ScalingStacks

Proof. [027I]

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Proof.

(1) follows from (1) in LemmaΒ 3.13.

(2) Since ΟƒLΒ―βŠ—(n+nβ€²)​(x)≀σLΒ―βŠ—n​(x)+ΟƒLΒ―βŠ—n′​(x)\sigma_{\overline{L}^{\otimes(n+n^{\prime})}}(x)\leq\sigma_{\overline{L}^{\otimes n}}(x)+\sigma_{\overline{L}^{\otimes n^{\prime}}}(x) for n,nβ€²βˆˆβ„•β‘(L)n,n^{\prime}\in{\mathbb{N}}(L) by (2) in LemmaΒ 3.13, the assertion follows from Fekete’s lemma.

(3) and (4) follow from (2) and (3) in LemmaΒ 3.13 together with (2), respectively.

(5) If n=0n=0, then the assertion is obvious, so that we may assume that nβ‰₯1n\geq 1. We fix n0βˆˆβ„•β‘(L)n_{0}\in{\mathbb{N}}(L). Then n0βˆˆβ„•β‘(LβŠ—n)n_{0}\in{\mathbb{N}}(L^{\otimes n}). Thus, by (2),

ΞΌLΒ―βŠ—n​(x)=limmβ†’βˆžΟƒLβŠ—m​n0​n​(x)m​n0=n​limmβ†’βˆžΟƒLβŠ—m​n0​n​(x)m​n0​n=n​μL¯​(x).\mu_{\overline{L}^{\otimes n}}(x)=\lim_{m\to\infty}\frac{\sigma_{L^{\otimes mn_{0}n}}(x)}{mn_{0}}=n\lim_{m\to\infty}\frac{\sigma_{L^{\otimes mn_{0}n}}(x)}{mn_{0}n}=n\mu_{\overline{L}}(x).

∎

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