Re: Confused with uniform/pointwise convergence

From: Isaac (isharu_at_yahoo.com)
Date: 07/13/04 

Date: Tue, 13 Jul 2004 09:50:57 -0400

"David C. Ullrich" <ullrich@math.okstate.edu> wrote in message
news:81c7f0lgftibsehojq4sfhm3o40i1a61ei@4ax.com...
> On Mon, 12 Jul 2004 19:35:27 -0400, "Isaac" <isharu@yahoo.com> wrote:
>
> >
> >"David C. Ullrich" <ullrich@math.okstate.edu> wrote in message
> >news:0dc5f05ejl46r72e8n00qpsp3701b8lhl1@4ax.com...
> >> On Mon, 12 Jul 2004 10:41:20 -0400, "Isaac" <isharu@yahoo.com> wrote:
> >>
> >> >Dear all,
> >> >
> >> >
> >> >
> >> >I think I am confused about the differences between pointwise and
uniform
> >> >converence, but I may not be, and would highly appreciate it if
someone
> >> >could help me.
> >> >
> >> >
> >> >
> >> >The difference between the two is that with uniform convergence, there
is
> >a
> >> >single N for all x's such that |f_n(x) - f(x) | < eps for all n > N.
For
> >> >pointwise, the N may depend on x. Fine, I get that.
> >> >
> >> >
> >> >
> >> >However, I am trying to read some theorems. In particular, let f_n be
> >> >real-valued functions that are riemann integrable on [a,b] and that
> >f_n -> f
> >> >uniformly on [a,b]. Define g_n(x) = integral (from a to x) f_n(t)dt
if
> >x
> >> >is in [a,b]. Then f is riemann integrable on [a,b] and g_n -> g
> >uniformly
> >> >on [a,b] where g(x) = integral (from a to x) f(t)dt.
> >> >
> >> >
> >> >
> >> >Well, it goes on to say that the "conclusion" implies that for each x
in
> >> >[a,b], we can write
> >> >
> >> >Limit integral (a to x) f_n(t) dt = integral (from a to x) limit
> >f_n(t)
> >> >dt
> >> >
> >> >
> >> >
> >> >Ok, no problem. But my question is, if g_n -> g pointwise, wouldn't
this
> >> >conclusion be implied as well?
> >>
> >> If everything's as you wrote it then yes, saying g_n -> g pointwise is
> >> exactly the same as saying "Limit integral (a to x) f_n(t) dt
> >> = integral (from a to x) limit f_n(t) dt", by the definition of
> >> the g_n and g.
> >>
> >> Maybe they meant something else, like that the integral of the g_n
> >> converges to the integral of g?
> >>
> >> >In the "conclusion", we are saying that for
> >> >every x, the equality is satisfied. So , if you give me an x in
[a,b], I
> >> >can show you that Limit integral (a to x) f_n(t) dt = integral
(from
> >a
> >> >to x) limit f_n(t) dt. In other words, I can show you that |g_n(x) -
> >g(x)|
> >> >< eps for all n > N and for some N (maybe depending on x). That is,
> >g_n(x)
> >> >converges to g(x) if you give me an x. Why do we need a single N that
> >does
> >> >it for all x in [a,b]? The "conclusion" seems to tell me "ok, give me
an
> >x,
> >> >any x in [a,b], and this equality holds". Well that says to me give
me
> >an x
> >> >and sure I can make the equality hold, depending on what x is of
course.
> >So
> >> >if g_n(x) -> g(x) pointwise, we'd have the same conclusion. I just
don't
> >> >see why we need to say that we need to be able to have a single N that
> >makes
> >> >the equality in the "conclusion" hold.
> >> >
> >> >
> >> >
> >> >What the "conclusion" says is that a uniformly convergent sequence can
be
> >> >integrated term by term. But if g_n -> g pointwise, that would happen
as
> >> >well (which is what I am claiming), because like I said, if you give
me
> >any
> >> >x in [a,b], I can make |g_n(x) - g(x)| as small as I want, and so
we're
> >> >fine.
> >> >
> >> >
> >> >
> >> >I'm not saying we need to change the assumptions of the theorem.
Those
> >are
> >> >needed to prove this g_n -> g uniformly. But if I'm right (which I'm
