Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let ${a_k}$, ${b_k}$ be two sequences in $\mathbb{R}$. Suppose ${a_k}$ converges to $a$ s.t. $\liminf_{k \to \infty} {a_k} = \lim_{k \to \infty} {a_k} = a$.

Is it then true that $\liminf_{k \to \infty} (a_k - b_k) = \lim_{k \to \infty} a_k + \liminf_{k \to \infty} -b_k$?

share|improve this question

3 Answers 3

up vote 3 down vote accepted

Yes. In general you only have subadditivity of the limit inferior (that is, in general you only have $\liminf (a_n+b_n) \geq \liminf(a_n) + \liminf(b_n)$) but if at least one sequence converges, then you have equality.

To see why you have equality, think of $\liminf$ as the infimum of all limits of converging subsequences. If $a_{j_k}+b_{j_k}$ is a converging subsequence of $\{a_n+b_n\}$, then $b_{j_k}$ must also converge (since $a_{j_k}$ converges: every subsequence of a converging sequence converges); and likewise, if $b_{j_k}$ is a converging subsequence of $b_n$, then $a_{j_k}+b_{j_k}$ is a converging subsequence of $a_n+b_n$. Moreover, if $b_{j_k}$ converges to $s$, then $a_{j_k}+b_{j_k}$ converges to $a+s$. Conversely, if $a_{j_k} + b_{j_k}$ converges to $a+t$, then $b_{j_k}$ must converge to $t$. So if $L$ is the set of all points that are limits of subsequences of $b_n$, then the set of all points that are limits of subsequences of $a_n+b_n$ is $a+L = \{a+t \mid t\in L\}$. Therefore \[ \liminf (a_n+b_n) = \inf (a+L) = a + \inf L = a + \liminf b_n = \lim_{n\to\infty}a_n + \liminf b_n.\] To get the difference, just apply this to $a_n$ and $(-b_n)$.

(Or you can try the same argument with $\limsup$, and use the fact that $\liminf -b_n = -\limsup b_n$; you would think of $\limsup$ as the supremum of all limits of converging subsequences of $b_n$).

Now see if you can figure out why you only have an inequality in the general case...

share|improve this answer
    
Do you mean, "Conversely, if $a_{j_k}+b_{j_k}$ converges to $a+t$, then $b_{j_k}$ converges to $t$," or some equivalent reformulation of that part of your answer? (It is clear what you mean, but the way it is worded you seem to state the same implication twice.) –  Jonas Meyer Nov 1 '10 at 3:28
    
+1: It's a nice approach. –  Jonas Meyer Nov 1 '10 at 3:28
    
@Jonas: Yes; thank you. –  Arturo Magidin Nov 1 '10 at 3:33

Yes. Here's an approach to proving this, without full details. Given $\epsilon\gt0$, take $N$ sufficiently large so that $a_k-b_k$ is within $\epsilon$ of $a-b_k$ for all $k\geq N$. Then the $\liminf$ of the sequence $a_N-b_N,a_{N+1}-b_{N+1},\ldots$ must lie within $\epsilon$ of that of the sequence $a-b_N,a-b_{N+1},\ldots$. Removing a finite number of terms from the sequence doesn't change the $\liminf$.

share|improve this answer

Yes. One possible proof outline:

  1. If there is a $N$ such that $c_k \geq d_k$ for all $k > N$ then $\liminf c_k \geq \liminf d_k$
  2. If $C$ is a constant then $\liminf (C+a_k)=C+\liminf a_k$
  3. Let $\lim a_k=a$. For any $\epsilon > 0$ we can find a $N$ such that for all $k > N$ $$a-\epsilon - b_k \leq a_k - b_k \leq a+\epsilon -b_k$$ Apply (1) and (2), then make use of the fact that the result is true for all $\epsilon > 0$.
share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.