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Let $f$ be a real valued function on $(a,b)$. Prove $f$ is uniformly continuous on $(a,b)$ iff $f$ can be extended to a continuous function $F$ on $[a,b]$.

This direction $\Leftarrow$ is OK! I have trouble in proving first implication!

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Sketch of proof:

  • We have to give values to $F(a)$ and $F(b)$. Take a sequence $\{a_n\}\subset (a,b)$ converging to $a$. Using uniform continuity of $f$, show that $\{f(a_n)\}\subset\Bbb R$ is Cauchy.

  • Then we can define $F(a):=\lim_{n\to +\infty}f(a_n)$. Show that it doesn't depend on the choice of the converging sequence $\{a_n\}$.

  • Do the same for $b$.

  • Show that $f$ is continuous on $[a,b]$ (actually, it's quite an obvious step if we know that continuity is equivalent to sequential continuity for functions from $[a,b]$ to $\Bbb R$).

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We will show that $\lim_{x \to a^+} f(x)$ exists, the existence of $\lim_{x\to b^-} f(x)$ can be shown along the same lines. Then define $F(a)$ and $F(b)$ as these limits and you found your extension:

So let $x_n \to a$ with $x_n > a$ for all $a$, we have to prove that $\bigl(f(x_n)\bigr)$ converges. As $\mathbb R$ is complete, it suffices to prove that it is a Cauchy sequence, so let $\epsilon > 0$. Then, as $f$ is uniformly continuous, there is some $\delta > 0$, such that $|f(x) - f(y)| < \epsilon$ for $|x-y| < \delta$. As $(x_n)$ is convergent, hence Cauchy, there is an $N$ such that $|x_n - x_m| < \delta$ for $n, m \ge N$. But then, by uniform continuity $|f(x_n) - f(x_m)| < \epsilon$ for $n, m \ge N$.

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