Sum of limit inferior: $\liminf s_n + \liminf t_n \le \liminf (s_n+t_n)$

Question

I'm seeking to prove the following statement:

$$\lim \inf s_{n} + \lim \inf t_{n} \le \lim \inf (s_{n}+t_{n})$$ provided that $s_{n}$ and $t_{n}$ are bounded.

My solution so far:

Given that $s_{n}$ and $t_{n}$ are bounded, $\lim \inf s_{n}$ and $\lim \inf t_{n}$ both exist and are real numbers. Call them $s$ and $t$ respectively. Now since $\lim \inf s_{n} = s$, given some $\epsilon > 0$, the inequality $s_{n} > s + \epsilon$ fails for a finite amount of $n$'s. These $n$'s then have the property that $s_{n} + t_{n} \le s + t_{n} + \epsilon$...

So here I was able to find an inequality relating $s_{n}$ and $t_{n}$ but I seem to have it in the reverse order. Nor do I understand how to preserve the inequality if I argue about applying $\lim \inf$ to $s_{n}$ and $t_{n}$ in the inequality.

Any help would be appreciated.

Answer 1

So, I will do the analogous problem involving $\limsup$, leaving $\liminf$ up to you! :) I think that will be more beneficial to you anyway. (Study this problem first. Understand it. Then go back to your problem.)

Since both $(s_n)$ and $(t_n)$ are bounded, $(s_n + t_n)$ is bounded. Set $s = \limsup s_n$ and $t = \limsup t_n$, both of which are in $\mathbb{R}$. Let $\varepsilon > 0$. Since $s = \limsup s_n$, there exists an $N_1 \in \mathbb{N}$ such that, for every $n \in \mathbb{N}$, $n \geq N_1$ implies $s_n < s + \varepsilon/2$. Since $t = \limsup t_n$, there exists an $N_2 \in \mathbb{N}$ such that, for every $n \in \mathbb{N}$, $n \geq N_2$ implies $t_n < t + \varepsilon/2$.

Set $N = \max \{ N_1, N_2 \}$. Let $n \in \mathbb{N}$, and suppose $n \geq N$. Then $n \geq N_1$ and $n \geq N_2$, which implies $$ s_n + t_n < \biggl( s + \dfrac{\varepsilon}{2} \biggr) + \biggl( t + \dfrac{\varepsilon}{2} \biggr) = s + t + \varepsilon. $$ Since $\varepsilon$ was arbitrary, we have shown that there are (at most) finitely many $n \in \mathbb{N}$ for which $s_n + t_n \geq s + t + \varepsilon$; no subsequence of $(s_n + t_n)$ can converge to a number greater than $s + t$. That is, $\limsup {(s_n + t_n)} \leq s + t$.

Answer 2

First we understand the two crucial properties of $\liminf$. Let $m = \liminf x_{n}$ where $x_{n}$ is a bounded sequence. Then

• Given any $\epsilon > 0$ we have $x_{n} > m - \epsilon$ for all sufficiently large values of $n$.
• Given any $\epsilon > 0$ we have $x_{n} < m + \epsilon$ for infinitely many values of $n$.

Now consider your problem where $$s = \liminf s_{n}, t = \liminf t_{n}, u = \liminf (s_{n} + t_{n})$$ and assume on the contrary that $u < s + t$. We will derive a contradiction under this assumption. To do so we need to use the above properties of $\liminf$ and for that we choose a suitable $\epsilon$ namely $\epsilon = (s + t - u)/3$. Thus we have $s_{n} > s - \epsilon, t_{n} > t - \epsilon$ for all sufficiently large values of $n$ so that $s_{n} + t_{n} > u + \epsilon$ for all sufficiently large values of $n$. Since $u = \liminf s_{n} + t_{n}$ this contradicts the second property of $\liminf$ applied to $s_{n} + t_{n}$. Hence we must have $u \geq s + t$.

Answer 3

Your trying to prove the wrong inequality. The liminf of a sum is greater than the sum of the liminf's https://people.math.aau.dk/~cornean/analyse2_F14/limsup-liminf.pdf

Answer 4

Another way to approach it is the following:

\begin{align} \liminf_n t_n + \liminf_n s_n &= \lim_{n\to \infty} \inf_{k\ge n} t_k + \lim_{n\to \infty} \inf_{k\ge n} s_k\\ &= \lim_{n\to \infty} \left(\inf_{k\ge n} t_k + \inf_{k\ge n} s_k\right)\\ &\le \lim_{n\to \infty} \inf_{k\ge n} \left(t_k+s_k\right)\\ &= \liminf_n \left(t_n+s_n\right) \end{align}