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I got this proof that I can't show for some hours now. Does anyone have a hint?

If $a_n, b_n$ are bounded sequences and $\lim_{n\to\infty} a_n = a < 0$ then $$\liminf_{n\to\infty} a_nb_n = a \cdot \limsup_{n\to\infty} b_n$$

Thanks!

Martin Sleziak
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craft
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  • Related posts: http://math.stackexchange.com/questions/768011/product-of-lim-sups http://math.stackexchange.com/questions/776517/product-of-limitsuperior-of-bounded-sequences http://math.stackexchange.com/questions/947822/if-limsup-x-n-x-lim-y-n-y-x-n-y-n-0-then-does-limsup-x-n-y – Martin Sleziak Dec 09 '14 at 09:32

1 Answers1

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I would try something like this:

Let $\varepsilon > 0$. Then there exists $n_0 \in \Bbb{N}$ s.t. $$ n \geq n_0 \implies a - \varepsilon < a_n < a + \varepsilon $$ Now if $n \geq k_0$ \begin{align*} (a - \varepsilon) \sup_{k \geq n} b_k & \overset{(*)}{=} \inf_{k \geq n} ( (a - \varepsilon)b_k) \\ &\leq \inf_{k \geq n} (a_k b_k) \\ & \leq \inf_{k \geq n} ((a+\varepsilon)b_k) \\ &\overset{(*)}{=} (a+\varepsilon) \sup_{k \geq n} b_k \end{align*} Then by taking the limit $n \to \infty$ we'd get \begin{align*} (a-\varepsilon) \limsup_{n \to \infty} b_n &\leq \liminf_{n \to \infty} (a_n b_n) \leq (a + \varepsilon) \limsup_{n \to \infty} b_n \\ \iff -\varepsilon \limsup_{n \to \infty} b_n &\leq \liminf (a_nb_n) - a \limsup_{n \to \infty} b_n \leq \varepsilon \limsup_{n \to \infty} b_n \\ \iff \left| \liminf (a_nb_n) - a \limsup_{n \to \infty} b_n \right| &\leq \varepsilon |\limsup_{n \to \infty} b_n| \end{align*} The sequence $(b_n)$ is bounded, so $|\limsup b_n| < \infty$, and the result follows.

The $(*)$ equalities feel right, but need some more justification.

desos
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  • I don't quite see how the last line is equal to what I want to prove? – craft Dec 07 '14 at 21:47
  • Oops, had an error. Fixed it now. – desos Dec 07 '14 at 21:50
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    Thanks! I guess for the () equalties I could use that -sup(A)=inf(-A) and sup(cA)=csup(A). However I have to choose the E such that a+E<0 so that the second () equation is right. – craft Dec 07 '14 at 22:34
  • I am also not quite sure by which theorem the last line is equal to what I want to prove. How do I see the equalit yin this inequality? – craft Dec 07 '14 at 23:01
  • It's basically the squeeze theorem. Our $\varepsilon$ was arbitrary, so we can take the limit $\varepsilon \to 0$. – desos Dec 07 '14 at 23:21
  • Can we really do this since we said that E>0? – craft Dec 08 '14 at 08:38
  • Define $A = \limsup b_n$ and $B = \liminf a_n b_n$ Then for all $\varepsilon > 0$ we have $aA - \varepsilon a \leq B \leq aA + \varepsilon A$. Equivalently $- \varepsilon a \leq B - aA \leq \varepsilon a \iff |B-aA| \leq \varepsilon a$. – desos Dec 08 '14 at 11:40