> [!definition] > > Let $X$ and $Y$ be [[Metric Space|metric spaces]], $f: X \to Y$ and $\gamma \in (0, 1]$. If there exists $C > 0$ such that > $ > d(f(x), f(y)) \le Cd(x, y)^{\gamma} \quad \forall x, y \in X > $ > then $f$ is **Hölder continuous with exponent** $\gamma$. > [!definition] > > Let $X$ be a metric space and $f \in BC(X)$, then the $\gamma$-th Hölder [[Seminorm|seminorm]] is defined as > $ > [f]_{C^{0, \gamma}} = \sup\bracs{\frac{d(f(x), f(y))}{d(x, y)^\gamma}: x, y \in X, x \ne y} > $ > and the $\gamma$-th Hölder norm is > $ > \norm{f}_{C^{0, \gamma}(X)} = \norm{f}_u + [f]_{C^{0, \gamma}} > $ > [!theorem] > > Let $U \subset \real^d$ be [[Open Set|open]], $k \in \nat$ and $\gamma \in (0, 1]$. Define the [[Normed Vector Space|norm]] on the [[Space of Uniformly Continuously Differentiable Functions|uniformly continuously differentiable functions]] $UC^k(U)$ as > $ > \norm{f}_{C^{k, \gamma}(U)} = \sum_{\abs{\alpha} \le k}\norm{\partial^\alpha f}_{C^{0, \gamma}(U)} = \sum_{\abs{\alpha} \le k}\paren{\norm{\partial^\alpha f}_u + [\partial^\alpha f]_{C^{0, \gamma}}} > $ > and let the **Hölder space** as > $ > C^{k, \gamma}(U) = \bracs{f \in UC^k(U): \norm{f}_{C^{k, \gamma}(U)} < \infty} > $ > then $C^{k, \gamma}(U)$ is a [[Banach Space|Banach space]]. > > *Proof*. Let $\seq{f_n}$ be an absolutely convergent series and let $F_n = \sum_{k \le n}f_k$. By completeness of $UC^k(U)$, there exists $F = \sum_{n \in \nat}f_n \in UC^k(U)$ such that $F_n \to F$ in $UC^k(U)$. It's sufficient to verify that $F \in C^{k, \gamma}(U)$ and $F_n \to F$ in $C^{k, \gamma}(U)$. Let $\alpha$ be any multi-index and $x, y \in U$, then > $ > \begin{align*} > \frac{\abs{\partial^\alpha F(x) - \partial^\alpha F(y)}}{\abs{x - y}^\gamma} &\le \sum_{n \in \nat}\frac{\abs{\partial^\alpha f_n(x) - \partial^\alpha f_n(y)}}{\abs{x - y}^\gamma} \\ > &\le \sum_{n \in \nat}[f_n]_{C^{0, \gamma}} < \infty > \end{align*} > $ > so $F \in C^{k, \gamma}(U)$. With the pointwise (weak) convergence of the derivatives to a continuous function, we get that $\partial^\alpha (F - F_n) = \sum_{k > n}\partial^\alpha f_k$. From here, we note that the set > $ > D = \bracs{\sum_{k = n + 1}^N \partial^\alpha f_k(x) - \sum_{k = n + 1}^N \partial^\alpha f_k(y): N \in \nat} > $ > is bounded by the sum > $ > \sum_{k > n}\abs{\partial^\alpha f_k(x) - \partial^\alpha f_k(y)} \le \sum_{k > n}\norm{f_k}_{C^{k, \gamma}} \cdot \abs{x - y}^\gamma > $ > which goes to $0$ as $n \to \infty$. Therefore $F_n \to F$ in $C^{k, \gamma}$ as well.