I have a question which is potentially very basic but has me stumped. Let us suppose we have two real sequences $a_n,b_n$ which are non-negative and bounded by $1$. Is this sufficient to conclude
$\limsup{a_n}+\limsup{b_n}=\limsup{a_n+b_n}$?
$\geq$ is classical but I can't find anything claiming $\leq$. Any help would be much appreciated!
Hint: What if$$(\forall n\in\mathbb{N}):a_n=\frac12(1+(-1)^n)\text{ and }b_n=\frac12(1-(-1)^n)?$$
$a_n=\frac{1}{2}+(-1)^n\cdot\frac{1}{4}$
$ b_n=1-a_n$
Then $\lim \sup (a_n+b_n)=1$
However, $\limsup a_n = \limsup b_n = \frac{3}{4}$
This means that the equality doesn't always hold
This question has already been answered by asdf and José Carlos Santos, but the following stronger version might still be of interest.
In general, the following holds for sequences $\{a_n\}$ and $\{b_n\}$ of real numbers:
$$\liminf a_n \; + \; \liminf b_n \;\; \leq \;\; \liminf(a_n + b_n)$$
$$ \leq \;\; \min\left\{\,\liminf a_n + \limsup b_n, \;\; \limsup a_n + \liminf b_n\,\right\}$$
$$ \leq \;\; \max\left\{\,\liminf a_n + \limsup b_n, \;\; \limsup a_n + \liminf b_n\,\right\}$$
$$ \leq \;\; \limsup(a_n+b_n) \;\; \leq \;\; \limsup a_n \; + \; \limsup b_n $$
I'm pretty sure that strict inequality can hold throughout for the same two sequences, each of which is nonnegative and bounded by $1.$ For an example in which all but one of the inequalities is strict, let
$$ a_n \; = \; \begin{cases} 0 & \text{if $n$ is even} \\ 1 & \text{if $n$ is odd} \end{cases}$$
$$ b_n \; = \; \begin{cases} \frac{1}{2} & \text{if} \; n \equiv 0 \mod 4 \\ 0 & \text{if} \; n \equiv 1 \mod 4 \\ 1 & \text{if} \; n \equiv 2 \mod 4 \\ \frac{1}{2} & \text{if} \; n \equiv 3 \mod 4 \\ \end{cases}$$
That is,
$$ a_n \; = \; 1, \; 0, \; 1, \; 0, \; 1, \; 0, \; 1, \; 0, \; 1, \; 0, \; 1, \; 0, \; \ldots $$
$$b_n \; = \; 0, \; 1, \; \frac{1}{2}, \; \frac{1}{2}, \; 0, \; 1, \; \frac{1}{2}, \; \frac{1}{2}, \; 0, \; 1, \; \frac{1}{2}, \; \frac{1}{2}, \ldots $$
$$a_n + b_n \; = \; 1, \; 1, \; \frac{3}{2}, \; \frac{1}{2},\; 1, \; 1, \; \frac{3}{2}, \; \frac{1}{2},\; 1, \; 1, \; \frac{3}{2}, \; \frac{1}{2}, \; \ldots $$
For these two sequences the original inequality chain becomes
$$ 0 \;\; < \;\; \frac{1}{2} \;\; < \;\; 1 \;\; = \;\; 1 \;\; < \;\; \frac{3}{2} \;\; < \;\; 2 $$
Incidentally, for multiplication replacing addition, a similar inequality chain holds, as well as a similar example for strict inequality throughout.
