[proofplan] We prove completeness of $C^{k,\alpha}(\overline{\Omega})$ by induction on $k$. The base case $k = 0$ uses the completeness of $C^0(\overline{\Omega})$ (uniform convergence) together with a pointwise-limit argument on the Holder seminorm to show that the limit function inherits Holder regularity and that convergence holds in the Holder norm. For $k \ge 1$, we show that every Cauchy sequence in $C^{k,\alpha}(\overline{\Omega})$ produces uniform Cauchy sequences of all derivatives up to order $k$, identify the limits via a derivative-closure lemma (passing difference quotients through the limit using the Fundamental Theorem of Calculus), and verify that the top-order derivatives converge in $C^{0,\alpha}$. [/proofplan] [step:Prove completeness of the base space $C^{0,\alpha}(\overline{\Omega})$] Let $(g_m)_{m \in \mathbb{N}}$ be a Cauchy sequence in $C^{0,\alpha}(\overline{\Omega})$. Since the $C^{0,\alpha}$ norm dominates the uniform norm $\|g_m\|_{L^\infty}$, the sequence is Cauchy in $C^0(\overline{\Omega})$. The space $C^0(\overline{\Omega})$ is a [Banach space](/page/Banach%20Space) (completeness of the uniform norm on a compact domain), so there exists a continuous $g: \overline{\Omega} \to \mathbb{R}$ with $g_m \to g$ uniformly. Since $(g_m)$ is Cauchy in the Holder norm, for every $\varepsilon > 0$ there exists $M \in \mathbb{N}$ such that for all $m, \ell \ge M$: \begin{align*} [g_m - g_\ell]_{C^{0,\alpha}} \le \varepsilon. \end{align*} Fix distinct $x, y \in \overline{\Omega}$ and $m \ge M$. Taking the limit $\ell \to \infty$ and using uniform convergence $g_\ell \to g$: \begin{align*} \frac{|(g_m - g)(x) - (g_m - g)(y)|}{|x - y|^\alpha} = \lim_{\ell \to \infty} \frac{|(g_m - g_\ell)(x) - (g_m - g_\ell)(y)|}{|x - y|^\alpha} \le \varepsilon. \end{align*} Taking the supremum over all distinct $x, y$: $[g_m - g]_{C^{0,\alpha}} \le \varepsilon$ for all $m \ge M$. This yields two conclusions. First, $g = g_m - (g_m - g)$ is the difference of two Holder continuous functions, so $g \in C^{0,\alpha}(\overline{\Omega})$. Second, $[g_m - g]_{C^{0,\alpha}} \to 0$, which combined with $\|g_m - g\|_{L^\infty} \to 0$ gives $\|g_m - g\|_{C^{0,\alpha}} \to 0$. [guided] We establish the base case: $C^{0,\alpha}(\overline{\Omega})$ is complete. Let $(g_m)$ be Cauchy in the $C^{0,\alpha}$ norm. The norm has two components: the sup norm and the Holder seminorm $[g]_{C^{0,\alpha}} = \sup_{x \ne y} \frac{|g(x) - g(y)|}{|x - y|^\alpha}$. Since $\|g_m - g_\ell\|_{C^{0,\alpha}} \ge \|g_m - g_\ell\|_{L^\infty}$, the sequence is Cauchy in the sup norm. The space $C^0(\overline{\Omega})$ with the sup norm is complete (this is a standard result: $\overline{\Omega}$ is compact, so uniform limits of continuous functions are continuous). Hence $g_m \to g$ uniformly for some continuous $g: \overline{\Omega} \to \mathbb{R}$. The question is: does $g$ inherit the Holder regularity, and does $g_m \to g$ in the Holder norm? Since $(g_m)$ is