# Uniform Continuity

1. Feb 12, 2008

### varygoode

[SOLVED] Uniform Continuity

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

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

3. The attempt at a solution

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

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!

2. Feb 13, 2008

### EnumaElish

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. Feb 13, 2008

### HallsofIvy

Staff Emeritus
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. Feb 13, 2008

### varygoode

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.

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

If $$x \in \bar{A}, \ \ \exists$$ some sequence $$(x_k) \in A$$ s.t. $$\lim_{\substack{k \rightarrow \infty}} (x_k) = x$$. Then $$f(x) = \lim_{\substack{k \rightarrow \infty}} f(x_k)$$ by continuity. So I'm thinking I can let $$g(x) = \lim_{\substack{k \rightarrow \infty}} f(x_k), \$$ where $$(x_k) \rightarrow x$$. Then that is continuous since:
$$\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*}$$
(by uniform continuity of $$f$$). Thus $$g(x)$$ is a continuous function satisfying the criteria.

5. Feb 13, 2008

### EnumaElish

I think you should use the triangle inequality.

6. Feb 13, 2008

### varygoode

Use it where? Are you responding to my response to you or my response to Ivy? (Or both?)

7. Feb 13, 2008

### EnumaElish

It's even simpler than that:

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

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

And since by definition f(xk) ---> g(x), I know that for any $\epsilon$, 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: Feb 13, 2008
8. Feb 13, 2008

### varygoode

Thanks so much EnumaElish, this problem is solved.

9. Feb 13, 2008

### EnumaElish

Do you mind posting your proof so it will be a reference to others?