0
$\begingroup$

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!

$\endgroup$
2
$\begingroup$

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

$ 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

$\endgroup$
3
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – Mr Martingale Jun 29 '18 at 10:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.