From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik) Subject: Re: subadditive function Date: 7 Sep 1999 05:18:44 -0400 Newsgroups: sci.math Keywords: Subadditive functions on R reach a limiting multiple of x. In article <7r088b$m8n$1@news.si.fct.unl.pt>, manuel wrote: > I need help for an analysis problem: > > Let f be a subadditive real function ( f(m+n)<=f(m)+f(n)) >Prove that the limit of f(n)/n exist when n tends to infinity. Considering the strange ways the posts propagate, I might be duplicating someone else's proof. Here is my take: The plan is to show that lim sup [n to inf.] f(n)/n <= lim inf [n to inf.] f(n)/n because the reverse inequality is valid generally, and the equality of lim inf and lim sup is equivalent to the existence of the limit. In addition, we can show that this limit actually equals inf { f(n)/n | n >= 1 } Since this infimum can be -infinity, I will avoid epsilons. Consider an integer p >= 2, and denote M = max(f(1),...,f(p)). Every n >= 1 can be expressed (by a modified division) as m*p+r where r is one of 1, 2, ..., p, and m is integer. So, for n>p, we have by induction f(m*p+r) <= m*f(p) + f(r) <= m*f(p) + M Divide by m*p+r and use m*p+r > m*p: f(m*p+r)/(m*p+r) <= (m/(m*p+r)) * f(p) + (1/(m*p+r)) * M so f(m*p+r)/(m*p+r) - M/(m*p+r) <= f(p)/p Take lim sup on the left side - remember n=m*p+r runs through all positive integers: lim sup[n to inf.] f(n)/n <= f(p)/p The left side is now independent of p, so we can take infimum on the right: lim sup[n to inf.] f(n)/n <= inf[p>=1] f(p)/p and of course inf[p>=1] f(p)/p <= lim inf[p to inf.] f(p)/p The plan has been carried out, and so lim[n to inf.] f(n)/n = inf[p>=1] f(p)/p. Of course, the course you are studying might not have covered lim sup and lim inf yet. Good luck, ZVK(Slavek).