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 <mnas@mail.fct.unl.pt> 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).