Sequence of functions; uniform convergence and integrating

Click For Summary
SUMMARY

The sequence of functions defined as fn(x) = nxe^(-nx²) converges pointwise to f(x) = 0 for x in the interval [-1, 1]. The discussion centers on verifying the equality lim(n->∞) (∫₀¹ fn(x) dx) = ∫₀¹ (lim(n->∞) fn(x)) dx. It is established that fn is continuous for every n, but the challenge lies in demonstrating uniform convergence on the interval [0, 1]. The approach involves showing that |fn(x) - f(x)| can be made arbitrarily small for sufficiently large n, which requires further exploration of the behavior of fn as n increases.

PREREQUISITES
  • Understanding of pointwise and uniform convergence of functions
  • Familiarity with the properties of continuous functions on closed intervals
  • Knowledge of integration techniques, particularly with limits
  • Basic calculus, including differentiation and finding extrema
NEXT STEPS
  • Study the criteria for uniform convergence in detail
  • Learn how to apply the Weierstrass M-test for uniform convergence
  • Explore techniques for evaluating improper integrals involving limits
  • Investigate the relationship between pointwise convergence and integration
USEFUL FOR

Mathematics students, particularly those studying real analysis, calculus, and anyone interested in the convergence of sequences of functions and their implications for integration.

Appa
Messages
15
Reaction score
0

Homework Statement


The sequence fn: [-1,1] -> R, fn(x)= nxe-nx2 converges pointwise to f(x)= 0, x in [-1,1]. Can you verify the following:

limn->\infty (\int^{1}_{0}fn(x)dx) = \int^{1}_{0} (limn->\infty fn(x))dx

Homework Equations


Theorem: If fn is continuous on the interval D for every n and fn converges uniformly to f on D=[a,b], then

limn->\infty (\int^{a}_{x}fn(t)dt) = \int^{a}_{x} (limn->\infty fn(t))dt = \int^{a}_{x}f(t)dt) for every x in D.

The Attempt at a Solution


The main idea is to find out whether the sequence is uniformly convergent on D= [0,1]. fn = nxe-nx2 is continuous for every n because et is always continuous. I tried something like this to verify uniform convergence:
fn(x) is uniformly convergent on D if there is a positive number \epsilon >0 and an index N, so that
|fn(x) -f(x)|<\epsilon for every n\geqN and every x in [0,1]
\Leftrightarrow |nxe-nx2-0|<\epsilon
\Leftrightarrow nxe-nx2 <\epsilon

But I don't know how to go forward. How can I show that the greater the index n gets, the closer fn gets to 0? I tried comparison but couldn't find any sequence, the term of which would be greater than fn so that the sequence would still converge.
 
Physics news on Phys.org
Finding a uniform limit for f_n isn't that hard. You just need to find the extrema of the functions as a function of n. Take the derivative, set it equal to zero etc. But I'm going to guess that what you are really expected to do is actually calculate the integral of f_n and verify that the limit of the integrals is zero.
 

Similar threads

  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
26
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 34 ·
2
Replies
34
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K