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Suppose $\{s_n\}$ and $\{t_n\}$ are bounded sequences of nonnegative numbers. Show that as $n\rightarrow \infty$ $$\limsup(s_n.t_n) ≤ \limsup(s_n) · \limsup(t_n)$$

My attempt:

Let $\limsup(s_n)=l$ and $\limsup(t_n)=m$ then $s_n<l+\epsilon$ for $n>N_1$ and $t_n<m+\epsilon$ for $n>N_2$. Also $\{s_n\}$ and $\{t_n\}$ are bounded sequences, sequence $\{s_n.t_n\}$ is also bounded. But $$s_nt_n<lm+l\epsilon +m\epsilon +\epsilon^2$$

How to prove that. I have no idea how to prove. Please help.

user1942348
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3 Answers3

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$$ \begin{align} \limsup_{n\to\infty}s_nt_n &=\lim_{n\to\infty}\sup_{k\ge n}s_kt_k\tag{1}\\ &=\lim_{n\to\infty}\sup_{\substack{j\ge n\\k\ge n\\j=k}}s_jt_k\tag{2}\\ &\le\lim_{n\to\infty}\sup_{\substack{j\ge n\\k\ge n}}s_jt_k\tag{3}\\ &=\lim_{n\to\infty}\sup_{j\ge n}s_j\sup_{k\ge n}t_k\tag{4}\\ &=\lim_{n\to\infty}\sup_{j\ge n}s_j\lim_{n\to\infty}\sup_{k\ge n}t_k\tag{5}\\ &=\limsup_{n\to\infty}s_n\limsup_{n\to\infty}t_n\tag{6} \end{align} $$ Explanation:
$(1)$: definition
$(2)$: rewrite in terms of two variables
$(3)$: the sup over a smaller set is less than the sup over a larger set
$(4)$: separate the sups
$(5)$: limit of a product is the product of the limits
$(6)$: definition

Clarification

$$ \begin{align} \sup_{\substack{j\ge n\\k\ge n}}s_jt_k &=\lim_{m_1,m_2\to\infty}\max_{\substack{m_1\ge j\ge n\\m_2\ge k\ge n}}s_jt_k\tag{7}\\ &=\lim_{m\to\infty}\max_{\substack{m\ge j\ge n\\m\ge k\ge n}}s_jt_k\tag{8}\\ &=\lim_{m\to\infty}\max_{m\ge j\ge n}s_j\max_{m\ge k\ge n}t_k\tag{9}\\ &=\lim_{m\to\infty}\max_{m\ge j\ge n}s_j\lim_{m\to\infty}\max_{m\ge k\ge n}t_k\tag{10}\\ &=\sup_{j\ge n}s_j\sup_{k\ge n}t_k\tag{11} \end{align} $$ Explanation:
$\phantom{1}(7)$: write sup as a limit of max
$\phantom{1}(8)$: limit of any subsequence is the limit of the sequence
$\phantom{1}(9)$: the greatest product is the product of the greatest elements
$(10)$: limit of a product is the product of the limits
$(11)$: definition

robjohn
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You've shown that for any $\epsilon > 0$: there exists a sufficiently large $N$ such that $n>N$ implies that $$ s_nt_n < lm + (l+m)\epsilon + \epsilon^2 $$ Thus, for any such $n$, we have $\sup_{k \geq n} s_k t_k < lm + (l+m)\epsilon + \epsilon^2$. Thus, $$ \limsup_{n \to \infty} s_nt_n = \lim_{n \to \infty} \sup_{k \geq n} s_k t_k \leq lm + (l+m)\epsilon + \epsilon^2 $$ Howevever, since $\epsilon>0$ is arbitrary, we may conclude that $$ \limsup_{n \to \infty} s_nt_n \leq lm + (l+m)(0) + (0)^2 = lm $$ as desired.

Ben Grossmann
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  • how epsilon is replaced by 0 – user1942348 Dec 21 '16 at 13:53
  • $L = \limsup_{n \to \infty} s_nt_n$ satisfies $L \leq lm + (l+m)\epsilon + \epsilon^2$ for every $\epsilon > 0$. This can only be true if $L \leq lm + (l+m)(0) + (0)^2$, since $f(x) = lm + (l+m)x + x^2$ is a continuous increasing function over $[0,\infty)$. – Ben Grossmann Dec 21 '16 at 14:00
  • Or, if you prefer: $L \leq lm + (l+m)\epsilon + \epsilon^2$ for every $\epsilon > 0$, which means that $L \leq \lim_{\epsilon \to 0^+} lm + (l+m)\epsilon + \epsilon^2$ – Ben Grossmann Dec 21 '16 at 14:03
  • I got it. Thanks a lot. – user1942348 Dec 21 '16 at 14:16
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Let $s_{k_n}t_{k_n}$ be the subsequence of $s_nt_n$ which converges to $\limsup s_nt_n$.

As $s_{k_n}$ is bounded, it possesses a converging subsequence $s_{k'_n}\to s$, and as $t_{k'_n}$ is bounded, it possesses a converging subsequence $t_{k''_n}\to s$.

Clearly, $$ st=\lim_{n\to\infty}s_{k''_n}t_{k''_n} = \lim_{n\to\infty}s_nt_n= \limsup_{n\to\infty}s_nt_n . $$ At the same time $$ s=\lim_{n\to\infty}s_{k''_n}\le\limsup_{n\to\infty}s_n \quad\text{and}\quad t=\lim_{n\to\infty}t_{k''_n}\le\limsup_{n\to\infty}t_n. $$ Hence $$ \lim_{n\to\infty}s_nt_n=st\le \big(\limsup_{n\to\infty}s_n\big)\big( \limsup_{n\to\infty}t_n\big). $$

Yiorgos S. Smyrlis
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