Disproving Uniform Convergence of f_n(x) on [0,1]

In summary, the conversation discusses the topic of uniform convergence and how to disprove it using the example of the function f_n(x)=x^n on the closed interval [0,1]. The conversation explores the definition of uniform convergence and how it does not depend on x. It is concluded that the sequence is not uniformly convergent because there exists a counterexample that does not satisfy the definition. Additionally, it is mentioned that if the domain was changed to (0,1), the sequence would be uniformly convergent. The conversation also briefly mentions the concept of a limit function in this context.
  • #1
philosophking
175
0
I haven't done uniform convergence since last year when I took analysis, and now I have this problem for topology (we're studying metric spaces right now) and I can't remember how to disprove uniform convergence:

f_n: [0,1] -> R , f_n(x)=x^n

Show the sequence f_n(x) converges for all x in [0,1] but that the sequence does not converge uniformily.

I think I have a pretty good concept of uniform convergence, that is, a sequence of functions is uniformily convergent on some domain if the closeness of the function values (in the range) does not depend on the closeness of the values (in thedomain). To disprove uniform convergence, should I show that there exists some epsilon such that |f_n(x) - f_n(y)| >/= epsilon? But I don't know how I would do that.

Thanks for the help!
 
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  • #2
um... what happens @ x=1?
 
  • #3
You get the sequence 1,1,1,1,... which converges? I guess I don't know where you're going with that.
 
  • #4
If the sequence converges uniformly, the limit function is continuous. What's the limit function in this case?
 
  • #5
I'm sorry, I'm not sure what you mean by "limit function". Although I think I now know where Fourier was going.

The definition of uniform convergence does not depend on x. That means I can take any two x in the domain and their images must necessarily converge to less than epsilon. But say in this example I take x=0 and y=1. Then I can find a counterexample epsilon (.5) such that 1^n - 0^n is not less than or equal to 1/2 as n approaches infinity. I think I can now say that hence, the sequence is not uniform convergence.

EDIT: Also, just as a side-question, if the domain were (0,1) instead of [0,1], this would imply that f_n is uniformly convergent on the domain, right?
 
  • #6
yes, that's right :)
 
  • #7
oh yeah, & the "limit function" would be f(x) where:

[tex]f(x) = \lim_{n\rightarrow\infty} f_{n}(x)[/tex]
 
  • #8
The open interval at 1 creates a problem, even though it does not include 1. Every N you choose will only create uniform convergence for a certain closed interval [0,b] where b<1. Once you get to the right of this b the distance to zero will be greater than epsilon.
 

FAQ: Disproving Uniform Convergence of f_n(x) on [0,1]

What is uniform convergence?

Uniform convergence is a type of convergence in which a sequence of functions converges to a single function in such a way that the convergence is independent of the value of the input variable. In other words, the functions in the sequence approach the limit function at the same rate for all values of the input variable.

How is uniform convergence different from pointwise convergence?

In pointwise convergence, the functions in the sequence may approach the limit function at different rates for different values of the input variable. This means that the convergence is dependent on the value of the input variable. Uniform convergence, on the other hand, ensures that the convergence is independent of the input variable.

Why is the concept of uniform convergence important?

Uniform convergence is important because it allows us to make stronger statements about the convergence of a sequence of functions. It also has many applications in analysis and is a fundamental concept in the study of real analysis and functional analysis.

How is uniform convergence of f_n(x) on [0,1] determined?

To determine uniform convergence of f_n(x) on [0,1], we need to check if the sequence of functions satisfies the definition of uniform convergence. This means that for any given epsilon, there exists an N such that for all n greater than or equal to N, the distance between the nth function and the limit function is less than epsilon for all values of the input variable in the interval [0,1]. If this condition is met, then the sequence of functions is uniformly convergent on [0,1].

How can we disprove uniform convergence of f_n(x) on [0,1]?

To disprove uniform convergence of f_n(x) on [0,1], we need to find a counterexample that violates the definition of uniform convergence. This means finding an epsilon for which the condition of the definition is not satisfied, or finding an input value in the interval [0,1] for which the distance between the nth function and the limit function is not less than epsilon for all n greater than or equal to some N. If we can find such a counterexample, then we can conclude that the sequence of functions does not converge uniformly on [0,1].

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