Uniform convergence of sequence of functions

In summary, for the sequence f_n(x) = x/(1+x^n) on the interval [0,∞), the pointwise limit f is given by 0 if x = 0, x if 0 < x < 1, 1/2 if x = 1, and 0 if x > 1. However, this sequence does not uniformly converge to f on the given interval as shown by the fact that for any natural number N, there exists an x close to, but not equal to, 1 such that |f_n(x) - f(x)| is greater than or equal to x.
  • #1
phosgene
146
1

Homework Statement



Let [itex]f_{n}(x)=\frac{x}{1+x^n}[/itex] for [itex]x \in [0,∞)[/itex] and [itex]n \in N[/itex]. Find the pointwise limit f of this sequence on the given interval and show that [itex](f_{n})[/itex] does not uniformly converge to f on the given interval.

Homework Equations


The Attempt at a Solution



I found that the pointwise limit f is :

[itex]0[/itex] if [itex]x=0[/itex]
[itex]x[/itex] if [itex]0<x<1[/itex]
[itex]1/2[/itex] if [itex]x=1[/itex]
[itex]0[/itex] if [itex]x>1[/itex]

But I'm stuck on proving that it's not uniformly convergent. I know that for any natural number N, I need to find some x such that [itex]|f_{n}(x) - f(x)| ≥ ε[/itex], but I'm not sure how to go about this.
 
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  • #2
phosgene said:

Homework Statement



Let [itex]f_{n}(x)=\frac{x}{1+x^n}[/itex] for [itex]x \in [0,∞)[/itex] and [itex]n \in N[/itex]. Find the pointwise limit f of this sequence on the given interval and show that [itex](f_{n})[/itex] does not uniformly converge to f on the given interval.

Homework Equations





The Attempt at a Solution



I found that the pointwise limit f is :

[itex]0[/itex] if [itex]x=0[/itex]
[itex]x[/itex] if [itex]0<x<1[/itex]
[itex]1/2[/itex] if [itex]x=1[/itex]
[itex]0[/itex] if [itex]x>1[/itex]

But I'm stuck on proving that it's not uniformly convergent. I know that for any natural number N, I need to find some x such that [itex]|f_{n}(x) - f(x)| ≥ ε[/itex], but I'm not sure how to go about this.

Look at [itex]|f_n(x) - f(x)|[/itex] for [itex]x[/itex] close to, but not equal to, 1.
 
  • #3
[tex] \frac{x}{1+x^n}-x = \frac{x^{n+1} }{1+x^n} \geq \frac{x^{N+1} }{1+x^N} \geq x [/tex]

Now choose x = e +1?
 

1. What is uniform convergence of a sequence of functions?

Uniform convergence of a sequence of functions is a type of convergence where the rate of convergence is independent of the point in the domain. This means that the functions in the sequence approach the same limit at the same rate, regardless of the input value.

2. How is uniform convergence different from pointwise convergence?

Pointwise convergence of a sequence of functions means that for every point in the domain, the sequence of functions approaches the same limit. In contrast, uniform convergence requires that the rate of convergence is independent of the point in the domain, which is not necessarily true for pointwise convergence.

3. What is the significance of uniform convergence in analysis?

Uniform convergence is important in analysis because it allows for stronger conclusions about the behavior of a sequence of functions. It guarantees that the limit function is continuous and that the sequence of functions can be integrated term by term. This allows for the use of more powerful tools and techniques in mathematical analysis.

4. How is uniform convergence related to the uniform norm?

The uniform norm is a metric used to measure the distance between functions. Uniform convergence is defined in terms of this norm, where a sequence of functions converges uniformly if the distance between each function and the limit function approaches 0 as the sequence progresses.

5. Can a sequence of continuous functions converge uniformly to a non-continuous function?

No, a sequence of continuous functions can only converge uniformly to a continuous function. This is because the uniform limit of continuous functions is always continuous. If a sequence of functions were to converge uniformly to a non-continuous function, it would violate this property of uniform convergence.

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