The question is the one in the title. I was wondering if $f(x) = \frac{1}{\sqrt{x}}$ is Hölder continuous because I know that for $x^{\beta}$ with $\beta \in (0,1]$ we have Hölder continuity. But when does this break?
I think the $f$ I provided it is not because one can check numerically over a grid of $\alpha$'s that the quantity (fixing one $\alpha$ at each simulation): $$\frac{|f(x) - f(y)|}{|x-y|^{\alpha}}$$ is unbounded as we let $x \rightarrow y$. In this way I was able to show numerically that no matter how you choose $\alpha$ you will not be able to find a bound for the quantity above. The same simulations for $f(x) = \sqrt{x}$ show that for this function we are actually able to find a bound for the quantity specified above (using $\alpha = 1/2$). As expected because $\sqrt{x}$ is $1/2$ Holder. This method I think constitutes a numerical counterexample to the statement in the title.
I was curious to know if anyone else ever wondered about this and if your conclusions coincide with mine.
All Hölder continuous functions are uniformly continuous. So any such function on $(0,1]$ will have a continuous extension to $[0,1]$. In particular, it will be bounded. Since $f$ is not bounded, it is not Hölder continuous on $(0,1]$ (certainly not on its natural domain $(0,\infty)$ either).