$\lim\sup_{n \to \infty} (a_n-b_n) \geq \lim\sup_{n \to \infty} (a_n) - \lim\sup_{n \to \infty} (b_n)$?

Question

$\lim\sup_{n \to \infty} (a_n-b_n) \geq \lim\sup_{n \to \infty} (a_n) - \lim\sup_{n \to \infty} (b_n)$? One possible condition is $a_n \geq b_n$, but I dont know if it is possible without this condition. Thank you.

Answer 1

In steps: each step after the first follows from the previous one.

$1).\ a_n=a_n-b_n+b_n=(a_n-b_n)+b_n$

$2).\ a_n\le \sup_{k\ge n}(a_n-b_n)+b_n\le \sup_{k\ge n}(a_n-b_n)+\sup_{k\ge n}b_n$

$3).\ \sup_{k\ge n}a_n\le \sup_{k\ge n}(a_n-b_n)+\sup_{k\ge n}b_n$

$4).\ \underset{n\to \infty}\lim\sup_{k\ge n}a_n\le \underset{n\to \infty}\lim\sup_{k\ge n}(a_n-b_n)+\underset{n\to \infty}\lim\sup_{k\ge n}b_n$

$5).\ \underset{n\to \infty}\lim\sup_{k\ge n}a_n-\underset{n\to \infty}\lim\sup_{k\ge n}b_n\le \underset{n\to \infty}\lim\sup_{k\ge n}(a_n-b_n)$

Answer 2

We know that $\limsup x_n + \liminf y_n \leq \limsup (x_n + y_{n})$. But, $\liminf y_{n} =-\limsup -y_{n}$. Then choose $y_{n} = -b_{n}$ and $x_{n}=a_{n}$

We have $\limsup x_n + \liminf y_n \leq \limsup (x_n + y_n) \Rightarrow \limsup a_n -\limsup b_{n} \leq \limsup (a_n -b_n)$.

Look this link https://math.la.asu.edu/~quigg/teach/courses/371/2005fall/lectures/limsup.pdf