# Real Analysis alpha Holder condition

Holder condition A function $f:(a,b)\rightarrow R$ satisfies a Holder condition of $\alpha$ order if $\alpha > 0$, and for some constant $H$ and for all $u,x \in (a,b)$, $$|f(u)-f(x)| \leq H|u-x|^\alpha.$$

Prove that an $\alpha$ - Holder function defined on $(a,b)$ is uniformly continuous and infer that it extends uniquely to a continuous function defined on $[a,b]$. Is the extended function $\alpha$ - Holder?

I understand how to show uniform continuity. Just choose $\delta = ( \frac{\epsilon}{H})^ {1/ \alpha}$ and that works out nicely , but I don't understand extending uniquely to a continuous function. Any help with the second part? Thanks

• Sorry guys about the script. – Jerry Oct 22 '14 at 1:37

Take a Cauchy sequence that converges to $a$, say $\{a_n\} \subset (a,b)$. By uniform continuity (or Holder continuity if you prefer), $\{f(a_n)\}$ is also a Cauchy sequence, hence it has a limit, say $f_a$.
It is easy to prove that this limit does not depend on the Cauchy sequence $\{a_n\}$, hence we can say that $\lim_{x \to a^+}f(x)$ exists and equals $f_a$.
Define your function on $[a,b)$ by setting $f(a) = f_a$. (This also implies that your function is continuous at $a$.)
Do the same for $b$ :)