Is there a condition to characterize when limsup is additive?

Question

I know that in general $\limsup(a_{n} + b_{n}) \leq \limsup(a_{n}) + \limsup(b_{n})$. I was wondering when would the equality holds i.e. $\limsup(a_{n} + b_{n}) = \limsup(a_{n}) + \limsup(b_{n})$.

Answer 1

[I assume in this answer that $\limsup a_n$ and $\limsup b_n$ are both finite; indeed, if they could be infinite, then $\limsup a_n+\limsup b_n$ might not even be defined, if one is $\infty$ and the other is $-\infty$.]

One characterization is that $$\limsup (a_n+b_n)=\limsup a_n+\limsup b_n$$ iff both limsups can be realized by a common subsequence, in other words there exists a sequence $(n_k)$ going to $\infty$ such that $$\lim a_{n_k}=\limsup a_n$$ and $$\lim b_{n_k}=\limsup b_n.$$ Indeed, if such $(n_k)$ exists, then $\lim (a_{n_k}+b_{n_k})=\limsup a_n+\limsup b_n$ which implies $\limsup (a_n+b_n)\geq \limsup a_n+\limsup b_n.$

Conversely, suppose $\limsup (a_n+b_n)= \limsup a_n+\limsup b_n$ and call this common value $L$. Then there exists a sequence $(n_k)$ going to $\infty$ such that $\lim (a_{n_k}+b_{n_k})=L$. We then have $$L=\lim (a_{n_k}+b_{n_k})\leq \limsup a_{n_k}+\limsup b_{n_k}\leq \limsup a_n+\limsup b_n=L.$$ This implies we must have equality in the second $\leq$, which can only happen if $\limsup a_{n_k}=\limsup a_n$ and $\limsup b_{n_k}=\limsup b_n$ (here we are using the assumption that these limsups are finite). But then there is a subsubsequence $(a_{n_{k_i}})$ such that $\lim a_{n_{k_i}}=\limsup a_n$ and a subsubsubsequence $(b_{n_{k_{i_j}}})$ such that $\lim b_{n_{k_{i_j}}}=\limsup b_n$. This subsubsubsequence then has the desired properties.

(In fact, getting your hands a bit more dirty with epsilons, you can show that the original subsequences $(a_{n_k})$ and $(b_{n_k})$ themselves converge, so the subsubsequence and subsubsubsequence are unnecessary.)

Answer 2

If $a_n$ and $b_n$ are increasing, then their limsups are additive.

If they are oscillating, then basically, the two sequences have to be oscillating "in sync".

If you have an increasing subsequence a'$_k$ of a$_n$ such that for each n, there exists a k such that $a_n \leq a'_k$, then the limsup of $a_n$ is equal to $lim_{k \to \infty }{a'_k}$. If you can find $a'_k$ and $b'_k$ with the same indices, then the limsups are additive. That is, if there is a sequence $n_k$ such that both $a_{n_k}$ and $b_{n_k}$ have the above property with respect to their original sequences, then the limsups are additive.

If they are "out of sync", you can have one of them approaching the limsup on indices for which the other has arbitrary values, and vice versa. For instance, suppose $a_n = 1-\frac{1}{n}$ for even n, and $b_n = 1-\frac{1}{n}$ for odd n. Clearly both sequences have a limsup of at least 1, regardless of what their other values are, so the sum of their limsups is at least 2. Now suppose $a_n =\frac{1}{n}$ for odd n, and $b_n = \frac{1}{n}$ for even n. Then $a_n+b_n$ = 1 for all n, and so limsup($a_n+b_n$) = 1.

The reason their limsups don't add is that although, for all $\epsilon$, you can find an n for which 1-$a_n < \epsilon$, and an n for which 1-$b_n < \epsilon$, these aren't the same n.

It's analogous to sine waves; if they have the same period and phase, then their amplitudes add. If they are out of phase, then their amplitudes are subadditive.