liminf of products of sequences

Dec. 17, 2013

I needed a result about the liminf\liminf of a product of two sequences that I could not find in the references I had immediately at hand. Noting it down for future reference.


Let ana_n and bnb_n be two sequences such that liminfan=A\liminf a_n = A and limbn=B,0<B<\lim b_n = B, 0<B<\infty.

Then, liminfanbn=AB.\liminf a_nb_n = AB.


Since B>0B>0 it must be the case for all sufficiently large nn that 2B>bn>B/2>02B>b_n>B/2>0. Since all the properties asserted in the theorem depend only on the tail behaviour of the sequences there is no harm in assuming that this inequality holds for all nn.

We make use of the definition of liminf\liminf as the infimum of subsequential limits (Rudin, Principles of Mathematical Analysis, 3rd edition, Definition 3.16).

The theorem will be proved if we can show that any subsequence anka_{n_k} converges to a limit α\alpha if and only if the corresponding subsequence ankbnka_{n_k}b_{n_k} converges to αB\alpha B. If we can establish this fact, then, since B>0B>0, the infimum of the subsequential limits of anbna_nb_n will be BB times the infimum of the subsequential limits of ana_n.

When α\alpha is finite, the required fact is a direct consequence of the usual theorems on limits of products and quotients.

For α\alpha infinite we use the fact that bnb_n has positive lower and upper bounds, so ankbnka_{n_k}b_{n_k} will be unbounded above or below if and only if anka_{n_k} is unbounded in the same way.


The condition B>0B>0 is essential in the above theorem. Consider the following example: an=2,2,2,4,2,6,2,8,2,10,a_n = 2,2,2,4,2,6,2,8,2,10,\ldots. bn=1,1/2,1/3,1/4,b_n = -1,-1/2,-1/3,-1/4,\ldots. Then liminfan=2\liminf a_n = 2, limbn=0\lim b_n = 0, but liminfanbn=1\liminf a_nb_n = -1.

We can get by with B0B \ge 0 if we are willing to assume that ana_n is bounded. But not having to check boundedness when B>0B>0 is a big help.