> >> >probably not), can't we relax the assumptions so that we can just aim
for
> >> >g_n(x) -> g(x) pointwise?
> >> >
> >> >
> >> >
> >> >Any help is highly appreciated,
> >> >
> >> >
> >> >
> >> >Isaac
> >> >
> >>
> >>
> >> ************************
> >>
> >> David C. Ullrich
> >
> >Dear David,
> >
> >
> >
> >The text from which I quote is the standard "Mathematical Analysis" by
Tom
> >Apostol on page 225 at the bottom. I quote: "Note. The conclusion
implies
> >that, for each x in [a,b], we can write lim integral (from a to x)
f_n(t)dt
> >= integral (from a to x) lim f_n(t)dt. This property is often described
by
> >saying that a uniformly convergent sequence can be interated term by
term."
>
> Well, probably they were just stating this for emphasis, in case you
> missed this point.
>
> >I have asked this question elsewhere, and another poster seems to think
that
> >I am not right also.
>
> My _conjecture_ is that the people who are saying you're wrong are not
> reading the question carefully.
>
> > Like I said, g_n -> g uniformly of course does imply
> >the above "conclusion". However my claim is (which you agree with) that
> >g_n -> g pointwise also implies the above "conclusion".
> >
> >
> >
> >You also said "Maybe they meant something else, like that the integral of
> >the g_n converges to the integral of g?" Well what do you mean by that?
Do
> >you just mean
> >
> > g_n converges uniformly to g (well I guess you have to mean that by what
> >the theorem says) ? And, why would one care about this? It seems to me
> >that we should much more care about the "conclusion" (a specific case of
> >g_n -> g uniformly as you agree), which tells us when we can integrate
> >sequences (or as an easy corollary, series) term by term, and which is
> >equivalent to the statement that g_n -> g pointwise (as you agree).
> >
> >
> >
> >In my very youthful and naïve opinion, shouldn't we just care about how
we
> >can arrive at lim integral (from a to x) f_n(t)dt = integral (from a to
x)
> >lim f_n(t)dt ? If so, then we care about how g_n -> g pointwise. And
as
> >the original theorem said, we need a bunch of assumptions (like f_n -> f
> >uniformly, f_n riemann integrable, etc.) in order to arrive at g_n -> g
> >uniformly, which of course implies the "conclusion" which is what we're
> >after. But maybe we can relax our assumptions a bit to arrive at g_n ->
g
> >pointwise so that we can still have the "conclusion"?
> >
> >
> >
> >Yikes, I hope I made some sense and really hope you can help,
> >
> >
> >
> >Isaac
> >
>
>
> ************************
>
> David C. Ullrich

Dear David,

Thanks for your responses! Last question : I was wondering, because it
doesn't say in my text,
are there less restrictive requirements to arrive at the "conclusion"?
Because like I mentioned before,
it seems to me that the "conclusion" is the very important (i.e. when can we
integrate term
by term?) part/consequence of this theorem. So, maybe if we relax the
assumptions of the theorem a bit, like f_n -> f pointwise, then we can
still have g_n -> g pointwise (which is the same as the "conclusion"), but
not uniformly. Because now if someone asks me when can you integrate
a sequence term by term, I will say that the sequence must be riemann
integrable and uniformly convergent, but these aren't necessary conditions
it seems to me, they are sufficient. Well actually, I believe that f_n must
be riemann integrable, I believe that has to be necessary. But do f_n have
to converge uniformly to f? Maybe a less restrictive convergence?

Thanks,

Isaac

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