When do we have $\liminf_{n\to\infty}(a_n+b_n)=\liminf_{n\to\infty}(a_n)+\liminf_{n\to\infty}(b_n)$?

Question

It's not hard to show that $$\liminf_{n\to\infty}(a_n+b_n)\geq\liminf_{n\to\infty}(a_n)+\liminf_{n\to\infty}(b_n)$$ for any $\{a_n\},\{b_n\}\subset{\Bbb R}$ such that the right hand side is defined (i.e. no $\infty-\infty$ or $-\infty+\infty$). Also, if both $\lim a_n$ and $\lim b_n$ exist, then we have $$ \liminf_{n\to\infty}(a_n+b_n)=\liminf_{n\to\infty}(a_n)+\liminf_{n\to\infty}(b_n). $$

The wikipedia article about limit inferior and limit superior gives a sufficient conditions for "$=$" to hold which I don't see how to prove it:

if one of $\lim a_n$ and $\lim b_n$ exists, then we have "$=$".

Here are my questions:

How can I show the statement above?

Is this condition also necesarry?

Assume for example $\lim a_n=a$. To show $$ \liminf_{n\to\infty}(a_n+b_n)=a+\liminf_{n\to\infty}(b_n), $$ it suffices to show that $$ \liminf_{n\to\infty}(a_n+b_n)\leq\liminf_{n\to\infty}(a_n)+\liminf_{n\to\infty}(b_n) $$ I think somehow I would need to use $\liminf_{n\to\infty}(a_n)=\limsup_{n\to\infty}(a_n)=a$. But I don't see how this works.

Julien · Accepted Answer · 2013-03-22T02:17:17.803

Equality: I will use the most convenient characterization of $\liminf x_n$ as the least limit in $[-\infty,+\infty]$ of all converging subsequences of $x_n$ (including the infinity cases, where I would say, tend, rather than converge).

Take a subsequence $b_{n_k}$ of $b_n$ which tends to $\liminf b_n$. Then $$\lim_k a_{n_k}+b_{n_k}= \lim_k a_{n_k}+\lim_k b_{n_k}=\lim a_n +\liminf b_n.$$ So $\liminf (a_n+b_n)\leq \lim a_n +\liminf b_n$.

Now take a subsequence $a_{n_k}+b_{n_k}$ of $a_n+b_n$ which tends to $\liminf (a_n+b_n)$. Then $$\lim_k b_{n_k}=\lim_k (a_{n_k}+b_{n_k})-a_{n_k}=\liminf (a_n+b_n)-\lim a_n.$$ So $\liminf b_n\leq \liminf (a_n+b_n)-\lim a_n$.

This proves the desired equality.

Non equivalence: For instance $$ \liminf (-1)^n+(-1)^n=-2=(-1)+(-1)=\liminf (-1)^n+\liminf (-1)^n. $$

Halil Duru · Answer 2 · 2013-03-25T22:22:23.320

0

Simply $$\liminf_{n\to\infty} (b_n)= \liminf_{n\to\infty}(a_n-a+b_n)=\liminf_{n\to\infty} (a_n+b_n)-a $$

Note that in the first equality $|a_n-a|$ can be made less than $\varepsilon$ for sufficiently

large $n$ $\clubsuit$

edited Mar 25 '13 at 22:22

answered Mar 22 '13 at 01:49

Halil Duru

1,347

How do you justify the first equality? That's essentially what your asked to prove. – Julien Mar 22 '13 at 01:52
$a_n-a $ converges to $0$.. – Halil Duru Mar 22 '13 at 01:54
So LHS can't be even epsilon greater.. – Halil Duru Mar 22 '13 at 01:59
1

So? $a_n$ tends to $a$...That's the question. I agree it suffices to prove it when $a_n$ tends to $0$. But that's all. You still have to prove something. And the proof is not easier when the limit of $a_n$ is $0$. – Julien Mar 22 '13 at 01:59
You can make $a_n-a $ less than epsilon. – Halil Duru Mar 22 '13 at 02:01
That's what I say. You have to prove something. And again, it does not make it easier to bring this back to the $0$ case. – Julien Mar 22 '13 at 02:02
For instance, proving that $f(x)$ is continuous at $x_0$ is equivalent to proving that $f(x+x_0)$ at $0$. – Julien Mar 22 '13 at 02:04
Well, I don't. Personal tastes...that's not really important. The fact that you haven't proven anything, sorry to say that, is more annoying. – Julien Mar 22 '13 at 02:07
Much simpler, really? The translation from $a$ to $0$ is trivial. The proof of the equality in either case is easy, but not trivial. – Julien Mar 22 '13 at 02:14
? What particular notation did I use? Try to read through. Every step is easy. – Julien Mar 22 '13 at 02:18
Ok, you have a proof but extracting the meaning could be hard.But well-done!!+1 – Halil Duru Mar 22 '13 at 02:20
Thanks, what do you suggest I make more explicit? I like using this characterization of $\liminf$ and $\limsup$, because there is no more tedious $\epsilon$ manipulation. – Julien Mar 22 '13 at 02:22
it is fine in fact.. – Halil Duru Mar 22 '13 at 02:28

When do we have $\liminf_{n\to\infty}(a_n+b_n)=\liminf_{n\to\infty}(a_n)+\liminf_{n\to\infty}(b_n)$?

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