$\sup_{n >k}a_n \sup_{n>k}b_n = \sup_{m,n > k} a_nb_m \geq \sup_{n > k} a_nb_n$

Question

I'm considering the proof written by robjohn in the following post:

lim sup inequality $\limsup ( a_n b_n ) \leq \limsup a_n \limsup b_n $

Now, in that proof, he says that $$\sup_{n >k}a_n \sup_{n>k}b_n = \sup_{m,n > k} a_nb_m \geq \sup_{n > k} a_nb_n$$ by monotonicity of the sup function. How does he get the first equality sign and why the change of variable to $m$? And how does he get the second inequality?

Answer 1

The second inequality is true because the size of the set over which the supremum is searched, has been reduced, i.e. $$ \{a_nb_n, {n > k}\} \subseteq\{a_nb_m, {m,n > k} \} $$ so the supremum of first set is less than or equal to the supremum of the second set.

To see the first equality, see that (assuming positive values) for $n,m>k$ $$ a_nb_m\leq \sup_{n >k}a_n \sup_{n>k}b_n $$ Therefore: $$ \sup_{m,n > k} a_nb_m \leq\sup_{n >k}a_n \sup_{n>k}b_n$$ On the other hand for each $a_i$ we have: $$ \frac{ \sup_{m,n > k} a_nb_m} {\sup_{n>k}b_n} \geq \frac{ \sup_{n=i,m > k} a_ib_m} {\sup_{n>k}b_n} =a_i $$ and hence: $$ \sup_{n >k}a_n\leq \frac{ \sup_{m,n > k} a_nb_m} {\sup_{n>k}b_n} $$ which proves that: $$ \sup_{n >k}a_n \sup_{n>k}b_n = \sup_{m,n > k} a_nb_m $$