Solved: Proving Uniform Continuity of Unique Continuous Function on A

In summary, the uniform continuity theorem states that there exists a unique continuous function such that for any x in the closure of A, f(x) = g(x).
  • #1
varygoode
45
0
[SOLVED] Uniform Continuity

Homework Statement



Let [tex] A \subset \mathbb{R}^n [/tex] and let [tex] f: A \mapsto \mathbb{R}^m [/tex] be uniformly continuous. Show that there exists a unique continuous function [tex] g: \bar{A} \mapsto \mathbb{R}^m [/tex] such that [tex] g(x)=f(x) \ \forall \ x \in A [/tex].

Homework Equations



The definition for uniform continuity I am using is as follows:

Let [tex] f: A \subset \mathbb{R}^n \mapsto \mathbb{R}^m [/tex]. Then [tex] f [/tex] is uniformly continuous on [tex] A [/tex] if [tex] \forall \ \ \varepsilon > 0 \ \ \exists \ \ \delta > 0 \ \ s.t. \ \ \Vert f(x) \ - \ f(y) \Vert < \varepsilon \ \ \forall \ \ x, \ y \ \ \in A \ \ s.t. \ \ \Vert x - y \Vert < \delta [/tex].

The Attempt at a Solution



I've got to admit, I think I'm pretty clueless about this one. I was thinking about some function

[tex] $ g(x)=\left\{\begin{array}{cc}x,&\mbox{ if }
x\in \bar{A} \backslash A\\f(x), & \mbox{ if } x\in A\end{array}\right $ [/tex].​

But I'm not even sure if I'm allowed to do that or if it meets the criteria. Any help leading to a correct proof would be superb, thanks!
 
Physics news on Phys.org
  • #2
g(x) = x (on the boundary) does not satisfy the criteria. Suppose A = [0,1), x = 1, f(1) = 10 vs. g(1) = 1.
 
  • #3
If x is in the closure of A, then there exist a sequence of points in A converging to x. What should f(x) be?
 
  • #4
EnumaElish said:
g(x) = x (on the boundary) does not satisfy the criteria. Suppose A = [0,1), x = 1, f(1) = 10 vs. g(1) = 1.

I'm not exactly sure where you got the value for f(1) since I never gave f(x) a definition, but I'll go with you on it.

HallsofIvy said:
If x is in the closure of A, then there exist a sequence of points in A converging to x. What should f(x) be?

Hmm... here's what I'm thinking now:

If [tex] x \in \bar{A}, \ \ \exists [/tex] some sequence [tex] (x_k) \in A [/tex] s.t. [tex] \lim_{\substack{k \rightarrow \infty}} (x_k) = x [/tex]. Then [tex] f(x) = \lim_{\substack{k \rightarrow \infty}} f(x_k) [/tex] by continuity. So I'm thinking I can let [tex] g(x) = \lim_{\substack{k \rightarrow \infty}} f(x_k), \ [/tex] where [tex] (x_k) \rightarrow x [/tex]. Then that is continuous since:
[tex]
\begin{equation*}
\begin{split}
\Vert g(x) - g(y) \Vert &= \Vert \left( \lim_{\substack{k \rightarrow \infty}} f(x_k) \right) - \left( \lim_{\substack{k \rightarrow \infty}} f(y_k) \right) \Vert \\
&= \Vert f(x) - f(y) \Vert \\ &\leq \varepsilon \ \ \forall \ \Vert x - y \Vert \leq \delta
\end{split}
\end{equation*}
[/tex]
(by uniform continuity of [tex] f [/tex]). Thus [tex] g(x) [/tex] is a continuous function satisfying the criteria.
 
  • #5
I think you should use the triangle inequality.
 
  • #6
EnumaElish said:
I think you should use the triangle inequality.

Use it where? Are you responding to my response to you or my response to Ivy? (Or both?)
 
  • #7
It's even simpler than that:

g is continuous at x if for every [itex]\epsilon[/itex] > 0 there is [itex]\delta[/itex] > 0 such that |x-xk| < [itex]\delta[/itex] implies |g(x)-g(xk)| = |g(x)-f(xk)| < [itex]\epsilon[/itex].

Since xk ---> x, I know that for any [itex]\delta[/itex], I can find N1 such that k > N1 implies ...

And since by definition f(xk) ---> g(x), I know that for any [itex]\epsilon[/itex], I can find N2 such that k > N2 implies ...

Let N = max{N1, N2}, then for k > N, ...

Uniqueness follows from uniqueness of limit.
 
Last edited:
  • #8
Thanks so much EnumaElish, this problem is solved.
 
  • #9
Do you mind posting your proof so it will be a reference to others?
 

Related to Solved: Proving Uniform Continuity of Unique Continuous Function on A

1. What is uniform continuity?

Uniform continuity is a mathematical concept that describes the behavior of a function over its entire domain. It means that for any given value of epsilon, there exists a corresponding delta such that the distance between any two points in the domain is less than delta, then the distance between their corresponding function values is less than epsilon. In other words, a function is uniformly continuous if its rate of change is consistent throughout its domain.

2. How is uniform continuity different from regular continuity?

The main difference between uniform continuity and regular continuity is that regular continuity only requires the function to be continuous at each point in its domain, while uniform continuity requires the function to be continuous over the entire domain. This means that a uniformly continuous function will have a consistent rate of change throughout its domain, whereas a continuously differentiable function may have different rates of change at different points.

3. How can you prove uniform continuity?

To prove uniform continuity, we can use the epsilon-delta definition of uniform continuity. This means that for any given epsilon, there exists a delta such that the distance between any two points in the domain is less than delta, then the distance between their corresponding function values is less than epsilon. We can also use the concept of compactness, which states that a continuous function on a compact set is uniformly continuous.

4. Can all continuous functions be uniformly continuous?

No, not all continuous functions can be uniformly continuous. For a function to be uniformly continuous, it must have a bounded rate of change over its entire domain. This means that functions with sharp corners or vertical asymptotes, such as the absolute value function, are not uniformly continuous.

5. Why is uniform continuity important in mathematics?

Uniform continuity is important in mathematics because it helps us understand the behavior of functions over their entire domain. It allows us to make precise statements about the continuity of a function and its rate of change, which can be useful in various applications such as optimization, differential equations, and numerical analysis. It also helps us establish the existence of certain limits and derivatives in calculus.

Similar threads

  • Calculus and Beyond Homework Help
Replies
8
Views
2K
  • Calculus and Beyond Homework Help
Replies
26
Views
1K
  • Calculus and Beyond Homework Help
Replies
22
Views
635
  • Calculus and Beyond Homework Help
Replies
19
Views
395
  • Calculus and Beyond Homework Help
Replies
4
Views
577
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
2K
  • Calculus and Beyond Homework Help
Replies
17
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
12
Views
2K
Back
Top