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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 $\liminf_{k \to \infty} (a_k - b_k) = \lim_{k \to \infty} a_k + \liminf_{k \to \infty} -b_k$?

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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...

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  • $\begingroup$ 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.) $\endgroup$ – Jonas Meyer Nov 1 '10 at 3:28
  • $\begingroup$ @Jonas: Yes; thank you. $\endgroup$ – Arturo Magidin Nov 1 '10 at 3:33
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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 $\liminf$.

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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 $\epsilon > 0$.
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