# Show that $\liminf_{n \to \infty} x_n+ \liminf_{n \to \infty} y_n \leq \liminf_{n \to \infty}(x_n + y_n)$

Let $\{x_n\}$ and $\{y_n\}$ be bounded sequences.

Show that $\liminf_{n \to \infty} x_n+ \liminf_{n \to \infty} y_n \leq \liminf_{n \to \infty}(x_n + y_n)$

Hint: Find a subsequence ${x_{n_i} +y_{n_i} }$ of $x_n + y_n$ that converges. Then find a subsequence ${x_{n_{m_i}}}$ of ${x_{n_i}}$ that converges. then apply what you know about limits.

Even though there is a hint, hint doesn't help me ..

I have Let ${x_{n_i} +y_{n_i}}$ be subsequence of ${x_n + y_n}$ $\lim \inf (x_n + y_n) = \lim (x_{n_i} + y_{n_i})$ then $x_{n_i}$ is sequence in R a nd it has monoton sequence $x_{n_{m_i}}$ . This subsequence is bounded Therefore ${x_n}$ is bounded, so it is convergent.

This is all I can do..also, I am not sure wether what I did is right or not..

• You can't show that inequality if $\lim\limits_{n\to\infty} x_n+y_n$ does not exist. – user228113 Mar 10 '17 at 4:36
• I believe you meant to write $\liminf_{n\to\infty} (x_n+y_n)$. – Yes Mar 10 '17 at 4:54
• sorry I fixed ! – user421044 Mar 10 '17 at 4:56

Note that for $k\geq n,$ we have the following inequality,
$$\inf\limits_{k\geq n}x_k+\inf\limits_{k\geq n}y_k \leq x_j+y_j \mbox{ for all }j\geq n.$$ Hence, we obtain $$\inf\limits_{k\geq n}x_k+\inf\limits_{k\geq n}y_k \leq \inf\limits_{j\geq n}(x_j+y_j).$$ That is, $$\inf\limits_{k\geq n}x_k+\inf\limits_{k\geq n}y_k \leq \inf\limits_{k\geq n}(x_k+y_k).$$ All this three sequence are monotone and bounded, so all limits exist. Taking $n\rightarrow \infty.$ We complete the proof.
Suppose $\liminf\limits_{n\to\infty} (x_n+y_n)=L$. There exists a subseqence $x_{n_k}+y_{n_k}$ s.t. $\lim\limits_{k\to\infty} (x_{n_k}+y_{n_k})=L$. $x_{n_k}$ is bounded, so there exists a subsequence $x_{n_{k_l}}$ s.t. $\lim\limits_{l\to\infty} x_{n_{k_l}} = K$. From limits arithmetics we have: $$\lim\limits_{l\to\infty} y_{n_{k_l}}=\lim\limits_{l\to\infty} (x_{n_{k_l}}+y_{n_{k_l}}-x_{n_{k_l}})=\lim\limits_{l\to\infty} (x_{k_l}+y_{n_{k_l}}) - \lim\limits_{l\to\infty} x_{n_{k_l}}=L-K$$ Meaning, $y_{n_{k_l}}$ converges.
Clearly, $L-K\geq\liminf\limits_{n\to\infty} y_n$, $K\geq\liminf\limits_{n\to\infty}x_n$ Therefore we have: $$\liminf\limits_{n\to\infty}(x_n+y_n)=L=K+(L-K)\geq\liminf\limits_{n\to\infty} x_n+\liminf\limits_{n\to\infty} y_n$$
• How do we know that L - K is greater than or equl to $\lim \inf y_n$ also K is greater than or equl to $\lim \inf x_n$ – user421044 Mar 10 '17 at 6:26
• $\liminf x_n$ is the smallest limit of any subsequence of $x_n$ (try to prove it - it's a good exercise). Same goes for $x_n+y_n$. – Yes Mar 10 '17 at 6:39