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For real sequences $a_n$ and $b_n$, how to show that $$\limsup \max (a_n, b_n) \leq \max(\limsup a_n, \limsup b_n)?$$

It seems to me that $\sup(\max (a_n,b_n),\max(a_{n+1},b_{n+1}), \dots)$ is always greater than (or equal to) $\sup(a_n,a_{n+1}, \dots)$, so how do we get that inequality in reverse?

Yiorgos S. Smyrlis
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Silent
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3 Answers3

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Assume that $\,c_n=\max\{a_n,b_n\}$ and choose a subsequence $$ c_{n_k}\to \limsup_{n\to\infty}c_n= c\in[-\infty,\infty].\,\,$$

Clearly, either $c_{n_k}=a_{n_k}$, for infinitely many $k\in\mathbb N\,$ or $\,c_{n_k}=b_{n_k}$, for infinitely many $k\in\mathbb N$.

Hence, either a subsequence of $\{a_n\}$ converges to $c$ or a subsequence of $\{b_n\}$ converges to $c$.

Thus, either $$c\le \limsup_{n\to\infty}a_n \qquad \text{or}\qquad c\le \limsup_{n\to\infty}b_n, $$ and therefore $$ \limsup_{n\to\infty}c_n=c \le \max\big\{\limsup_{n\to\infty}a_n,\limsup_{n\to\infty}b_n\big\}. $$

Altogether $$ \limsup_{n\to\infty}\max\{a_n,b_n\} \le \max\{\limsup_{n\to\infty} a_n, \limsup_{n\to\infty} b_n\}. $$

Yiorgos S. Smyrlis
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  • How does the $c_{n_k}$ and the $a_{n_k}$ being equal for infinitely many indicies imply that $a_{n_k}$ must converge to $c$? $y_n = 1$ and $x_n = (-1)^n$ are equal for infinitely many indices, but $x_n$ does not even converge. – frito_mosquito Oct 11 '22 at 15:21
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    @frito_mosquito Because $c_{n_k}\to c$, and hence every subsequence of ${c_{n_k}}$ converges to $c$. Thus, as $a_{n_k}=c_{n_k}$, for infinitely many terms, then $a_{n_k}$ possesses a subsequence converging to $c$. – Yiorgos S. Smyrlis Oct 13 '22 at 07:30
  • Here an equality is considered which goes in the opposite direction https://math.stackexchange.com/questions/3089365/interchanging-limsup-and-sup I wonder if it is also correct – user12703198 Jul 31 '23 at 07:34
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Let $M\in\mathbb R$ be such that $M>\max(\limsup a_n,\limsup B_n)$ (if $\limsup a_n=+\infty$ or $\limsup b_n=+\infty$, there is no such $M$, but then then the statement is trivial). Take $\varepsilon>0$ such that $M-\varepsilon>\max(\limsup a_n,\limsup B_n)$. Since $M-\varepsilon>\limsup a_n$ there is a $p_a\in\mathbb N$ such that $n\geqslant p_a\implies a_n\leqslant M-\varepsilon$. And, since $M-\varepsilon>\limsup b_n$ there is a $p_b\in\mathbb N$ such that $n\geqslant p_b\implies b_n\leqslant M-\varepsilon$. So, if $p=\max\{p_a,p_b\}$, then$$n\geqslant p\implies\max\{a_n,b_n\}\leqslant M-\varepsilon.$$Therefore, $\limsup\max\{a_n,b_n\}\leqslant M-\varepsilon<M$. Since this occurs for each $M>\max(\limsup a_n,\limsup B_n)$,$$\limsup\max\{a_n,b_n\}\leqslant\max(\limsup a_n,\limsup B_n).$$

Silent
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José Carlos Santos
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First, note how $\lim\sup(a_n)$ is nothing more than the set of all subsequetial limits of $a_n$. A set of numbers, that's it. Let us call this set of numbers $A$ and consider a similar set $B$ for $b_n$. We'll also let $c_n =\max(a_n,b_n)$ so we may use $C$ as the collection of all subsequential limits of $\{c_n\}$

We then wish to show that $$\sup C \leq \max(\sup A, \sup B).$$

So suppose we have some subsequence $\{c_{n_k}\}$, this can continue in 3 different ways:

  1. For some $N$, $c_{n_k} = a_{n_k}$, so $c_{n_k}\leq \sup(A)$ whenever $n_k>N$ .
  2. For some $N$, $c_{n_k} = b_{n_k}$, so $c_{n_k}\leq \sup(B)$ whenever $n_k>N$.
  3. There exists no such $N$ for which this is the case.

In case 1 and case 2, the inequality trivially holds.

For case 3 we're now able to assume more information: For every $a_{n_i}\in\{c_{n_k}\}$, there exists some $b_{n_j}$ that comes after $a_{n_i}$, and the other way around is equally true.

We therefore know that if we take the subsequences $\{a_{n_i}\}, \{b_{n_j}\}$ of $c_{n_k}$, that they converge to the same number.* This means that $$\sup C\leq \sup A,\qquad\sup C\leq \sup B$$ and so definitely $$\sup C \leq \max(\sup A, \sup B).$$

*: Of course they may also diverge, but the inequality trivially holds when $\infty$ lies in $A$ or $B$.

Mitchell Faas
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