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SiddharthM
#2
Oct17-07, 10:08 AM
P: 176
Theorem: A function that is uniformly continous on a bounded subset of the line,E, has a continous extension on the closure of E.

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

Taking any other [tex] y_n \rightarrow x [/tex] then by the triangle inequality:

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


[tex] d(f(y_n),A) \leq d(f(y_n),f(x_n)) + d(f(x_n),A) [/tex].

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

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

So define our continous extension as [tex] g(x)=f(x) [/tex] for [tex] x \in E [/tex] and [tex] g(y)=\lim_{\substack{x\rightarrow y}} f(x) [/tex] for [tex] y \in E' [/tex].

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

By the triangle inequality [tex]

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)) [/tex]

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

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

[tex] d(g(x_m),g(s_m)) \leq d(g(y_i^m),g(s_m)) [/tex].

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

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

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

Hence g is continuous on the closure of E.

QED