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

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?

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

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 $k\in\mathbb N$.

Without loss of generality, assume that $c_{n_k}=a_{n_k}$, for infinitely many $k$'s.

Then, $a_{n_k}\to c,\,$ and thus $c$ is a subsequential limit of $\{a_n\}$ and hence $$c\le \limsup_{n\to\infty} a_n.$$ Altogether $$\limsup_{n\to\infty}\max\{a_n,b_n\}=c\le \limsup_{n\to\infty} a_n \le \max\{\limsup_{n\to\infty} a_n, \limsup_{n\to\infty} b_n\}.$$

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 $M>\max(\limsup a_n,\limsup B_n)$,$$\limsup\max\{a_n,b_n\}\leqslant\max(\limsup a_n,\limsup B_n).$$

• Wow ! thank you so much! – Silent Jan 10 '18 at 12:23

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 $B$.