Lim inf and product commutation

  • Thread starter jostpuur
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  • #1
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Main Question or Discussion Point

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:

Answers and Replies

  • #2
2,111
16
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),
\quad\quad
(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.
 

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