lim sup inequality $\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 $$ a_n\ge\lim_{n\to\infty}a_n-\epsilon\tag{4} $$ For the time being, assume that $\lim\limits_{n\to\infty}a_n\gt0$. We are interested in small $\epsilon$, so it doesn't hurt to assume $\epsilon\lt\lim\limits_{n\to\infty}a_n$. For $k\gt N$, 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 $$ \limsup_{n\to\infty}a_nb_n\ge\lim_{n\to\infty}a_n\limsup_{n\to\infty}b_n\tag{7} $$ If $\lim\limits_{n\to\infty}a_n=0$, then $(7)$ is trivial, so we can remove the assumption that $\lim\limits_{n\to\infty}a_n\gt0$.

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

Answer 2

$$\{{a_m}\cdot {b_m}:m\geqslant n\}\subseteq \{{a_m}\cdot {b_k}:m,k\geqslant n\}$$

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.

Answer 3

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.

Answer 4

• 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.
• R. Bartle: A modern theory of integration (AMS, 2001), Theorem A.3(f) in Appendix A (which is devoted to limit superior and limit inferior).
• Problem 42 in Section 1.5 of H. L. Royden and P. M. Fitzpatrick: Real Analysis, (Fourth Edition). (This book does not include solutions to problems.)

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

Answer 5

Let's assume that $(a_n)$ and $(b_n)$ are non-negative.

Suppose that both the quantities on RHS are finite. $$\limsup a_n\limsup b_n=(\lim \sup_{m\ge n}a_m) (\lim\sup_{m\ge n}b_m)=\lim(\sup_{m\ge n}a_m\sup_{m\ge n}b_m)\tag 1$$ where the second equality follows by limit rules (product of limits of two sequences is equal to limit of product of the sequences, if both the limits under product exist).

We have $\sup_{m\ge n}a_m\sup_{m\ge n}b_m\ge a_kb_k$ for every $k\ge n$ (note that this is where we require the non-negativity of the sequences) and hence by completeness property of the reals: $\sup_{k\ge n} a_kb_k\le \sup_{m\ge n}a_m\sup_{m\ge n}b_m$. Taking limit as $n\to \infty$ on both sides, we get: $$\lim\sup_{k\ge n} a_kb_k\le \lim (\sup_{m\ge n}a_m\sup_{m\ge n}b_m)\tag 2$$

It follows by $(1)$ and $(2)$ that $$\limsup a_n\limsup b_n\ge \limsup (a_nb_n)\tag 3$$

$(3)$ does not hold if the sequence $(a_n)$ is defined as $a_{2k}=0$ and $a_{2k+1}=-1$ and $b_n:=a_n$ for all $n\in \mathbb N$. (Note that $\limsup a_nb_n=1$ and $\limsup a_n=0=\limsup b_n$).

It can be shown using the same ideas as above that: $$\liminf a_n\limsup b_n\le \limsup (a_nb_n)\le \limsup a_n \limsup b_n$$, where the second inequality has already been proven above. The second result in OP follows by noting that $\lim a_n$ exists $\iff \limsup a_n=\liminf a_n$.

The case when either of the quantity inside parentheses in RHS of first equality of $(1)$ is $+\infty$ (and the other quantity is $+\infty$ or a positive number) can be handled separately noting that in that case the corresponding sequence will be unbounded.

Answer 6

[Comment: This is, frankly, one of the most surreal and bizarre proofs I've ever written. Please let me know if you identify a mistake].

We provide a very simple proof which makes use of the following facts. First, that any sequence $s_n$ admits a convergent subsequence which converges to $\limsup s_n$. Second, any subsequence of a convergent sequence converges to the same value. Third, the limsup of a sequence is at least the limsup of any subsequence. Each of these claims hold even when the limits or limsups are $+ \infty$.

There is a subsequence $n_j$ such that $\lim_{j} a_{n_j}b_{n_j} = \limsup_{n} a_{n}b_{n}$. There is a further subsequence ${n_{j_{k}}}$ such that $\lim_{k} a_{{n_{j_{k}}}} = \limsup_{j} a_{n_j}$. There is yet a further subsequence $n_{j_{k_q}}$ such that $\lim_{q} b_{n_{j_{k_q}}} = \limsup_{k} b_{n_{j_{k}}}$.

Hence $\limsup_{n} a_{n}b_{n} = \lim_{j} a_{n_j}b_{n_j} = \lim_{q} a_{n_{j_{k_q}}}b_{n_{j_{k_q}}} \overbrace{=}^{*} \lim_{q} a_{n_{j_{k_q}}}\lim_{q} b_{n_{j_{k_q}}} = \limsup_{q} a_{n_{j_{k_q}}}\limsup_{q} b_{n_{j_{k_q}}} \overbrace{\leq}^{**} \limsup_{n} a_n \limsup_{n} b_n$

At ($*$), we use the fact that limits commute with multiplication, provided that we don't have a $0 \cdot \infty$ situation. At ($**$), we use the monotonicity of the $\limsup$ --- that is, the limsup of the whole sequence is at least the limsup of any subsequence. That, and the fact that $a \leq b \land c \leq d \to ac \leq cd$ provided $a,b,c,d \in [0, +\infty]$, and none of the products are of the form $0 \cdot \infty$.

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A similar argument can show we have equality if $\alpha = \lim_{n} a_n$ exists. We again avoid the $0 \cdot \infty$ case.

If $\alpha = 0$ and $\limsup_{n} b_n<C<\infty$, then use that $b_n<C+2$ for sufficiently large $n$. Hence $0 \leq \limsup a_nb_n \leq (C+2)\limsup a_n = 0$. So we have equality in this case.

Otherwise assume $\alpha>0$.

We start as before. There is a subsequence $n_j$ such that $\lim_{j} a_{n_j}b_{n_j} = \limsup_{n} a_{n}b_{n}$. There is a further subsequence $n_{j_{k}}$ such that $\lim_{k} b_{n_{j_{k}}} = \limsup_{j} b_{n_{j}}$. Thus $$\limsup_{n} a_nb_n = \lim_{j} a_{n_j}b_{n_j} = \lim_{k} a_{n_{j_{k}}} b_{n_{j_{k}}} = (\lim_{k} a_{n_{j_{k}}})(\lim_{k} b_{n_{j_{k}}}) = \alpha \limsup_{j} b_{n_j} \ \ \ \ \ \ \ \ (1)$$

We claim that $\limsup_{j} b_{n_j} = \limsup_{n} b_n$. If this isn't true, then there is another subsequence $m_{j}$ such that $\limsup_{j} b_{n_j} < \lim_{j} b_{m_j} = \limsup_{n} b_n$. But, using that $\lim_{n} a_n$ exists, $\limsup_{n} a_nb_n \geq \lim_{j} a_{m_j}b_{m_j} = \alpha \lim_{j} b_{m_j}$. By $(1)$, this means $\alpha \limsup_{j} b_{n_j} \geq \alpha \lim_{j} b_{m_j} = \alpha\limsup_{j} b_{m_j}$, and dividing through we get $\limsup_{j} b_{n_j} \geq \limsup_{j} b_{m_j} = \lim_{j} b_{m_j}$ contradicting our choice of $m_j$.