Solved: Proving Uniform Continuity of Unique Continuous Function on A

[varygoode]

[SOLVED] Uniform Continuity

Homework Statement

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 $$g(x)=f(x) \ \forall \ x \in A$$.

Homework 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 $$\forall \ \ \varepsilon > 0 \ \ \exists \ \ \delta > 0 \ \ s.t. \ \ \Vert f(x) \ - \ f(y) \Vert < \varepsilon \ \ \forall \ \ x, \ y \ \ \in A \ \ s.t. \ \ \Vert x - y \Vert < \delta$$.

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

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

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!

 

[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.

 

[HallsofIvy]

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?

 

[varygoode]

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.

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 $$x \in \bar{A}, \ \ \exists$$ 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*}<br /> \begin{split}<br /> \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 \\<br /> &= \Vert f(x) - f(y) \Vert \\ &\leq \varepsilon \ \ \forall \ \Vert x - y \Vert \leq \delta<br /> \end{split}<br /> \end{equation*}$$
(by uniform continuity of $$f$$). Thus $$g(x)$$ is a continuous function satisfying the criteria.

 

[EnumaElish]

I think you should use the triangle inequality.

 

[varygoode]

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?)

 

[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.

 

[varygoode]

Thanks so much EnumaElish, this problem is solved.

 

[EnumaElish]

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