Uniform convergence, Lp convergence

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SUMMARY

The discussion centers on the convergence of the sequence of functions fn(x) defined as fn(x)=1 for 0≤x≤1/n and fn(x)=0 for 1/n PREREQUISITES

  • Understanding of measure theory concepts, particularly Lebesgue measure.
  • Familiarity with Dominated Convergence Theorem (DCT).
  • Knowledge of Lp spaces and their properties.
  • Basic calculus and integration techniques.
NEXT STEPS
  • Study the Dominated Convergence Theorem in detail, focusing on its applications in measure theory.
  • Explore Lp convergence and its implications in functional analysis.
  • Review examples of uniform convergence and the differences between uniform and almost uniform convergence.
  • Investigate Lebesgue measure and its properties in relation to convergence of functions.
USEFUL FOR

Mathematicians, students of analysis, and anyone studying measure theory and convergence concepts in functional analysis.

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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