Uniform convergence, Lp convergence

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Homework Help Overview

The discussion revolves around the convergence of a sequence of functions defined in the context of measure theory, specifically examining the behavior of the functions fn(x) and f(x) under the Lebesgue measure. The original poster poses several types of convergence to analyze, including pointwise, almost everywhere, uniform, and Lp convergence.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore various types of convergence and provide their interpretations of the convergence behavior of the sequence of functions. Questions arise regarding the definitions of "uniformly almost everywhere" and "almost uniformly," as well as the application of the Dominated Convergence Theorem (DCT) in this context.

Discussion Status

There is an ongoing exploration of the convergence types, with some participants providing answers and others questioning the reasoning behind certain conclusions. The discussion includes attempts to clarify the implications of the DCT and its application to the problem, indicating a productive exchange of ideas.

Contextual Notes

Participants note the potential confusion surrounding the definitions of different convergence types and the assumptions involved in applying the DCT. There is also mention of the behavior of norms as p varies, suggesting a need for further clarification on these points.

onthetopo
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Homework Statement


A stereotypical problem in measure theory.
fn(x)=1 when 0<=x<=1/n
=0 when 1/n<x<=1
f(x)=0 0<=x<=1

Under the lebesgue measure
Does fn converge to f?
a. for all x
b. almost everywhere
c. uniformly on [0,1]
d. uniformly almost everywhere on [0,1]
e. almost uniformly
f. in measure
g. in Lp

Homework Equations


The Attempt at a Solution


My guess of the answer would be:
a. for all x: No, doesn't converge at x=0
b. almost everywhere: Yes, x=0 has lesbesgue measure 0
c. uniformly on [0,1]: No
d. uniformly almost everywhere on [0,1]: No
e. almost uniformly: Yes
f. in measure: Yes
g. in Lp: shouln't it depend on p? no idea.
 
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My answers:
a) No
b) Yes
c) No
d) What do you mean by uniformly almost everywhere? How does this differ from almost uniformly? I've never seen this concept before.
e) Yes
f) Yes
g) Yes

To see this last one, note that for any p > 0, |f_n|^p \to 0 a.e., and |f_n|^p \leq 1, so the DCT implies that \int |f_n|^p dm \to 0.
 
Sorry how does dominated convergence theorem implies that \int |f_n|^p dm \to 0.?

All DCT says is that it is less than the integral of 1 right?
 
onthetopo said:
Sorry how does dominated convergence theorem implies that \int |f_n|^p dm \to 0.?

All DCT says is that it is less than the integral of 1 right?

I think so, but you don't need it anyway. ||f_n||_p=(1/n)^(1/p). Doesn't that go to zero for all p<infinity?
 
The http://en.wikipedia.org/wiki/Dominated_convergence_theorem" allows you to pull the limit inside the integral, so, since |f_n|^p is dominated by an integrable function (f\equiv1),
\begin{align*}<br /> \lim \int |f_n|^p dm = \int ( \lim |f_n|^p ) dm = \int 0 dm = 0.<br /> \end{align*}

Dick is of course correct in his evaluation of ||f_n||_p both methods work.
 
Last edited by a moderator:
rochfor1 said:
The http://en.wikipedia.org/wiki/Dominated_convergence_theorem" allows you to pull the limit inside the integral, so, since |f_n|^p is dominated by an integrable function (f\equiv1),
\begin{align*}<br /> \lim \int |f_n|^p dm = \int ( \lim |f_n|^p ) dm = \int 0 dm = 0.<br /> \end{align*}

Dick is of course correct in his evaluation of ||f_n||_p both methods work.

Right. Sorry, I didn't bother to look it up the DCT because you don't need it. But rochfor1 is correct.
 
Last edited by a moderator:
No worries! Your way is actually easier than mine...I'm just so used to doing these types of problems without being able to explicitly compute the norms, so DCT and MCT are the first things I go to.
 
Last edited:

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