upper bound for the $$k_0$-$ th tail. Since $$(s_k)$$ is monotonically decreasing, $n\ge k_0$ implies $$x_n<y$.$ (2) Let $$T$$ be the set of every infimum $$t_k$$ for $$\{x_n:n\ge k\}$.$ Since $$y<\sup T$,$ $$y$$ is not an upper bound for $$T$.$ So there exists $$k_0$$ such that $$t_{k_0}>y$.$ Then $$y$$ is a lower bound for the $$k_0$-$ th tail. Since $$(t_k)$$ is monotonically increasing, $n\ge k_0$ implies $$x_n>y$.$ $\square$

In particular, if the limit superior and limit inferior coincide at a finite value, these eventual bounds squeeze every sufficiently late term toward that value.

Theorem 3.4.6. Let $$(x_n)$$ be a sequence in $\mathbb R$ and let $$x\in \mathbb R$.$ Then $$(x_n)$$ converges to $$x$$ if and only if

\[\limsup_{n\to \infty}{x_n} = \liminf_{n\to \infty}{x_n} = x.\]

Proof. Let $$A$$ be the set of all extended subsequential limits of $$(x_n)$.$ If $x_n\to x$ then every subsequential limit of $$(x_n)$$ is $$x$.$ Hence $$A = \{x\}$,$ so that $$\max A = \min A = x$.$ Suppose, conversely, that the limit superior and limit inferior of $$(x_n)$$ are both $$x$.$ Then for any $$\varepsilon>0$,$ eventually

\[x-\varepsilon < x_n < x + \varepsilon\]

by Theorem 3.4.5. That is, $$| x_n - x| <\varepsilon$.$ Hence $$x_n \to x$.$ $\square$

Finally, because tail suprema and tail infima preserve order, the limit superior and limit inferior are also compatible with eventual inequalities between sequences.

Theorem 3.4.7. Let $$(x_n)$$ and $$(y_n)$$ be sequences in $$\mathbb R$.$ If $x_n\le y_n$ eventually, then the following hold.

$$\limsup_{n\to \infty}{x_n}\le \limsup_{n\to \infty}{y_n}$.$
$$\liminf_{n\to \infty}{x_n} \le \liminf_{n\to \infty}{y_n}$.$

Proof. Suppose $n\ge N$ implies $$x_n\le y_n$.$ Let $$k\ge N$,$ and let $s_k = \sup {\{x_n : n\ge k\}}$ and $$s_k' = \sup {\{y_n : n\ge k\}}$.$ If $n\ge k$ then $$x_n\le y_n\le s_k'$.$ Hence $$s_k'$$ is an upper bound for $$\{x_n : n\ge k\}$,$ so

\[s_k\le s_k'.\]

It follows that $$\inf_{k\ge N}{s_k}\le s_k\le s_k'$.$ Hence $\inf_{k\ge N}{s_k}$ is a lower bound for $$\{s_k' : k\ge N\}$,$ so $$\inf_{k\ge N}{s_k} \le \inf_{k\ge N}{s_k'}$.$ By Definition 3.4.2,

\[\limsup_{n\to \infty}{x_n} = \inf_{k\ge N}{s_k} \le \inf_{k\ge N}{s_k'} = \limsup_{n\to \infty}{y_n}.\]

The proof of (2) is analogous to (1). $\square$

References

Rudin, W. (1976). Principles of mathematical analysis (3rd ed.). McGraw-Hill Education.

External

3.4 Limit Superior and Limit Inferior

References