# Is the extension of an $\alpha$-Hölder function also $\alpha$-Hölder?

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 $$u \in (a,b]$$ $$\vert f_e(u) - f_e(a) \vert \leq H \vert u-a \vert ^{\alpha}.$$

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