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In Rudin, $\limsup$ is defined as follows:

Let $S$ be the set of numbers $x$ (in the extended real number system) such that $s_{n_k}\to x$ for some subsequence $\{s_{n_k}\}$. Then $$\limsup s_n = \sup S. \tag{1}$$

However, our real analysis instructor defined $\limsup$ in a different manner:

$$\limsup s_n = \lim_{n \to \infty} \sup_{m \ge n} s_m. \tag{2}$$

I am having trouble understanding how these two definitions are equivalent. It would be very helpful to me if somebody could provide a proof with some explanation.

My thoughts on the problem:

I have noticed that the usual trend with these sort of proofs is to prove the upper bound $(1) \le (2)$ and then the lower bound $(1) \ge (2)$ to get the desired conclusion. However, I am unsure how to even begin.

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2 Answers 2

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We first construct a sub-sequence in $S$ that has limit $\lim_{n\to\infty} \sup_{k>n} s_k$. We do this by observing that there's a $l(n)>n$ such that $\sup_{k>n}s_k \ge s_{l(n)} \ge \sup_{k>n}s_k-2^{-n}$, now $\lim_{n\to\infty}s_{l(n)} = \lim_{n \to \infty} \sup_{k>n} s_k$.

Now as $l(n)>n$ we have $l(n)\to\infty$ so $l(n)$ has an increasing subsequence $t_n$ and by that we can construct the desired subsequence of $s_n$.

That is we have proved that

$$\sup S \ge \lim t_n = \lim_{n\to\infty}\sup_{k>n}s_k$$

To prove the opposite lets consider "synchronized" subsequence $u_n$ (that $u_n=s_n$ or $u_n=u_{n-1}$ - just repeating the elements in the subsequence to keep them in sync) - this kind of construct doesn't alter the definition if $S$. Now $\{u_k: k>n\}\subseteq \{s_k: k>n\}$ and therefore $\sup_{k>n}u_k \le sup_{k_n}s_k$.

This means that:

$$\lim_{n\to\infty} u_k \le \lim_{n\to\infty}\sup_{k>n}u_k \le \lim_{n\to\infty}\sup_{k>n}s_k$$

That is for all subsequences in $S$ the limit is less than or equal the $\liminf$ of definition (2) that is:

$$\sup S\le\lim_{n\to\infty}\sup_{k>n}s_k$$

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    $\begingroup$ In your first line "we construct a subsequence in $S$..." I think you meant that a convergent subsequence of $s_n$ such that (obviously) it's limit lies in $S$. Is it so? $\endgroup$
    – Error 404
    Jul 21, 2018 at 11:48
  • $\begingroup$ Can you please elaborate on the "synchronized" subsequence? I couldn'e understand what you mean by that. Thanks a lot! $\endgroup$
    – Beerus
    Apr 7 at 16:19
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Let $s_n \in R$ and $S \subset R \cup \{+\infty,-\infty\}$. Define $s_n^* = \sup\{s_k \mid k\ge n\}$, $s^* = \lim_{n\to\infty}s_n^*$, and $\{s_k\}_{k\ge n} = s_n,s_{n+1},\dots$. If $s_1^* = +\infty$, then there exists a subsequence of $\{s_n\}$ that diverges to $+\infty$, so $\sup S = s^* = +\infty$. $s_1^* = -\infty$ is not possible because $s_n \in R$. Suppose $s_1^* < +\infty$. $\sup S \le s_n^*$ $(n\ge1)$ because $S$ is the set of subsequential limits of $\{s_k\}_{k\ge n}$ $(n\ge1)$. Therefore, $\sup S \le s^*$. Let $k_0 = 1$. For each $\{s_k\}_{k\ge n}$, there exists $k_n \ge n$ such that $k_n \ge k_{n-1}$ and $|s_n^* - s_{k_n}| < 1/2^n$ by the definition of the supremum. Construct a sequence $\{t_n\}$ by removing repeating elements from $s_{k_1},s_{k_2},\dots$. Then, $\{t_n\}$ is a subsequence of $\{s_n\}$ that converges to $s^*$. Therefore, $s^* \in S$ and $s^* \le \sup S$. It completes the proof.

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