Hölder space

Remarks

With this norm, $C^{k,\alpha}(\overline{U})$ is a Banach space. Condition 3 says that the derivatives of order $k$ are Hölder continuous with exponent $\alpha$ on $U$, so they extend continuously to the closure $\overline{U}$. The space $C^{0,1}(\overline{U})$ is the space of bounded Lipschitz functions. On a bounded set $U$, $C^{k,\beta}(\overline{U}) \subseteq C^{k,\alpha}(\overline{U})$ for $0 < \alpha < \beta \le 1$. The local space $C^{k,\alpha}(U)$ is the set of $u$ with $u \in C^{k,\alpha}(\overline{V})$ for every open $V$ with compact closure in $U$.

Example: $u(x) = \sqrt{\abs{x}}$ lies in $C^{0,1/2}([-1, 1])$. It is continuous and bounded by $1$, and $\abs{\sqrt{\abs{x}} - \sqrt{\abs{y}}} \le \sqrt{\abs{\abs{x} - \abs{y}}} \le \abs{x - y}^{1/2}$.

Non-example: the same $u$ is not in $C^{0,1}([-1, 1])$, so it fails condition 3 for $\alpha = 1$. Indeed $\abs{u(x) - u(0)} / \abs{x - 0} = \abs{x}^{-1/2}$ is unbounded as $x \to 0$.