Given $f$ Hoelder continuous on an open set $K \subset \mathbb{R}^d$, show that $f$ can be extended so that it is hoelder cont. on $\bar K$.
I only have to show hat for every $ x\in \bar K-K $ f can be extended. Since $\bar K$ is closed, every sequence $x_n \subset K$ with limit $x$ does convergence in $\bar K$ which means there exists $N$: $\forall n > N :$ $\|f(x_n)-f(x)\| < \epsilon$ ,$\epsilon >0$. My Problem: How can make use of the hoelder continuity? I know that the statement isn't true if $f$ would only be continuous.
Greetings.
OK here's a clean approach:
Take $x_n \in K$ with $x_n \to x \in \partial K$.
We have that $|f(x_n) - f(x_m)| \leq |x_n - x_m|^{\alpha}$, which is eventually less than $\epsilon$, once $n$ and $m$ are large enough. This shows that $\{f(x_n)\}_{n=1}^{\infty}$ is always Cauchy and hence convergent.
Now let's see that the limit does not depend on the sequence $x_n$. Can it be the case that given $x_n,\ y_n \to x$ we have $$\lim_{n \to \infty}f(x_n) = a \neq b = \lim_{n \to \infty} f(y_n)?$$ No, because eventually $|f(x_n) - f(y_n)|$ is close to $|a-b|$ and $$|f(x_n) - f(y_n)| \leq |x_n - y_n|^{\alpha}.$$ And this right-hand side can be chosen to be much smaller than $|a-b|$.