Difference between revisions of "The Continuous Extension Theorem"

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(Created page with " ==Resources== * [http://mathonline.wikidot.com/the-continuous-extension-theorem The Continuous Extension Theorem], mathonline.wikidot.com")
 
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<p>Recall that from <a href="/the-uniform-continuity-theorem">The Uniform Continuity Theorem</a> that if a function <math>I = [a, b]</math> is a closed and bounded interval and <math>f : I \to \mathbb{R}</math> is continuous on <math>I</math>, then <math>f</math> must also be uniformly continuous on <math>I</math>. The succeeding theorem will help us determine when a function <math>f</math> is uniformly continuous when <math>I</math> is instead a bounded open interval.</p>
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<p>Before we look at The Continuous Extension Theorem though, we will need to prove the following lemma.</p>
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<td><strong>Lemma 1:</strong> If <math>f : A \to \mathbb{R}</math> is a uniformly continuous function and if <math>(x_n)</math> is a <a class="newpage" href="/cauchy-sequences">Cauchy Sequence</a> from <math>A</math>, then <math>(f(x_n))</math> is a Cauchy sequence from <math>\mathbb{R}</math>.</td>
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<li><strong>Proof:</strong> Let <math>f : A \to \mathbb{R}</math> be a uniformly continuous function and let <math>(x_n)</math> be a Cauchy sequence from <math>A</math>. We want to show that <math>(f(x_n))</math> is also a Cauchy sequence. Recall that to show that <math>(f(x_n))</math> is a Cauchy sequence we must show that <math>\forall \varepsilon > 0</math> then <math>\exists N \in \mathbb{N}</math> such that <math>\forall m, n \in \mathbb{N}</math>, if <math>m, n \geq N</math> then <math>\mid f(x_n) - f(x_m) \mid < \varepsilon</math>.</li>
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<li>Since <math>f</math> is uniformly continuous on <math>A</math>, then for any <math>\varepsilon > 0</math>, <math>\exists \delta_{\varepsilon} . 0</math> such that for all <math>x, y \in A</math> where <math>\mid x - y \mid < \delta_{\varepsilon}</math> we have that <math>\mid f(x) - f(y) \mid < \varepsilon</math>.</li>
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<li>Now for <math>\delta_{\varepsilon} > 0</math>, since <math>(x_n)</math> is a Cauchy sequence then <math>\exists N_{\delta_{\varepsilon}} \in \mathbb{N}</math> such that <math>\forall m, n \geq N_{\delta_{\varepsilon}}</math> we have that <math>\mid x_n - x_M \mid < \delta_{\varepsilon}</math>. So this <math>N_{\delta_{\varepsilon}}</math> will do for the sequence <math>(f(x_n))</math>. So for all <math>n, m \geq N_{\delta_{\varepsilon}}</math> we have that <math>\mid x_n - x_m \mid < \delta_{\varepsilon}</math> and from the continuity of <math>f</math> this implies that <math>\mid f(x_n) - f(x_m) \mid < \varepsilon</math> and so <math>(f(x_n))</math> is a Cauchy sequence. <math>\blacksquare</math></li>
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<p>We are now ready to look at The Continuous Extension Theorem.</p>
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<td><strong>Theorem 1 (The Continuous Extension Theorem):</strong> If <math>I = (a,b)</math> is an interval, then <math>f : I \to \mathbb{R}</math> is a uniformly continuous function on <math>I</math> if and only if <math>f</math> can be defined at the endpoints <math>a</math> and <math>b</math> such that <math>f</math> is continuous on <math>[a, b]</math>.</td>
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<li><strong>Proof:</strong> <math>\Rightarrow</math> Suppose that <math>f</math> is uniformly continuous on <math>I = (a, b)</math>. Let <math>(x_n)</math> be a sequence in <math>(a,b)</math> that converges to <math>a</math>. Then since <math>(x_n)</math> is a convergent sequence, it must also be a Cauchy sequence. By lemma 1, since <math>(x_n)</math> is a Cauchy sequence then <math>(f(x_n))</math> is also a Cauchy sequence, and so <math>(f(x_n))</math> must converge in <math>\mathbb{R}</math>, that is <math>\lim_{n \to \infty} f(x_n) = L</math> for some <math>L \in \mathbb{R}</math>.</li>
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<li>Now suppose that <math>(y_n)</math> is another sequence in <math>(a, b)</math> that converges to <math>a</math>. Then <math>\lim_{n \to \infty} (x_n - y_n) = a - a = 0</math>, and so by the uniform continuity of <math>f</math>:</li>
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<span class="equation-number">(1)
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<div class="math-equation" id="equation-1">\begin{align} \lim_{n \to \infty} f(y_n) = \lim_{n \to \infty} [f(y_n) - f(x_n) + f(x_n)] \\ \lim_{n \to \infty} f(y_n) = \lim_{n \to \infty} [f(y_n) - f(x_n) ] + \lim_{n \to \infty} f(x_n) \\ \lim_{n \to \infty} f(y_n) = 0 + L = L \end{align}</div>
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<li>So for every sequence <math>(y_n)</math> in <math>(a, b)</math> that converges to <math>a</math>, we have that <math>(f(y_n))</math> converges to <math>L</math>. Therefore by the Sequential Criterion for Limits, we have that <math>f</math> has the limit <math>L</math> at the point <math>a</math>. Therefore, define <math>f(a) = L</math> and so <math>f</math> is continuous at <math>a</math>. We use the same argument for the endpoint <math>b</math>, and so <math>f</math> is can be extended so that <math>f</math> is continuous on <math>[a, b]</math>.</li>
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<li><math>\Leftarrow</math> Suppose that <math>f</math> is continuous on <math>[a, b]</math>. By <a href="/the-uniform-continuity-theorem">The Uniform Continuity Theorem</a>, since <math>[a, b]</math> is a closed and bounded interval then <math>f</math> is uniformly continuous. <math>\blacksquare</math></li>
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==Resources==
 
