Characterization of Uniform Continuity on the line

[SiddharthM]

It began with my trying to prove that a uniformly continuous function on a bounded subset of the line is bounded. I took the hard route cause I couldn't figure out how to do this directly. I prove that if a real function is uniformly continuous on a bounded set E then there exists a continuous extension on the closure of E. My first post will be this lemma.

Lemma: Let $$E \subset X$$ If $$x_n \Rightarrow x$$ and for each n there is a sequence $$y_k \Rightarrow x_n$$ with $$(y_k) \in E$$ then $$s_p=y_p^p$$ (the latter being the pth term of the pth sequence) is a sequence contained within E that converges to x.

Proof: Consider subsequences of each $$y_k^n$$ to obtain a new collection of $$y_m^n$$ with the following property:

$$d(y_m^n,y_j^n) \leq \frac{1}{n}$$ for all m,j

Put $$s_p = y_p^p$$ so that $$s_p \in E$$ and let $$\epsilon > 0$$. Then by the triangle inequality

$$d(s_p,x) \leq d(s_p,y_m^p) + d(y_m^p, x_n) + d(x_n,x) < \frac{1}{p} + d(y_m^p,x_n) + d(x_n,x)$$
Because the LHS is independent of n, m and $$x_n \rightarrow x$$
taking limits as $$n,m \rightarrow \infty$$ we obtain

$$d(s_p,x) \leq \frac{1}{p} + d(x_p,x_n)$$. Taking n and p large enough we easily see that $$s_p \rightarrow x$$

PS: this is my first go at Latex. I will be slow to post the rest! Please give me feedback on both readability and the content.

 

[SiddharthM]

Theorem: A function that is uniformly continuous on a bounded subset of the line,E, has a continuous extension on the closure of E.

Proof: We begin by showing that for each $$y\in E' , \lim_{\substack{x\rightarrow y}} f(x)$$ exists. Consider $$x_n \rightarrow y$$ with $$x_n \in E$$ for all n. We show $$f(x_n)$$ is Cauchy and therefore convergent b/c the line is complete. Fix $$\epsilon > 0$$ because f is uniformly continuous there is a $$\delta > 0$$ so that when $$d(x_n,x_m) < \delta$$ we also have $$d(f(x_n),f(x_m)) < \epsilon$$. Because $$(x_n)$$ is convergent it is cauchy so we see that $$f(x_n)$$ is cauchy and therefore convergent.

Taking any other $$y_n \rightarrow x$$ then by the triangle inequality:

$$d(y_n,x_m) \leq d(y_n,x) + d(x_n,x)$$. So the LHS can be made small for all but finitely many terms. Put $$\lim_{\substack{n\rightarrow\infty}} f(x_n) = A$$. Using the triangle inequality again


$$d(f(y_n),A) \leq d(f(y_n),f(x_n)) + d(f(x_n),A)$$.

Fix $$\epsilon > 0$$ then there is a $$\delta > 0$$ so that when $$d(x_n,y_m)<\delta$$ we also have $$d(f(x_n),f(y_n))<\frac{\epsilon}{2}$$. It is clear that $$d(f(y_n),A) < \epsilon$$ for all but finitely many n.

We have shown that $$\lim_{\substack{x\rightarrow y}} f(x)$$ exists for all $$y \in E'$$.

So define our continuous extension as $$g(x)=f(x)$$ for $$x \in E$$ and $$g(y)=\lim_{\substack{x\rightarrow y}} f(x)$$ for $$y \in E'$$.

We now prove g is continuous on E. Take a sequence $$x_n \rightarrow x$$ and WLOG $$x_n \in E'$$ for all n. For each of these $$x_n$$ there is a $$y_k \rightarrow x_n$$ with $$y_k \in E$$. Define $$(s_p)$$ as in the lemma above. Then $$s_p \rightarrow x$$, so $$g(s_p) \rightarrow g(x)$$ because $$s_p \in E$$.

By the triangle inequality $$d(g(x_m),g(x_n)) \leq d(g(x_m),g(s_m)) + d(g(s_m),g(s_n)) + d(g(s_n),g(x_n))$$

Using the triangle inequality on the first term on the RHS we obtain

$$d(g(x_m),g(s_m)) \leq d(g(y_i^m),g(s_m)) + d(g(y_i^m),g(x_m))$$ and since $$y_i^m \rightarrow x_m, y_i^m \in E$$ we also have that $$g(y_i^m) \rightarrow g(x_m)$$. Since the LHS is independent of i we see that

$$d(g(x_m),g(s_m)) \leq d(g(y_i^m),g(s_m))$$.

Now because g is uniformly continuous on E and $$y_i^m,s_m \in E$$ we can make the RHS and thus the LHS small if m is large enough. This is because the sequences $$(y_k^n)$$ were created as above in the lemma, that is $$d(y_k^n,y_j^n) \leq \frac{1}{n}$$ for all k,j. Similarly we see that $$d(g(s_n),g(x_n))$$ goes to 0. Since $$g(s_n)$$ converges it is cauchy and the middle term goes to zero as well.

Since these three values on the RHS goto zero as $$m,n \rightarrow \infty$$ so does the LHS and we have that $$g(x_n)$$ is cauchy.

Referring to the lemma above we see that $$g(x_n) \rightarrow \lim_{\substack{p\rightarrow \infty}} g(s_p)=g(x)$$

Hence g is continuous on the closure of E.

QED

 

[EnumaElish]

It began with my trying to prove that a uniformly continuous function on a bounded subset of the line is bounded. I took the hard route cause I couldn't figure out how to do this directly. I prove that if a real function is uniformly continuous on a bounded set E then there exists a continuous extension on the closure of E. My first post will be this lemma.

But your lemma does not address whether f is bounded (on E, or E'). Or is that implicit in uniform continuity?

 

[EnumaElish]

Ah. Well if f is continuous on E, and $$\lim_{\substack{x\rightarrow y}} f(x)$$ exists. Then if f is extended at this single point appropriately (that is $$\lim_{\substack{x\rightarrow y}} f(x) = f(y)$$ for $$y \in E'$$. Then f is clearly still continous. Using induction I can then proof it for finitely many extensions of the sort. But what about an infinite number of extensions like this one?

It seems obvious to me that extending f this way should work, but I can't prove it when the set of limit points of E is infinite/dense.

Suggestions?

I think this is okay as it is.

 

[SiddharthM]

Corollary (domain and range are real, heine-borel is used implicitly): f uniformly continuous is bounded on a bounded set E. This is because there is a continuous extension onto a compact set containing E, so applying the max value theorem we see that f is bounded on E.

Lemma: Continuity implies uniform continuity on a compact set. (This is a true fact in general metric spaces)

Characterization of Uniform continuity on bounded subsets of the line: f is uniformly continuous on a bounded set E iff f can be continuously extended to the closure of E.

 

[mathwonk]

isnt this trivial? i.e. given epsilon equal to say 1, there is delta such that for any two x's closer togethger than delta, tgheir values are closer togetehr than 1,hence if the domain is contained in an interval of length b, then the function cannot grow by more than b/delta on that set. am i missing something?

 

[SiddharthM]

no that is correct! it's a shame i missed that, but not really - because it led me to the nontrivial theorems above!

did you read the proof? please do - boundedness of f on bounded E doesn't concern me as much as extending f onto the closure of E, i need feedback.