Cauchy in $C^{0,\alpha}$, for any $\varepsilon > 0$ there exists $M$ with $[g_m - g_\ell]_{C^{0,\alpha}} \le \varepsilon$ for all $m, \ell \ge M$. Fix any two distinct points $x, y \in \overline{\Omega}$ and any $m \ge M$. The key step is passing the Holder seminorm estimate through the limit. For each fixed pair $(x, y)$: \begin{align*} \frac{|(g_m - g)(x) - (g_m - g)(y)|}{|x - y|^\alpha} = \lim_{\ell \to \infty} \frac{|(g_m - g_\ell)(x) - (g_m - g_\ell)(y)|}{|x - y|^\alpha} \le \varepsilon. \end{align*} The equality holds because $g_\ell(x) \to g(x)$ and $g_\ell(y) \to g(y)$ pointwise (from uniform convergence). The inequality holds because $[g_m - g_\ell]_{C^{0,\alpha}} \le \varepsilon$ for $\ell \ge M$. Taking the supremum over all distinct pairs $(x,y)$ gives $[g_m - g]_{C^{0,\alpha}} \le \varepsilon$ for $m \ge M$. Two conclusions follow. First, $g = g_m - (g_m - g)$ is the difference of two functions in $C^{0,\alpha}(\overline{\Omega})$, so $g \in C^{0,\alpha}(\overline{\Omega})$. Second, $\|g_m - g\|_{C^{0,\alpha}} = \|g_m - g\|_{L^\infty} + [g_m - g]_{C^{0,\alpha}} \to 0$ as $m \to \infty$. This completes the base case. [/guided] [/step] [step:Establish the derivative-closure lemma via the Fundamental Theorem of Calculus] Let $U \subseteq \mathbb{R}^n$ be open. Suppose $(u_m) \subset C^1(U)$ satisfies $u_m \to u$ uniformly on compact subsets and $\partial_{x_i} u_m \to v_i$ uniformly on compact subsets for each $i \in \{1, \ldots, n\}$. We show $\partial_{x_i} u = v_i$. Fix $x \in U$ and choose $r > 0$ with $\overline{B}(x, r) \subset U$. For $h \in \mathbb{R} \setminus \{0\}$ with $|h| < r$, the Fundamental Theorem of Calculus along the segment from $x$ to $x + h e_i$ gives: \begin{align*} u_m(x + h e_i) - u_m(x) = \int_0^h \partial_{x_i} u_m(x + t e_i) \, d\mathcal{L}^1(t). \end{align*} Since $\partial_{x_i} u_m \to v_i$ uniformly on $\overline{B}(x, r)$, passing $m \to \infty$: \begin{align*} u(x + h e_i) - u(x) = \int_0^h v_i(x + t e_i) \, d\mathcal{L}^1(t). \end{align*} Dividing by $h$ and estimating: \begin{align*} \left| \frac{u(x + h e_i) - u(x)}{h} - v_i(x) \right| = \left| \frac{1}{h} \int_0^h \bigl(v_i(x + t e_i) - v_i(x)\bigr) \, d\mathcal{L}^1(t) \right| \le \sup_{|t| \le |h|} |v_i(x + t e_i) - v_i(x)|. \end{align*} Since $v_i$ is the uniform limit of continuous functions, $v_i$ is continuous. As $h \to 0$, the right-hand side tends to $0$, so $\partial_{x_i} u(x)$ exists and equals $v_i(x)$. [guided] This lemma justifies interchanging differentiation and limit-taking under uniform convergence of derivatives. The idea is clean: instead of working with the abstract definition of the derivative, we use the integral representation provided by the Fundamental Theorem of Calculus, which is easy to pass through a limit. Given $(u_m) \subset C^1(U)$ with $u_m \to u$ and $\partial_{x_i} u_m \to v_i$ uniformly on compacts, we fix $x \in U$, pick a ball $\overline{B}(x, r) \subset U$, and for small $h$ write: \begin{align*} u_m(x + h e_i) - u_m(x) = \int_0^h \partial_{x_i} u_m(x + t e_i) \, d\mathcal{L}^1(t). \end{align*} The integrand converges uniformly to $v_i(x + t e_i)$ on $\overline{B}(x, r)$, so we may pass $m \to \infty$ under the integral sign (uniform convergence on a compact interval justifies this): \begin{align*} u(x + h e_i) - u(x) = \int_0^h v_i(x + t e_i) \, d\mathcal{L}^1(t). \end{align*} Now we form the difference quotient and subtract $v_i(x)$: \begin{align*} \frac{u(x + h e_i) - u(x)}{h} - v_i(x) = \frac{1}{h} \int_0^h \bigl(v_i(x + t e_i) - v_i(x)\bigr) \, d\mathcal{L}^1(t). \end{align*} The absolute value is bounded by $\sup_{|t| \le |h|} |v_i(x + t e_i) - v_i(x)|$. Since $v_i$ is the uniform limit of the continuous functions $\partial_{x_i} u_m$, it is continuous. As $h \to 0$, the supremum tends to $0$ by continuity. Therefore $\partial_{x_i} u(x) = v_i(x)$. [/guided] [/step] [step:Prove completeness of $C^{k,\alpha}(\overline{\Omega})$ by induction on $k$] Let $(f_m)$ be a Cauchy sequence in $C^{k,\alpha}(\overline{\Omega})$. The norm definition implies that for every multi-index $\beta$ with $|\beta| \le k$, the sequence $(D^\beta f_m)$ is Cauchy in the uniform norm, so there exist continuous functions $g_\beta: \overline{\Omega} \to \mathbb{R}$ with: \begin{align*} D^\beta f_m \to g_\beta \quad \text{uniformly on } \overline{\Omega}. \end{align*} For $|\beta| = k$, the sequence $(D^\beta f_m)$ is also Cauchy in the $C^{0,\alpha}$ seminorm. By the completeness of $C^{0,\alpha}(\overline{\Omega})$ established in the first step: \begin{align*} D^\beta f_m \to g_\beta \quad \text{in } C^{0,\alpha}(\overline{\Omega}), \qquad g_\beta \in C^{0,\alpha}(\overline{\Omega}). \end{align*} Define $f := g_0$. We verify $D^\beta f = g_\beta$ for all $|\beta| \le k$ by induction on $|\beta|$. The base case $|\beta| = 0$ holds by definition. For the inductive step, suppose $D^\gamma f = g_\gamma$ for all $|\gamma| \le j < k$. For $|\beta| = j$ and any $i \in \{1, \ldots, n\}$, set $u_m := D^\beta f_m$. Then $u_m \to g_\beta$ uniformly and $\partial_{x_i} u_m = D^{\beta + e_i} f_m \to g_{\beta + e_i}$ uniformly. By the derivative-closure lemma, $\partial_{x_i} g_\beta = g_{\beta + e_i}$. By the inductive hypothesis $g_\beta = D^\beta f$, so $\partial_{x_i}(D^\beta f) = D^{\beta + e_i} f = g_{\beta + e_i}$. [/step] [step:Conclude that $f \in C^{k,\alpha}(\overline{\Omega})$ with norm convergence] The induction shows $f \in C^k(\overline{\Omega})$ with $D^\beta f = g_\beta$ for all $|\beta| \le k$. For $|\beta| = k$, $D^\beta f = g_\beta \in C^{0,\alpha}(\overline{\Omega})$, so $f \in C^{k,\alpha}(\overline{\Omega})$. For the norm convergence: for $|\beta| < k$, $\|D^\beta f_m - D^\beta f\|_{L^\infty} = \|D^\beta f_m - g_\beta\|_{L^\infty} \to 0$; for $|\beta| = k$, $\|D^\beta f_m - D^\beta f\|_{C^{0,\alpha}} \to 0$ by the $C^{0,\alpha}$ convergence. Summing over all multi-indices: $\|f_m - f\|_{C^{k,\alpha}} \to 0$. [/step]