Let $\alpha\in (0, 1)$ and $-\infty<a<b<+\infty$. Assume that $u\in C^{0, \alpha}((a, b), \mathbb R)$. Here $$|u|_{C^{0, \alpha}((a, b))} = \|u\|_{L^\infty((a, b))} +\sup_{x, y\in (a, b), x\neq y} \frac{|u(x)-u(y)|}{|x-y|^\alpha}.$$
The question is about if it is true that $$u\in C^{0, \alpha}((a, b), \mathbb R)\implies u\in L^\infty\big((a, b)\big).$$
I know that in general it is not true. In fact, for example, $u(x)=|x|^{\alpha}$ is Holder continuous with $\alpha\in (0, 1]$ but it is unbounded in $\mathbb R$. I am wondering if here, since we in a bounded subset of $\mathbb R$, I can conclude that $u\in L^\infty\big((a, b)\big)$.
In case, how to prove that? Thank you in advance.
As phrased, the question is trivial. The norm on $C^{0,\alpha}$ has the $L^{\infty}$ norm plus extra stuff, so if the Holder-norm is finite, so is the $L^{\infty}$ norm, so yes the function is bounded. In particular, we have a continuous inclusion $C^{0,\alpha}\hookrightarrow L^{\infty}$ (unbounded or bounded domain alike). The confusion might stem from the fact that usually when we speak of “$\alpha$-Holder continuous function” we only mean that the second piece of $C^{0,\alpha}$ norm, i.e the Holder seminorm, is finite.
A perhaps simpler illustration of this disconnect in terminology vs spaces of functions is when talking about continuous functions. Continuity has its own definition, while if we want to consider $C(X)$ as a Banach space with sup norm, one either restricts $X$ to be compact or implicitly considers bounded continuous functions, but still call the space $C(X)$ (though some authors use $C_b(X)$ or $BC(X)$, but still, $C(X)$ is the most prevalent).
Anyway, let us suppose only the second piece of the Holder norm is finite (and let’s work on a bounded open set $\Omega\subset\Bbb{R}^n$). This means there is an $M>0$ such that for all $x,y\in\Omega$, we have \begin{align} |u(x)-u(y)|\leq M\|x-y\|^{\alpha}. \end{align} In particular, $u$ is uniformly continuous on $\Omega$. Thus, by basic analysis, $u$ has a unique uniformly continuous extension $\tilde{u}$ to $\overline{\Omega}$, so $\tilde{u}$ is in particular bounded on $\overline{\Omega}$ and hence $u$ is bounded on $\Omega$. Also, by continuity of the extension, we have that these estimates hold on the boundary as well. So, $\|\tilde{u}\|_{C^{0,\alpha}(\overline{\Omega})}\leq \|u\|_{C^{0,\alpha}(\Omega)}$; note that the other inequality is obvious since we’re enlarging the domain, so we have equality.