Does Lim Inf of Product Equal Product of Lim Infs for Non-Negative Sequences?

  • Context: Graduate 
  • Thread starter Thread starter jostpuur
  • Start date Start date
  • Tags Tags
    Commutation Product
Click For Summary
SUMMARY

The discussion confirms that for arbitrary non-negative sequences \( f_n \) and \( g_n \), the inequality \( \liminf (f_n g_n) \geq (\liminf f_n)(\liminf g_n) \) holds true. However, it establishes that this is the best conclusion possible without additional constraints, as counterexamples demonstrate that equality does not necessarily follow. Specifically, sequences such as \( f_n = (1, \frac{1}{2}, 1, \frac{1}{2}, \ldots) \) and \( g_n = (\frac{1}{2}, 1, \frac{1}{2}, 1, \ldots) \) yield a limit inferior of \( \frac{1}{2} \), which is greater than \( \frac{1}{4} \), the product of their respective limit inferiors.

PREREQUISITES
  • Understanding of limit inferior in sequences
  • Familiarity with real analysis concepts
  • Knowledge of sequences and their properties
  • Basic mathematical proof techniques
NEXT STEPS
  • Study the properties of limit inferior in more complex sequences
  • Explore counterexamples in real analysis to understand limitations of inequalities
  • Learn about convergence and divergence of sequences
  • Investigate the implications of non-negativity in sequence analysis
USEFUL FOR

Mathematicians, students of real analysis, and anyone interested in the properties of sequences and limits will benefit from this discussion.

jostpuur
Messages
2,112
Reaction score
19
If [tex]f_1,f_2,f_3,\ldots[/tex] and [tex]g_1,g_2,g_3,\ldots[/tex] are some arbitrary real sequences, is it true that

[tex] \underset{n\to\infty}{\textrm{lim inf}}\;(f_n g_n) = (\underset{n\to\infty}{\textrm{lim inf}}\; f_n) (\underset{n\to\infty}{\textrm{lim inf}}\; g_n)?[/tex]

For arbitrary [tex]\epsilon >0[/tex] there exists [tex]N\in\mathbb{N}[/tex] so that

[tex] n > N\quad\implies\quad f_n > \underset{k\to\infty}{\textrm{lim inf}}\; f_k \;-\; \epsilon,[/tex]

so

[tex] \underset{n\to\infty}{\textrm{lim inf}}\;(f_n g_n) \geq \underset{n\to\infty}{\textrm{lim inf}}\big((\underset{k\to\infty}{\textrm{lim inf}}\; f_k \;-\; \epsilon) g_n\big) \;=\; (\underset{k\to\infty}{\textrm{lim inf}}\; f_k \;-\; \epsilon) (\underset{n\to\infty}{\textrm{lim inf}}\; g_n),[/tex]

which implies

[tex] \underset{n\to\infty}{\textrm{lim inf}}\;(f_n g_n) \geq (\underset{n\to\infty}{\textrm{lim inf}}\; f_n) (\underset{n\to\infty}{\textrm{lim inf}}\; g_n),[/tex]

but I don't know how to do the other direction.

edit: I just realized I'm assuming [tex]\underset{n\to\infty}{\textrm{lim inf}}\; g_n \geq 0[/tex] in the calculation, although it was not my original intention, but I think I'll try to not fix it in the remaining editing time. (Actually assuming [tex]\underset{k\to\infty}{\textrm{lim inf}}\; f_k - \epsilon \geq 0[/tex] too...)

edit edit: In fact I think I'll add the assumption that the sequences are non-negative, because otherwise I have a counter example [tex](f_n)_{n\in\mathbb{N}} = (1,-1,1,-1,\ldots)[/tex], [tex](g_n)_{n\in\mathbb{N}}=(-1,1,-1,1,\ldots)[/tex].
 
Last edited:
Physics news on Phys.org
I think I got this thing solved. If [tex]f_1,f_2,f_3,\ldots[/tex] and [tex]g_1,g_2,g_3,\ldots[/tex] are some arbitrary non-negative sequences, then

[tex] \underset{n\to\infty}{\textrm{lim inf}}\;(f_n g_n) \geq (\underset{n\to\infty}{\textrm{lim inf}}\; f_n) (\underset{n\to\infty}{\textrm{lim inf}}\; g_n),[/tex]

is the best one can say about the liminfs. One counter example for other direction is

[tex] (f_n)_{n\in\mathbb{N}} = (1,\frac{1}{2},1,\frac{1}{2},\ldots),<br /> \quad\quad<br /> (g_n)_{n\in\mathbb{N}} = (\frac{1}{2},1,\frac{1}{2},1,\ldots),[/tex]

because

[tex] \underset{n\to\infty}{\textrm{lim inf}}\;(f_n g_n) = \frac{1}{2} > \frac{1}{4} = (\underset{n\to\infty}{\textrm{lim inf}}\; f_n) (\underset{n\to\infty}{\textrm{lim inf}}\; g_n).[/tex]

Thank you very much.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 0 ·
Replies
0
Views
4K