liminf of a product

Printable View

December 30th 2010, 12:57 AM
amheissan

liminf of a product

I am trying to prove the following:
$\underbrace{(\lim \inf f_n)}_f\underbrace{(\lim\inf g_n)}_g\leq \underbrace{\lim\inf (f_ng_n)}_k$ .
Here's what I have so far:
FTSOC assume $fg>k$ . We know that $f_n<f-\epsilon$ and $g_n<g-\epsilon$ for only finitely many $n$ . Then $f_ng_n<fg-\epsilon(f+g-\epsilon)$ for only finitely many $n$ . This implies $f_ng_n\geq fg-\epsilon(f+g-\epsilon)$ almost always. So, $f_ng_n\geq k-\epsilon(f+g-\epsilon)$ almost always.
But this doesn't get me anywhere. Where am I going wrong? If I let $\epsilon=\min\{f,g\}$ , I just get $f_ng_n\geq 0$ . So, this tells me nothing.
Any help would be appreciated. (Even though it doesn't state it, I am assuming that all $f_n, g_n$ are nonnegative.)
December 30th 2010, 07:50 AM
snowtea

I don't think that is true.
Consider $f_n$ the sequence $-1,0,-1,0,...$
and $g_n$ the sequence $2,0,2,0,...$
so $f_ng_n$ is $-2, 0,-2, 0,...$

Now $(\lim \inf f_n)(\lim\inf g_n) = -1 \times 0 = 0$
$\lim\inf (f_ng_n) = -2$
December 30th 2010, 08:59 AM
amheissan

Quote:

Originally Posted by amheissan

...I am assuming that all $f_n, g_n$ are nonnegative.

You are right that it wouldn't hold if were allowed to have a function as you have described, but we are limited to non-negative functions.

Any more thoughts on proving this property? Thanks!
December 30th 2010, 09:18 AM
Drexel28

Quote:

Originally Posted by amheissan

You are right that it wouldn't hold if were allowed to have a function as you have described, but we are limited to non-negative functions.

Any more thoughts on proving this property? Thanks!

Note that by definition $\displaystyle \liminf_{n\to\infty}f_n(x)=\inf \left\{\lim_{n\to\infty}f_{n_k}(x):f_{n_k}(x)\text { is a subsequence of }f_n(x)\right\}$ and similarly for $\displaystyle \liminf_{n\to\infty}g_n(x)$ . Use the fact then that $\inf\left(A\right)\inf\left(B\right)\leqslant \inf\left(AB\right)$ .
December 30th 2010, 09:33 AM
amheissan

Okay, but how would you prove $\inf(A)\inf(B)\leq \inf (AB)$ ? I'm really just not seeing this property.
December 30th 2010, 12:17 PM
snowtea

For only nonnegative sequences, Drexel28 is giving you the hint to only look at the inf of the set of values at any subsequence (ignore the order of the sequence).

Let $AB = \{ab | a\in A, b\in B\}$ .
Consider any element $a\in A$ and $b\in B$ , $ab \geq \inf(A)\inf(B)$ (by definition since everything is nonnegative).

Also, remember that the inf of a subset of AB must be larger than the inf of AB