# Sufficient condition for equality of limsups

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!

$a_n=\frac{1}{2}+(-1)^n\cdot\frac{1}{4}$

$b_n=1-a_n$

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

Hint: What if$$(\forall n\in\mathbb{N}):a_n=\frac12(1+(-1)^n)\text{ and }b_n=\frac12(1-(-1)^n)?$$

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.

• Actually, I noticed just a short time after I posted the question that I needed the liminf+limsup\leq limsup of sum inequality. This is still a great answer, nonetheless – Mr Martingale Jun 29 '18 at 10:15