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Lim inf and product commutation

  1. May 19, 2008 #1
    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: May 19, 2008
  2. jcsd
  3. May 19, 2008 #2
    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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