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I want to show, that the equation in the headline is given, if $(x_n)_n$ is a real, non-negative sequence with $0\leq x_{n+m}\leq x_n x_m\quad \forall n,m\in\mathbb{N}.$

Daniel
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    What did you try? You need to show that $x_n^{1/n}$ is monotonically decreasing. Your hypothesis tells you that $x_n \le x_1^n$ for all $n$s. – Paolo Intuito Nov 23 '17 at 12:14
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    Try to use the additive problem: https://math.stackexchange.com/questions/760484/sequence-limit-problem-if-0-leq-x-mn-leq-x-n-x-m-then-limit-of-x-n-n?rq=1. Note that in general the sequence $x_n^{1/n}$ is not monotonic (take the additive example provided by Lucia on MO: in your situation, this is $x_{2n}=1$ and $x_{2n+1}=2$) – Kelenner Nov 23 '17 at 12:35
  • Right, we have the monotonic behaviour only on some subsequences. Fekete's lemma works wonders by taking $e^{a_n/n}$ and thus transforming the sum in a product. – Paolo Intuito Nov 23 '17 at 12:51

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