Proof of Fekete’s subadditive lemma

If there is a m such that am=-, then, by subadditivity, we have an=- for all n>m. Then, both sides of the equality are -, and the theorem holds. So, we suppose that an𝐑 for all n. Let L=infnann and let B be any number greater than L. Choose k1 such that


For n>k, we have, by the division algorithmPlanetmathPlanetmath there are integers pn and qn such that n=pnk+qn, and 0qnk-1. Applying the definition of subadditivity many times we obtain:


So, dividing by n we obtain:


When n goes to infinity, pnkn converges to 1 and aqnn converges to zero, because the numerator is bounded by the maximum of ai with 0ik-1. So, we have, for all B>L:


Finally, let B go to L and we obtain

Title Proof of Fekete’s subadditive lemma
Canonical name ProofOfFeketesSubadditiveLemma
Date of creation 2014-03-18 14:41:50
Last modified on 2014-03-18 14:41:50
Owner Filipe (28191)
Last modified by Filipe (28191)
Numerical id 3
Author Filipe (28191)
Entry type Proof