Real Analysis – Lim Sup Inequality Proof: $\limsup ( a_n b_n ) \leq \limsup a_n \limsup b_n $

Question

I´m not sure how to start with this proof, how can I do it?
$$
\limsup ( a_n b_n ) \leqslant \limsup a_n \limsup b_n
$$
I also have to prove, if $ \lim a_n $ exists then:
$$
\limsup ( a_n b_n ) = \limsup a_n \limsup b_n
$$
Help please, it´s not a homework I want to learn.

Answer 1

The basic idea is what could be called the monotonicity of $\sup$: the supremum over a set is at least as large as the supremum over a subset.

Of course, this only makes sense if the product of the $\limsup$s is not $0\cdot\infty$ or $\infty\cdot0$. We also make the assumption that $a_n,b_n\gt0$. To see that this is necessary, consider the sequences $a_n,b_n=(-1)^n-2$.

Recall the definition of $\limsup$: $$ \limsup_{n\to\infty}a_n=\lim_{k\to\infty\vphantom{d^{d^a}}}\sup_{n>k}a_n\tag{1} $$ The limit in $(1)$ exists since, by the monotonicity of $\sup$, $\sup\limits_{n>k}a_n$ is a decreasing sequence.

Furthermore, also by the monotonicity of $\sup$, if $a_n,b_n\gt0$, $$ \sup_{n>k}a_n \sup_{n>k}b_n=\sup_{m,n>k}a_nb_m\ge\sup_{n>k}a_nb_n\tag{2} $$ Taking the limit of $(2)$ as $k\to\infty$ yields $$ \limsup_{n\to\infty}a_n\limsup_{n\to\infty}b_n\ge\limsup_{n\to\infty}a_nb_n\tag{3} $$ since the limit of a product is the product of the limits.

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If the limit of $a_n$ exists, we have that for any $\epsilon>0$, there is an $N$, so that $n>N$ implies $$ a_n\ge\lim_{n\to\infty}a_n-\epsilon\tag{4} $$ We are interested in small $\epsilon$, so it doesn't hurt to assume $\epsilon\lt\lim\limits_{n\to\infty}a_n$.

Thus, for $k>N$, if $a_n,b_n\gt0$, $$ \sup_{n>k}a_nb_n\ge\left(\lim_{n\to\infty}a_n-\epsilon\right)\sup_{n>k}b_n\tag{5} $$ taking the limit of $(5)$ as $k\to\infty$ yields $$ \limsup_{n\to\infty}a_nb_n\ge\left(\lim_{n\to\infty}a_n-\epsilon\right)\limsup_{n\to\infty}b_n\tag{6} $$ Since $\epsilon$ is arbitrarily small, $(6)$ becomes $$ \limsup_{n\to\infty}a_nb_n\ge\lim_{n\to\infty}a_n\limsup_{n\to\infty}b_n\tag{7} $$ Combining $(3)$ and $(7)$ yields $$ \limsup_{n\to\infty}a_nb_n=\lim_{n\to\infty}a_n\limsup_{n\to\infty}b_n\tag{8} $$ since $\displaystyle\limsup_{n\to\infty}a_n=\lim_{n\to\infty}a_n$.