# Real Analysis, liminf/limsup inequality

• MHB
• joypav
In summary, the proof of the theorem is much improved by simplifying the assumption that the sequence is bounded.
joypav
I am working a bunch of problems for my Real Analysis course.. so I am sure there are more to come. I feel like I may have made this proof too complicated. Is it correct? And if so, is there a simpler method?

Problem:
Show that $liminfa_n \leq limsupa_n$.

Proof:
Consider a sequence of real numbers, $(a_n)$.
By the definitions of inf and sup, we know,

$\forall n \in N, infa_m \leq supa_m, m \in N$

Now consider the following sequences:

$(inf_{m \ge n}a_m)$ and $(sup_{m \ge n}a_m)$

We know that,

(1.) $inf_{m \ge n}a_m \le sup_{m \ge n}a_m$ for all $n > N$, some $N \in \Bbb{N}$

Claim 1: $lim_{n \rightarrow \infty}inf_{m \ge n}a_m \le lim_{n \rightarrow \infty}sup_{m \ge n}a_m$
Proof of Claim:
Let $\underline{a'}=lim_{n \rightarrow \infty}inf_{m \ge n}a_m$ and $\overline{a'}=lim_{n \rightarrow \infty}sup_{m \ge n}a_m$
Suppose the claim is false, then $\underline{a'}>\overline{a'}$.
Given $\epsilon>0, \exists N_1 \in \Bbb{N}, \vert inf_{m \ge n}a_m - \underline{a'} \vert < \frac{\epsilon}{2}$ and
$\exists N_2 \in \Bbb{N}, \vert sup_{m \ge n}a_m - \overline{a'} \vert < \frac{\epsilon}{2}$

Choose $\epsilon = \frac{\underline{a'}-\overline{a'}}{2} > 0$. (Because $\underline{a'} > \overline{a'} \implies \underline{a'}-\overline{a'} > 0$).
If $n > max{N,N_1,N_2}$, then
$inf_{m \ge n}a_m > \underline{a'} - \epsilon = \underline{a'} - \frac{\underline{a'}-\overline{a'}}{2} = \overline{a'} + \frac{\underline{a'}-\overline{a'}}{2} = \overline{a'} + \epsilon > sup_{m \ge n}a_m$
$\implies inf_{m \ge n}a_m > sup_{m \ge n}a_m$, a contradiction of (1.).
$\implies$ Claim 1 is true.

Then,
$lim_{n \rightarrow \infty}inf_{m \ge n}a_m \le lim_{n \rightarrow \infty}sup_{m \ge n}a_m$
$\implies lim_{n \rightarrow \infty}infa_n \le lim_{n \rightarrow \infty}supa_n$

Last edited:
Three things:

1. Based on your argument, it looks like there’s an implicit assumption that $(a_n)$ is bounded. Is that an assumption in the problem?

2. The inequality $\inf_{m\ge n} a_m \le \sup_{m\ge n}a_m$ is true for all positive integers $n$, not just all $n$ greater than some positive integer $N$.

3. In your argument, if you choose $N_1$ such that $\lvert \inf_{m\ge n} a_m - \underline{a’}\rvert < \epsilon$ for all $n > N_1$ and $\lvert \sup_{m\ge n} a_m - \overline{a’}\rvert < \epsilon$ for all $n > N_2$ (where $\epsilon = \frac{\underline{a’} - \overline{a’}}{2}$), then the string of inequalities you have at the end make sense for $n > \max\{N_1,N_2\}$.

The theorem itself can be proven directly as follows. First, suppose $(a_n)$ is unbounded. If $\limsup a_n =+\infty$, then the inequality is clear. If $\limsup a_n = -\infty$, given $\epsilon > 0$, $a_k < -\epsilon$ for all sufficiently large $k$. Thus $\liminf a_n \le -\epsilon$. Since $\epsilon$ was arbitrary, $\liminf a_n = -\infty$ and the inequality is satisfied.

Now suppose $(a_n)$ is bounded. For each $n$ and $k$ with $k > n$, $a_k \le \sup_{m \ge n} a_m$. Thus, for each $n$, $\inf_{k \ge n} a_k \le \sup_{m \ge n} a_m$. Taking limits as $n \to \infty$ results in $\liminf_{n \to \infty} a_n \le \limsup_{n\to \infty} a_n$.

That is... much better. Thanks!

## 1. What is Real Analysis?

Real Analysis is a branch of mathematics that deals with the study of real numbers and their properties. It involves the rigorous examination of concepts such as limits, continuity, derivatives, and integrals, and their applications in various fields of mathematics and science.

## 2. What is the difference between liminf and limsup?

Liminf and limsup are two important concepts in Real Analysis that describe the behavior of a sequence of real numbers. Liminf, or the limit inferior, is the smallest limit that a subsequence of the original sequence can converge to. On the other hand, limsup, or the limit superior, is the largest limit that a subsequence can converge to. In other words, liminf and limsup represent the lower and upper bounds of the possible limits of a sequence.

## 3. What is the liminf/limsup inequality?

The liminf/limsup inequality states that for any sequence of real numbers, the liminf is always less than or equal to the limsup. In mathematical notation, liminf ≤ limsup. This inequality is a fundamental property of real numbers and plays a crucial role in many proofs and theorems in Real Analysis.

## 4. How is the liminf/limsup inequality used in Real Analysis?

The liminf/limsup inequality is used in Real Analysis to establish the existence of limits, characterize the convergence of sequences, and prove important theorems such as the Bolzano-Weierstrass theorem and the Cauchy convergence criterion. It also helps in understanding the behavior of sequences and their limits in various contexts.

## 5. Can the liminf/limsup inequality be extended to infinite sequences?

Yes, the liminf/limsup inequality can be extended to infinite sequences in a similar way as it is applied to finite sequences. The only difference is that for infinite sequences, the liminf and limsup may be equal to each other, whereas for finite sequences, they are always different. This extension is important in the study of series and their convergence in Real Analysis.

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