Re: fractional part of the function at integer points+concave

In article
<15558845.1151055387707.JavaMail.jakarta@xxxxxxxxxxxxx
forum.org>,
eugene <jane1806@xxxxxxx> wrote:
Let f be a continious concave function from (0,
+infty) -> R such that lim_{x -> infty} f(x) = infty
and f(x) = o(x) when x -> +infty.

How can i prove that sup_{n -naturals} {f(n)} = 1,
where

{f(n)} is fractional part of f(n).

Hint: show that f(n+1) - f(n) tends to 0 as n tends
to infinity.
Thanks. I had the same idea, surely if we prove it, using the fact that f tends to infitiy everything becomes clear about the problem.
But, i'm not sure about my proof of that
lim_{n -> infy} (f(n+1)-f(n)) = 0.

So,using the properties of f i stated here

http://mathforum.org/kb/thread.jspa?threadID=1402907&tstart=0

f'(x) where it exists is decreasing and thus it may have finite limit L when x -> infty or it may have infinite limit. In both cases using the relation

f(x) = f(a) + int_[a,x] f'(t)dt we would have a

contradiction with that f(x) = o(x), x-> infty unless

L=0, whih means that lim f'(x) = 0 when x -> infty, whihch together with

f(n+1) - f(n) = f'(ksi) , ksi \in (n,n+1) gives the
desired lim_{n ->infty} (f(n+1)-f(n)) = 0.


I think these solutoon has some flaws and i'm not sure about it.
Is it in the whole near to be ok.

Thanks

Mike Guy
.

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