A function $f: (a,b) \rightarrow \mathbb{R}$ satisfies a $\alpha$-Hölder condition of order $\alpha$ if $\alpha > 0$ and for some constant $H$ and all $u,x \in (a,b),$ it follows that $$\vert f(u)-f(x) \vert \leq H \vert u-x \vert ^{\alpha}.$$ Prove that an $\alpha$-Hölder function defined on $(a,b)$ is uniformly constinuous and infer that it extends uniquely to a continuous function defined on $[a,b].$ Is the extended function $\alpha$-Hölder?

Consider any $\epsilon > 0.$ Given that $f$ is $\alpha$-Hölder, if $\delta = (\frac{\epsilon}{H})^{\frac{1}{\alpha}},$ then $$\vert f(u)-f(x) \vert \leq H \vert u-x \vert^{\alpha} < H\delta^{\alpha} = H\bigg(\bigg(\frac{\epsilon}{H}\bigg)^{\frac{1}{\alpha}}\bigg)^{\alpha} = H\bigg(\frac{\epsilon}{H}\bigg) = \epsilon$$ for all $u,x \in (a,b)$ such that $\vert u - x \vert < \delta.$ Hence, $f$ is uniformly continuous. Moreover, by Ch. 2, P. 54, a uniformly continuous function $\phi(x): S \rightarrow \mathbb{R}$ extends to a uniformly continuous function $\phi':\bar{S} \rightarrow \mathbb{R}.$ So, $f$ extends to a function $f_e: [a,b] \rightarrow \mathbb{R}$ that is uniformly continuous and, therefore, continuous.

TO SHOW THAT $f_e$ is $\alpha$-Hölder, I must show that the condition remains satisfied for $a,b \in [a,b].$ I wrote the following.

Consider any $u \in (a,b).$ For all $x \in [a,b]$ where $x \neq a,b,$ $$\vert f_e(u)-f_e(x) \vert \leq H \vert u-x \vert ^{\alpha}.$$

So, $$\lim_{x \rightarrow b} \vert f_e(u) - f_e(x) \vert \leq H \vert u-x \vert ^{\alpha},$$ and $$\vert f_e(u) - f_e(b) \vert \leq H \vert u-b \vert ^{\alpha}.$$ Hence, for all $u \in (a,b)$ $$\vert f_e(u) - f_e(b) \vert \leq H \vert u-b \vert ^{\alpha}.$$

I will show that the property holds for $a \in [a,b]$ via the same reasoning. Consider any $u \in (a,b].$ For all $x \in [a,b]$ where $x \neq a,$ it follows that $$\vert f_e(u) - f_e(x) \vert \leq H \vert u-x \vert ^{\alpha},$$ and so $$\lim_{x \rightarrow a} \vert f_e(u) - f_e(x) \vert \leq H \vert u-x \vert ^{\alpha},$$ and $$\vert f_e(u) - f_e(a) \vert \leq H \vert u-a \vert ^{\alpha}.$$ Hence, for all $u \in (a,b]$ $$\vert f_e(u) - f_e(a) \vert \leq H \vert u-a \vert ^{\alpha}.$$

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    $\begingroup$ Where did $f'$ come from? $\endgroup$ – Umberto P. Nov 8 '18 at 19:17
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    $\begingroup$ Your proof looks fine. $\endgroup$ – Sangchul Lee Nov 8 '18 at 19:33
  • $\begingroup$ Hey, Umberto. $f'$ is the extension of $f!$ I mean to write $f'$ instead of $g.$ Thanks for the catch! $\endgroup$ – Rafael Vergnaud Nov 8 '18 at 19:41
  • $\begingroup$ Thanks, Sangchul! :) $\endgroup$ – Rafael Vergnaud Nov 8 '18 at 19:41
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    $\begingroup$ Since $f'$ is nearly universally used to denote the derivative of $f$, using it to denote an extension is really not good notation. $\endgroup$ – Umberto P. Nov 8 '18 at 20:31

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