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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.

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In both cases you need the assumption $a_n\geq0$ and $b_n\geq0$ (for $n$ large enough). (Can you see some counterexamples?) –  AD. Feb 25 '12 at 6:48

3 Answers 3

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  • The inequality is a part of Problem 2.4.17 in the book Wieslawa J. Kaczor, Maria T. Nowak: Problems in mathematical analysis: Volume 1; Real Numbers, Sequences and Series, Problem 2.4.15. The problem is given on p.44 and solved on p.200-201. (AFAIK this book is also available in French and Polish.) See also this answer.

I am making this CW, feel free to add other references.

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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.

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.

If the limit of $a_n$ exists, we have that for any $\epsilon>0$, there is an $N$, so that $n>N$ implies $$ a_n\ge\lim_{n\to\infty}a_n-\epsilon\tag{4} $$ 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 arbitrary, $(6)$ 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 $\displaystyle\limsup_{n\to\infty}a_n=\lim_{n\to\infty}a_n$.

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You really need the positivity of $a_n$ for this to work, yet your proof never mentions that. –  Thomas Andrews Sep 23 '14 at 1:59
@ThomasAndrews: it was used implicitly in $(2)$ and $(5)$. I missed that this was not actually stated in the question, so I have now added this to those steps. –  robjohn Sep 23 '14 at 5:20

I assume all relevant values are positive as otherwise this is false. Note that whenever $\limsup(a_nb_n)$ exists, we have some subsequence $(a_n'b_n')$ of $(a_nb_n)$ which converges to $\limsup(a_nb_n)$. For any $\epsilon>0$, we have some $N$ such that $$k\geq N\implies a_k'b_k'>\limsup(a_nb_n)-\epsilon\text{ and } b_k'<\limsup(b_n)+\epsilon$$ and so we have $$k\geq N\implies a_k'>\frac{\limsup(a_nb_n)-\epsilon}{b_k'}>\frac{\limsup(a_nb_n)-\epsilon}{\limsup(b_n)+\epsilon}$$ and this goes to $\frac{\limsup(a_nb_n)}{\limsup(b_n)}$ as $\epsilon\to 0,k\to\infty$ giving us $\limsup(a_n)\geq \frac{\limsup(a_nb_n)}{\limsup(b_n)}$ so $$\limsup(a_n)\limsup(b_n)\geq \limsup(a_nb_n).$$ I will leave the case where $\lim\limits_{n\to\infty}(a_n)$ exists to you, as it is similar.

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@AD. Yes, you're quite right. Fixed. –  Alex Becker Feb 25 '12 at 6:58

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