# I have a question about limit supremum and limit infimum.

Let $\{x_n\}_{n=1}^{\infty}$ and $\{y_n\}_{n=1}^{\infty}$ be sequences of real numbers. Does the following hold:

$$\limsup x_n +\liminf y_n \le \limsup\,(x_n+y_n).$$

This is what I have tried but I am not quite sure if it is correct. $\text{Fix } K>1. \text{ Let }L=\inf_{1\le i \le k}y_i$. Now $$\sup_{1 \le i \le k}(x_i+y_i)\ge \sup_{1\le i \le k}(x_i+L)=L+\sup_{1\le i \le k}(x_i) = \inf_{1\le i \le k}(y_i)+\sup_{1\le i \le k}(x_i).$$ Now we take the limit as $k \rightarrow \infty$ and we have the desired result. Does this look OK?

• I'm not certain enough for this to be an answer, but it is certain that $\liminf(y_n) \le \limsup(y_n)$, and if $\limsup$ is also linear then it should follow. – Mark Fantini Jan 19 '14 at 10:11
• I think I have edited my question so that it fits the standard of the site. How can I get the "hold" removed? – tmpys Jan 19 '14 at 22:00

Let $$~\limsup x_n=\overline x~$$ and $$~\liminf y_n=\underline y~,$$ then $$~\overline x~,~\underline y~\in \mathbb R.$$
Let $$~\epsilon>0~$$ be given.
Then $$~\exists~~k\in \mathbb N~$$ such that
$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x_n>\overline x-\epsilon~/~2~,~~\forall~~n\ge k~$$ and
$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~y_n>\underline y-\epsilon~/~2~,$$ for infinitely many value of $$~n~.$$
Thus $$~x_n+y_n>\overline x+\underline y-\epsilon~,$$ for infinitely many value of $$~n~.$$
Therefore for given $$~m\in\mathbb N~,~~\exists~~n_0\ge m~$$ such that $$x_{n_0}+y_{n_0}>\overline x+\underline y-\epsilon$$ $$\implies \sup_{n\ge m}\left(x_n+y_n\right)\ge x_{n_0}+y_{n_0}>\overline x+\underline y-\epsilon~,$$ for each $$~m\in \mathbb N~.$$
Hence $$\overline x+\underline y-\epsilon\le\limsup\,(x_n+y_n)$$ Since $$~\epsilon > 0~$$ be arbitrary, $$\limsup x_n +\liminf y_n \le \limsup\,(x_n+y_n)~.$$

$$\limsup (x_n+y_n) = \lim_{n\to\infty} \left(\sup_{k>n}(x_k+y_k)\right) \leq \lim_{n\to\infty}\left(\sup_{k>n}x_k+\sup_{k>n}y_k\right)=\\=\limsup(x_n)+\limsup (y_n)$$

The inequality is because a property of supreme: $\sup (a+b) \leq \sup (a) + \sup (b)$.

• You forgot the dollar signs to turn them into LaTeX. You should also consider explaining the motivation to the OP, as the mere string of symbols does not lend itself to thought-process construction. – Mark Fantini Jan 19 '14 at 10:17
• ok well now the answer you wrote is gone, and there was a different equality the first time. – tmpys Jan 19 '14 at 10:35