$\liminf (a_n+b_n) = \liminf(a_n)+\liminf(b_n)$ provided that $\lim a_n$ exists
Question. Suppose $(a_n)$ and $(b_n)$ are two sequences of real numbers such that $\displaystyle \lim_{n\to\infty}a_n=a.$ Show that $$ \liminf_{n\to\infty}(a_n+b_n)=\liminf_{n\to\infty}(a_n)+\liminf_{n\to\infty}(b_n). $$ To avoid tedious discussion with infinity, assume in addition that both $a_n$ and $b_n$ are bounded sequences.
Notes. The question is based on the Wikipedia article on limit inferior and limit superior, where the statement above is mentioned explicitly in the article without any proof.
It follows immediately from the definition of $\liminf$ that $$ \liminf_{n\to\infty}(a_n+b_n)\geq\liminf_{n\to\infty}(a_n)+\liminf_{n\to\infty}(b_n) $$ See this section of the article.
So, the question reduces to showing the reversed inequality: $$ \liminf_{n\to\infty}(a_n+b_n)\leq\liminf_{n\to\infty}(a_n)+\liminf_{n\to\infty}(b_n) $$
There is one "algebraic" way to show this inequality, which I will write as my own answer below. Alternatives are welcome.
1 answer
We will use the following facts in the proof:
-
(1) If the limit of $(x_n)$ exists, so does that of $(-x_n)$
-
(2) If the limit of $x_n$ as $n\to\infty$ exists, then $$\liminf_nx_n=\limsup_nx_n=\lim_nx_n$$
-
(3) One obvious direction of the inequality is proved by definition: $$ \liminf_{n\to\infty}(a_n+b_n)\geq\liminf_{n\to\infty}(a_n)+\liminf_{n\to\infty}(b_n) $$
Proof.
$$ \begin{align} \liminf_{n\to\infty}b_n &\geq \liminf_{n\to\infty}(a_n+b_n)+\liminf_{n\to\infty} (-a_n)&\text{by fact (3)}\\ &=\liminf_{n\to\infty}(a_n+b_n)+\lim_{n\to\infty} (-a_n)&\text{by fact (1,2)}\\ &=\liminf_{n\to\infty}(a_n+b_n)-\lim_{n\to\infty} (a_n)\\ &=\liminf_{n\to\infty}(a_n+b_n)-\liminf_{n\to\infty} (a_n)\\ \end{align} $$which implies that $$ \liminf_{n\to\infty}(a_n+b_n)\leq\liminf_{n\to\infty}(a_n)+\liminf_{n\to\infty}(b_n) $$
0 comment threads
0 comment threads