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So I recently proved this inequality in my real analysis class: $$\lim\sup(a_n+b_n)\leq \lim\sup(a_n) + \lim\sup(b_n)$$ However I am wondering when this inequality is strictly less than. I've tried out a bunch of sequences and here is my thought process so far: $$$$ To get a view of when this inequality is unequal look at the two sequences: $$a_n=(0,1,0,1,0,....)$$ $$b_n=(1,0,1,0,1,....)$$ Clearly: $$\lim \sup (a_n+b_n)=1$$ However: $$\lim \sup (a_n)+\lim\sup(b_n)=2$$ This seems to imply that if the limit of a sequence does not exist then we have an inequality. Furthermore what if: $$a_n=(1,1,1,1,....)$$ $$b_n=(0,1,0,1,....)$$ Well then clearly: $$\lim \sup (a_n+b_n)=2$$ $$\lim \sup(a_n)+\lim \sup (b_n)=2$$ And we have equality, which contradicts our earlier hypothesis. Perhaps then it is if both limits do not exist, well let $b_n$ be defined as previously and $a_n=cb_n$, for some $c\in\mathbb{R}$ both these limits of the sequence do not exist however we obtain: $$\lim \sup (a_n+b_n)=c+1$$ $$\lim\sup(a_n)+\lim\sup(b_n)=c+1$$ And we have equality again. Well then perhaps it must be that both limits do not exist, and that the sequences cannot be scalar multiples of each other. Well then define: $$a_n=(0,1,0,1,0,....)$$ $$b_n-(0,0,0,1,0,0,0,1...)$$ Clearly: $$\lim \sup (a_n+b_n)=2$$ $$\lim \sup(a_n)+\lim \sup (b_n)=2$$ From here I tried testing some cases where $\sup(a_n)\neq \lim\sup(a_n)$ but I still continued to get equalities. So I really can't figure out for what conditions inequality holds. I guess if $a_n$ and $b_n$ look period but are shifted by a non even $n$ then that would make sense but I feel like there's more to it than that.

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    $\begingroup$ I suspect it's difficult to classify the cases with equality in any non-trivial, rigorous and useful way. But in some sense is when the "peaks" of the sequences (if they have oscillations) align often enough. $\endgroup$
    – Arthur
    Sep 25 '20 at 20:35
  • $\begingroup$ I as well think it is difficult to extract value from a characterization of the inequality. For example, equality holds iff there exists a sequence of indexes $n_k$ with a subsequence $n_{k_l}$ such that $a_{n_k} \to \limsup a_n$ and $ b_{n_{k_l}} \to \limsup b_n$. Broadly speaking, equality holds when they share a subsequence to go to their own limsup. $\endgroup$
    – Kolmo
    Oct 2 '20 at 23:33
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Set $$a=\limsup a_n, \quad b=\limsup b_n ,$$

$$ \limsup \ (a_n + b_n) \ \leq \ a+b . \tag{*} $$

Then, (*) holds with equality if and only if
$$ \exists \text{ a sequence of indexes } n_k \text{ such that} $$ $$ \lim_k a_{n_k}=a \quad \text{and} \quad \lim_k b_{n_k}=b. $$

Proof.

From RHS to LHS: $$ \limsup_n \ (a_n+b_n) \ \geq \ \lim_k \ (a_{n_k}+b_{n_k}) \ = \ a+b. $$ From LHS to RHS:

as $\limsup_n \ (a_n+b_n) \ = \ a+b$, $$ \text{it exists a subsequence of } a_n+b_n \text { such that} \\ \lim_k \ (a_{n_k}+b_{n_k}) \ = \ a+b . $$ Now, $\limsup_k a_{n_k} = a$. Otherwise: $$ a+b \ = \ \lim_k \ (a_{n_k}+b_{n_k}) \ = \ \limsup_k \ (a_{n_k}+b_{n_k}) \leq \limsup_k \ a_{n_k} + \limsup_k \ b_{n_k} \ < \ a+b $$ Therefore, it exists a sub-subsequence $a_{n_{k_l}}$ converging to $a$.

Finally, $b_{n_{k_l}}\ =\ (a_{n_{k_l}}+b_{n_{k_l}}) - a_{n_{k_l}} $, is the difference of two converging sequence with limits, respectively, $a+b$ and $a$.

Note that, if at least one of the sequence is convergent, then equality is achieved.

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