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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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    $\begingroup$ In both cases you need the assumption $a_n\geq0$ and $b_n\geq0$ (for $n$ large enough). (Can you see some counterexamples?) $\endgroup$ – AD. Feb 25 '12 at 6:48
  • $\begingroup$ Can you propose some? I was long under the impression (long before seeing this post) that if $lim\{a_{n}\}$ exists and $\varlimsup\{b_{n}\} \rightarrow B < 0$, then equality still holds. $\endgroup$ – Muno Oct 31 '16 at 21:47
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  • This inequality (for $a_n,b_n\ge0$ and excluding the indeterminate forms $0\cdot\infty$ and $\infty\cdot0$) 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.

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.


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 $\displaystyle\limsup_{n\to\infty}a_n=\lim_{n\to\infty}a_n$.

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  • $\begingroup$ You really need the positivity of $a_n$ for this to work, yet your proof never mentions that. $\endgroup$ – Thomas Andrews Sep 23 '14 at 1:59
  • $\begingroup$ @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. $\endgroup$ – robjohn Sep 23 '14 at 5:20
  • $\begingroup$ You also need the sequences to be bounded if you want to perform algebraic manipulations with them, right? Otherwise the lim sup's might not be finite $\endgroup$ – Arkya Nov 15 '16 at 14:23
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    $\begingroup$ It can be arbitrarily small is the point. $\endgroup$ – robjohn Feb 11 at 0:48
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    $\begingroup$ Sorry, I just realized that I didn't hit enter after my last change. I was looking on my phone and wondered what happened to my edit. $\endgroup$ – robjohn Feb 11 at 0:54
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$$\{{a_m}\cdot {b_m}:m\geqslant n\}\subseteq \{{a_m}\cdot {b_k}:m,k\geqslant n\}$$

since we are pairing elements from two sets together in the first set while drawing each elements at random from two sets in the second set. By taking the supremum we have:

$$\sup\{{a_m}\cdot {b_m}:m\geqslant n\}\leqslant\sup\{{a_m}\cdot {b_k}:m,k\geqslant n\}\\=\sup\{\{{a_m}:m\geqslant n\}\cdot\sup\{\{{b_m}:m\geqslant n\}$$

which is seen by using the $\textbf{lemma}$ : $\sup (A*B)=\sup A* \sup B$ , where $(A*B)=\{a*b:a\in A,b\in B\}$.

Taking limit in the above inequality gives:

$$\lim_{n\to\infty}\sup\{{a_m}\cdot{b_m}:m\geqslant n\}\leqslant \lim_{n\to\infty}\sup\{{a_m}\cdot{b_k}:m,k\geqslant n\}\\=\lim_{n\to\infty}\sup\{\{{a_m}:m\geqslant n\} \cdot\lim_{n\to\infty} \sup\{\{{b_m}:m\geqslant n\}$$ $$Q.E.D$$

Proof of $\textbf{lemma}$: First we note that for any $x,X,y,Y\in\mathbb R$, from the inequalities $$x\leq X\\y\leq Y$$ it follows that $xy\leq XY$ if either $x\ge 0$ and $Y\ge 0$ or if $y\ge 0$ and $X\ge 0$ (a sufficient condition).

Thus, if $a\ge 0,\,\forall a\in A$ and $\sup B\ge 0$ or if $b\ge 0,\forall b\in B$ and $\sup A\ge 0$, we have $$\forall c\in A*B,\exists a\in A,b\in B,s.t.c=a\cdot b\leqslant \sup A *\sup B$$ So $A*B$ is bounded by $\sup A *\sup B$.

Now, if $a\ge 0,\,\forall a\in A$ and $\sup B> 0$ or if $b\ge 0,\forall b\in B$ and $\sup A> 0$, for any small enough $\epsilon$, we have $$\forall \varepsilon \gt 0,\exists a \in A,b \in B ,s.t.a \gt \sup A-\varepsilon ,b \gt \sup B -\varepsilon ,\\a\cdot b\gt {\sup A }\cdot {\sup B}-\varepsilon\cdot \big(\sup A+\sup B)- {\varepsilon}^{2}$$

So any number less than $\sup A *\sup B $ is not an upper bound. Thus $\sup A +\sup B $ is the least upper bound.

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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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    $\begingroup$ @AD. Yes, you're quite right. Fixed. $\endgroup$ – Alex Becker Feb 25 '12 at 6:58
  • $\begingroup$ Wait, did you assume $\lim a'_n = limsup(a_n)$ ? How does that follow from the fact that $a'_n b'_n \to limsup(a_n b_n)$? $\endgroup$ – cppcoder Jul 18 '20 at 12:33

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