Holder functions with negative power

Question

Are there continuous unbounded functions which satisfy $$|f(x)-f(y)| \leq |x-y|^{-\alpha}, \text { for } \alpha>0 \quad x,y \in \Omega \subset \mathbb{R} $$
Clearly constant functions and piecewise constant functions satisfy the above conditions in $\mathbb{R}.$

Answer 1

You can not find in general. For example, if you consider your $\Omega$ to be unbounded, then there can not be any non-constant function which satisfies the given condition. This can be shown from the fact that, if so, then the inequality gives us $\lim_{|x|\rightarrow \infty}(f(x)-f(y))=0$ for any $y$.

But on the other hand, if you consider your $\Omega$ to be an unit length interval, then any Lipschitz function with slope bounded by $1$ (i.e. any function $f$ satisfies $|f(x)-f(y)|\leq C|x-y|\,\forall x,y$ where $C<1$) will automatically satisfies the condition.

Answer 2

To the stated question, the answer is No.

Let $x_0\in \Omega$, then there exists some small neighborhood $N_x$ such that $f|_{N_x}$ is bounded by continuity.

Suppose $f$ is an unbounded function on $\Omega$. Let $y_n$ be a sequence in $\Omega$ such that $|f(y_n)| \nearrow \infty$. Necessarily $y_n\not\in N_x$ for all sufficiently large $n$.

But this means that $|x_0 - y_n| \geq \delta$ for some fixed $\delta$, as long as $n$ is sufficiently large, and the Holder type assumption guarantees then $|f(y_n) - f(x_0)| \leq \delta^{-\alpha}$ or that $|f(y_n)| \leq |f(x_0)| + \delta^{-\alpha}$ is uniformly bounded. This contradicts the assumption that $y_n$ is a sequence along which $f$ blows up.