I am going through $C^{0, \alpha}(\Omega)$ regularity, $\alpha\in (0, 1)$, where $\Omega$ is a regular domain (not bounded, in general).
I can not find much material to study from, but I found this note in which is stated what follows. Let $f\in C^{0, \alpha}(\Omega)$ with $$|f|_{C^{0, \alpha}} = \|u\|_\infty +\sup_{x, y\in\Omega, x\neq y} \frac{|f(x)-f(y)|}{\|x-y\|^\alpha}.$$ An easy property is that there exists $\beta\in (0,1)$ (not necessarily equal to $\alpha$) and a constant $c>0$ such that $$|f(x)|\le c(1+|x|^\beta).$$
I am not interested in this property for a particular reason, but for the fact that it is stated that is "easy to check" and my trying to prove it since hours unsuccessfully.
I have tried writing $|f(x)|=|f(x)-f(y) +f(y)|$ and then trying to get an upper bound, but it did not work for me.
Does anyone have a better idea? Thank you in advance.
Using your line of reasoning, we’re free to choose $y$, so we might as well choose $y=0$. Of course, this assumes $0\in\Omega$; otherwise we can choose something like $y=cx$ for some fixed $c$ such that $cx\in \Omega$. The key is just that $y = O(x)$. Then,
$$\begin{align} |f(x)|&\leq |f(x)-f(0)|+|f(0)|\\ &\leq |f|_{C^{0,\alpha}} |x-0|^{\alpha} + \|f\|_\infty \\ &\leq |f|_{C^{0,\alpha}}(1+|x|^\alpha)\end{align}$$