# Lim inf and product commutation

• jostpuur

#### jostpuur

If $$f_1,f_2,f_3,\ldots$$ and $$g_1,g_2,g_3,\ldots$$ are some arbitrary real sequences, is it true that

$$\underset{n\to\infty}{\textrm{lim inf}}\;(f_n g_n) = (\underset{n\to\infty}{\textrm{lim inf}}\; f_n) (\underset{n\to\infty}{\textrm{lim inf}}\; g_n)?$$

For arbitrary $$\epsilon >0$$ there exists $$N\in\mathbb{N}$$ so that

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

so

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

which implies

$$\underset{n\to\infty}{\textrm{lim inf}}\;(f_n g_n) \geq (\underset{n\to\infty}{\textrm{lim inf}}\; f_n) (\underset{n\to\infty}{\textrm{lim inf}}\; g_n),$$

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

edit: I just realized I'm assuming $$\underset{n\to\infty}{\textrm{lim inf}}\; g_n \geq 0$$ 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 $$\underset{k\to\infty}{\textrm{lim inf}}\; f_k - \epsilon \geq 0$$ 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 $$(f_n)_{n\in\mathbb{N}} = (1,-1,1,-1,\ldots)$$, $$(g_n)_{n\in\mathbb{N}}=(-1,1,-1,1,\ldots)$$.

Last edited:
I think I got this thing solved. If $$f_1,f_2,f_3,\ldots$$ and $$g_1,g_2,g_3,\ldots$$ are some arbitrary non-negative sequences, then

$$\underset{n\to\infty}{\textrm{lim inf}}\;(f_n g_n) \geq (\underset{n\to\infty}{\textrm{lim inf}}\; f_n) (\underset{n\to\infty}{\textrm{lim inf}}\; g_n),$$

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

$$(f_n)_{n\in\mathbb{N}} = (1,\frac{1}{2},1,\frac{1}{2},\ldots), \quad\quad (g_n)_{n\in\mathbb{N}} = (\frac{1}{2},1,\frac{1}{2},1,\ldots),$$

because

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

Thank you very much.