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I have to prove the following statement, but I can't. If $\limsup x_{ n }=\, x,\lim y_{ n }=\, y, \, x_{ n },y_{ n }>0$, then $\limsup (x_{n}y_{n})=xy$.

Will you give me some hint or solution?

Martin Sleziak
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Yoon
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  • Could you explain the thoughts you already have about this problem? – Travis Willse Sep 27 '14 at 05:33
  • Since $limsup(x_{n}y_{n})<=limsup(x_{n})limsup(y_{n}),;limsup(x_{n}y_{n})<=xy.$ So I have to show that when $limsup(x_{n}y_{n})<xy$ a contradction occurs. But I can't for several hours. – Yoon Sep 27 '14 at 05:37
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    Try this little trick: $x_n y_n - xy= x_n y_n - xy_n + xy_n - xy$. – Jonas Gomes Sep 27 '14 at 05:41
  • using this trick and showing $x_{n}y_{n}$ converges to xy, then $limsup; (x_{n}y_{n}) = xy$ Am I right? – Yoon Sep 27 '14 at 05:47
  • This seems to be basically the same question as: http://math.stackexchange.com/questions/768011/product-of-lim-sups See also http://math.stackexchange.com/questions/776517/product-of-limitsuperior-of-bounded-sequences – Martin Sleziak Sep 27 '14 at 06:00

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Hint. If $\limsup x_n =x$ then you can find a subsequence $\langle x_{n_k}\rangle$ such that $\lim x_{n_k}=x$. In general, if $a$ is a limit point of $\langle x_n\rangle$ then you can find a subsequence of $\langle x_n\rangle$ which converges to $a$.

Hanul Jeon
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let $s_n = \sup_{k \ge n} x_n$. We have $s_n \uparrow x$. By continuity we have $s_n y_n \to xy$.

Find an upper bound for $\limsup_n x_n y_n$:

Clearly $x_n y_n \le s_n y_n$, hence $\limsup_n x_n y_n \le xy$.

Furthermore, we have a subsequence $x_{n_k} \to x$, and again, we have $x_{n_k}y_{n_k} \to xy$.

Find a lower bound for $\limsup_n x_n y_n$:

Since we have $\lim_n x_{n_k}y_{n_k} = xy$, we have $\limsup_n x_n y_n \ge xy$.

copper.hat
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