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Let $(x_n)_n$ and $(y_n)_n$ be sequences in $\mathbb{R}$, such that $(x_n)_n$ converges to $x \in \mathbb{R}^+$. Suppose that $(y_n)_n$ is bounded.

Prove that $$\limsup_{n \to +\infty} x_n \cdot y_n = x \cdot \limsup_{n \to +\infty} y_n$$

I know that $x=\limsup_{n \to +\infty} x_n$, so it basically comes down to proving $\limsup_{n \to +\infty} x_n \cdot y_n = \limsup_{n \to +\infty} x_n \cdot \limsup_{n \to +\infty} y_n$, but that's where I'm stuck...

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  • similar: http://math.stackexchange.com/questions/113121/lim-sup-inequality-limsup-a-n-b-n-leq-limsup-a-n-limsup-b-n – Yiyi Rao Nov 04 '16 at 11:46
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    https://math.stackexchange.com/questions/1244661/if-lim-n-to-inftya-n-a-in-mathbbr-prove-that-limsup-n-to-infty/201168 – Workaholic Nov 04 '16 at 12:08

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You can not prove your lemma because it is wrong. Let $(x_n)=(1,2,1,2,1,2,...)$ and $(y_n)=(2,1,2,1,2,1,...)$ then their product is constant $2$, the product of the lim sup is $4$.

Given an $0<ε<x/2$ there is an $N$ so that for all indices $n\ge N$ $$ x-ε\le x_n\le x+ε\text{ and } y_n\le \limsup_{k\to\infty} y_k+ε. $$ Additionally, for infinitely many of these indices $n$ one has $$ \limsup_{k\to\infty} y_k-ε\le y_n. $$ This can be combined to prove the claim.

Lutz Lehmann
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    Your $(x_n)_n$ does not converge to some $x \in \mathbb{R}^+$, though? – sTertooy Nov 04 '16 at 13:29
  • @SteamyRoot: This is the assumption of the claim and thus not repeated. It is implicit in $0<ε<x/2$. That $x>0$ is important for combining the inequalities. – Lutz Lehmann Nov 04 '16 at 13:34