Re: incresing functions on [0, inf)
From: Amanda (sca18_at_hotmail.com)
Date: 08/24/04
- Next message: Microcephalic Bob: "Re: Jeffrey Blair McAleer - February 9th 1984"
- Previous message: C. Bond: "Re: JSH: Checked by committee"
- In reply to: The World Wide Wade: "Re: incresing functions on [0, inf)"
- Next in thread: The World Wide Wade: "Re: incresing functions on [0, inf)"
- Reply: The World Wide Wade: "Re: incresing functions on [0, inf)"
- Messages sorted by: [ date ] [ thread ]
Date: 24 Aug 2004 08:15:01 -0700
The World Wide Wade <waderameyxiii@comcast.remove13.net> wrote in message news:<waderameyxiii-8840FD.14342223082004@news.supernews.com>...
> In article <waderameyxiii-EDE09B.13132323082004@news.supernews.com>,
> The World Wide Wade <waderameyxiii@comcast.remove13.net> wrote:
>
> > In article <6f75d9cf.0408231022.964b8ac@posting.google.com>,
> > sca18@hotmail.com (Amanda) wrote:
> >
> > > I'd like some hints on how to prove or disprove the following
> > > statement:
> > >
> > > f and g are real valued functions defined on [0, inf). f is strictly
> > > increasing and lim (x=> inf) g(x) =0. Then, there is a k>0 such that
> > > f+g is strictly increasing for x>k.
> > >
> > > I think this is false, but couldn't find a counter example.
> >
> > Try f(x) = x and g(x) = sin(x^2)/x. Then f(x) is str. incr. to oo, but
> > (f+g)'(x) < 0 for lots of x -> oo.
>
> In fact, for any differentiable f with f(x) -> oo as x -> oo, set g =
> [sin(f^2)]/f. Then g(x) -> 0 as x -> oo, and f + g is strictly decreasing
> on a sequence of intervals going out to oo.
Yes,that's right. Thank you all for your help.
Actually, my problem is a bit trickier than this. I have 2 sequences
of real valued functions, (f_n) and (g_n), both defined on [0, inf).
For every n, I know f_n is strictly increasing and lim (x=> inf)
g_n(x) =0. I also know that (f_n) converges on [0, inf) to a
continuous function f, which implies the convergence f_n -> f is
uniform on [0, M] for every M>0 (not sure if this conclusion is true
for [0, inf)). I know it's also true that g_n -> 0 (the identically
zero function) on [0, inf), but I dont know if uniformly or just
pointwise. Therefore, it's immediate that f_n+ g_n -> f on the whole
[0, inf), but the convergence may not be uniform.
For every n, there's a function F_n such that f_n = F'_n on [0, inf)
and I know F_n(0) = 0 for every n. So, I can conclude every f_n is
continuous (which, in this context, doesn't seem to be an interesting
conclusion) and F_n(0) goes trivially to 0. If I could assure each f_n
+ g_n would become ultimatelly strictly increasing on [0, inf), then I
could assure f_n + g_n -> f uniformly on [0, M}] for sufficiently
large M>0. And then I could have the conclusion I really want, that
is, f_n -> F', where F is the limit function of F_n. I want to get rid
of the functions g_n, cause they are a kind of residual function, a
kind of transient. By means of an algorithm I can compute f_n(x) very
nicely for every n and every x>=0, but, though I know |g_n(x) is small
when compared to |f_n(x|, I don't know how to compute g_n(x), except
when x=0.
Unfortunately, it seems the only consistent conclusion is that I have
no conclusion at all. I was told Lebesgue's Monotonic Convergence
Theorem might be useful here, but, though I know very little about
measure theory, I don't se how.
Thank you very much
Amanda
- Next message: Microcephalic Bob: "Re: Jeffrey Blair McAleer - February 9th 1984"
- Previous message: C. Bond: "Re: JSH: Checked by committee"
- In reply to: The World Wide Wade: "Re: incresing functions on [0, inf)"
- Next in thread: The World Wide Wade: "Re: incresing functions on [0, inf)"
- Reply: The World Wide Wade: "Re: incresing functions on [0, inf)"
- Messages sorted by: [ date ] [ thread ]
Relevant Pages
|