==Resources==
 
* [http://mathonline.wikidot.com/the-continuous-extension-theorem The Continuous Extension Theorem], mathonline.wikidot.com
 
* [http://mathonline.wikidot.com/the-continuous-extension-theorem The Continuous Extension Theorem], mathonline.wikidot.com

Revision as of 12:36, 20 October 2021

Recall that from <a href="/the-uniform-continuity-theorem">The Uniform Continuity Theorem</a> that if a function is a closed and bounded interval and is continuous on , then must also be uniformly continuous on . The succeeding theorem will help us determine when a function is uniformly continuous when is instead a bounded open interval.

Before we look at The Continuous Extension Theorem though, we will need to prove the following lemma.

Lemma 1: If is a uniformly continuous function and if is a <a class="newpage" href="/cauchy-sequences">Cauchy Sequence</a> from , then is a Cauchy sequence from .
  • Proof: Let be a uniformly continuous function and let be a Cauchy sequence from . We want to show that is also a Cauchy sequence. Recall that to show that is a Cauchy sequence we must show that then such that , if then .
  • Since is uniformly continuous on , then for any , such that for all where we have that .
  • Now for , since is a Cauchy sequence then such that we have that . So this will do for the sequence . So for all we have that and from the continuity of this implies that and so is a Cauchy sequence.

We are now ready to look at The Continuous Extension Theorem.

Theorem 1 (The Continuous Extension Theorem): If is an interval, then is a uniformly continuous function on if and only if can be defined at the endpoints and such that is continuous on .
  • Proof: Suppose that is uniformly continuous on . Let be a sequence in that converges to . Then since is a convergent sequence, it must also be a Cauchy sequence. By lemma 1, since is a Cauchy sequence then is also a Cauchy sequence, and so must converge in , that is for some .
  • Now suppose that is another sequence in that converges to . Then , and so by the uniform continuity of :

(1)

\begin{align} \lim_{n \to \infty} f(y_n) = \lim_{n \to \infty} [f(y_n) - f(x_n) + f(x_n)] \\ \lim_{n \to \infty} f(y_n) = \lim_{n \to \infty} [f(y_n) - f(x_n) ] + \lim_{n \to \infty} f(x_n) \\ \lim_{n \to \infty} f(y_n) = 0 + L = L \end{align}
  • So for every sequence in that converges to , we have that converges to . Therefore by the Sequential Criterion for Limits, we have that has the limit at the point . Therefore, define and so is continuous at . We use the same argument for the endpoint , and so is can be extended so that is continuous on .
  • Suppose that is continuous on . By <a href="/the-uniform-continuity-theorem">The Uniform Continuity Theorem</a>, since is a closed and bounded interval then is uniformly continuous.


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