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I have come across these three definitions of the limit superior (or upper limit) and the limit inferior (or lower limit) of a sequence of real numbers and I wonder how to establish the equivalence of these.

  1. Walter Rudin: PRINCIPLES OF MATHEMATICAL ANALYSIS, 3rd edition: Definition 3.16: Given a sequence $\{s_n\}$ of real numbers, let $E$ be the set of numbers $x$ (in the extended real number system) such that $s_{n_k} \to x$ for some subsequence $\{s_{n_k}\}$. This set contains all subsequential limits ..., plus possibly the numbers $+\infty$, $-\infty$. We now ... put $$s^{*} = \sup E,$$ $$s_{*} = \inf E.$$ The numbers $s^{*}$, $s_{*}$ are called the upper and lower limits of $\{s_n\}$; we use the notation $$ \lim_{n \to \infty} \sup s_n = s^{*}, \, \, \, \lim_{n \to \infty} \inf s_n = s_{*}.$$

  2. Tom M. Apostol: MATHEMATICAL ANALYSIS, 2nd edition: Sec. 8.3: Definition 8.2: Let $\{a_n\}$ be a sequence of real numbers. Suppose there is a real number $U$ satisfying the following conditions:

i} For every $\epsilon > 0$ there exists an integer $N$ such that $n > N$ implies $$a_n < U + \epsilon.$$ ii) Given $\epsilon > 0$ and given $m > 0$, there exists an integer $n> m$ such that $$a_n > U - \epsilon.$$ Then $U$ is called the limit superior (or upper limit) of $\{a_n\}$, and we write $$U = \lim_{n \to \infty} \sup a_n.$$ Statement (i) implies that the set $\{a_1, a_2, a_3, \ldots \}$ is bounded above. If this set is not bounded above, we define $$\lim_{n\to\infty}\sup a_n = +\infty.$$ If the set is bounded above but not bounded below and if $\{a_n\}$ has no finite limit superior, then we say $\lim \sup_{n\to\infty} a_n = -\infty$. The limit inferior (or lower limit) of $\{a_n\}$ is defined as follows: $$\lim_{n\to\infty}\inf a_n = -\lim_{n\to\infty}\sup b_n,$$ where $b_n = -a_n$ for $n= 1, 2, 3, \ldots$.

  1. Robert G. Bartle and Donald R. Sherbert: INTRODUCTION TO REAL ANALYSIS, 3rd edition: Exercises for Section 3.3: Problem 10: Let $(x_n)$ be a bounded sequence of real numbers, and for each $n\in \mathbb{N}$ let $s_n \colon= \sup \{x_k \colon k \geq n\}$ and let $t_n \colon= \inf \{x_k \colon k \geq n\}$. Prove that $(s_n)$ and $(t_n)$ are monotone and convergent. Also prove that if $\lim (s_n) = \lim (t_n)$, then $(x_n)$ is convergent. [One calls $\lim (s_n)$ the limit superior of $(x_n)$ and $\lim (t_n)$ the limit inferior of $(x_n)$.]

Now how can one show that the above three definitions are equivalent (i.e. these three definitions are of the same pair of numbers)?

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  • $\begingroup$ This is a related post. $\endgroup$ – Bijesh K.S Jan 7 '18 at 11:36
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I'm going to drop the boundedness condition (it is a legitimate difference) and use the extended reals. Also, I'm doing it for $\limsup$ ($\liminf$ is analogous).

Let $\{a_n\}_{n\geq 0}$ be a sequence in $\mathbb{R}$.


Let $E$ be the set of sub-sequential limits as in $(1)$. I will show that it is closed.

Let $x$ be a limit point of $E$. Then there are $x_n\in E$ such that $x_n\to x$ as $n\to\infty$. Each $x_n$ is the limit of a subsequence $a^n_{k_j}$.

Suppose $x\in\mathbb{R}$. For all $\epsilon>0$ there is $N$ such that $\vert x-x_n\vert<\frac{\epsilon}{2}$ if $n>N$. Furthermore, there is some index $j_n$ such that $\vert x_n-a^n_{k_{j_n}}\vert<\frac{\epsilon}{2}$. We will choose the indices $j_n$ so that $k_{j_{n+1}}>k_{j_{n}}$ (they can be chosen inductively, starting with $j_1$). Now we have $a^n_{k_{j_n}}$, a subsequence of $a_k$.

Given $\epsilon>0$, we get $N$ (since $x_n$ is convergent) and if $n>N$ we have

$\vert x-a^n_{k_{j_n}}\vert\leq\vert x-x_n\vert+\vert x_n-a^n_{k_{j_n}}\vert<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$

Consequently, $x$ is a sub-sequential limit and $x\in E$.

For the cases that $x=\pm\infty$ you reword the same idea into bounding below or bounding above respectively.

So $E$ contains its limit points and is closed.

(1) is equivalent to (2)

Since $E$ is closed, $\sup E\in E$ and there is a subsequence with $\sup E$ as its limit.

Let $U=\sup E$. If there were $\epsilon>0$ such that for every $N>0$ there is $n>N$ with $a_n\geq U+\epsilon$, then there would be a subsequence with limit greater than $U$, but $U$ is the supremum of the sub-sequential limits. This proves (i).

Since there is a subsequence converging to $U$ we get (ii) by letting the $n$ be the index of an element in this subsequence.

So $\sup E$ satisfies the criteria of $U$ in (2).

(2) is equivalent to (3)

Note that $s_n$ is a weakly decreasing sequence. Since it is monotone, it has a limit (it's bounded below by $-\infty$ and we are in the extended reals).

Let $U$ satisfy the criteria in (2). By (ii), we can see that for every $\epsilon>0$, $s_m>U-\epsilon$. So $\lim s_n\geq U$.

By (i) we can see that for every $\epsilon>0$ there is $N>0$ such that $s_N\leq U+\epsilon$ for some $N$. Since $s_n$ is decreasing, we actually get $s_n\leq U+\epsilon$ for all $n>N$. So $\lim s_n\leq U$.

Consequently, $\lim s_n=U$, and we see that the definitions are equivalent.

Finally, note that since $s_n$ is decreasing, $\lim_{n\to\infty} s_n=\inf_{n\geq 0} s_n=\inf_{n\geq 0}\sup_{k\geq n}a_k$. This last expression is an alternative definition.


The only significance of the boundedness is to ensure that monotone sequences converge, working in the extended reals has the same effect with less need to separate into cases.

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