**Step 1: Completeness of the Base Space $C^{0,\alpha}(\overline{\Omega})$** We first establish the result for $k=0$. Let $\{g_m\}_{m \in \mathbb{N}}$ be a Cauchy sequence in $C^{0,\alpha}(\overline{\Omega})$. Since the norm includes the uniform norm $\|g_m\|_{L^\infty}$, the sequence is Cauchy in $C^0(\overline{\Omega})$. Because $C^0(\overline{\Omega})$ is a Banach space, there exists a continuous function $g: \overline{\Omega} \to \mathbb{R}$ such that: \begin{align*} g_m \to g \quad \text{uniformly on } \overline{\Omega} \end{align*} We now demonstrate that $g \in C^{0,\alpha}(\overline{\Omega})$ and that convergence holds in the Hölder norm. Since $\{g_m\}$ is Cauchy in the Hölder 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 $x, y \in \overline{\Omega}$ with $x \neq y$. For $m \ge M$, we take the limit as $\ell \to \infty$. Using the 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 \sup_{x \neq y} \frac{|(g_m - g_\ell)(x) - (g_m - g_\ell)(y)|}{|x - y|^\alpha} \le \varepsilon \end{align*} Taking the supremum over all distinct $x, y$: \begin{align*} [g_m - g]_{C^{0,\alpha}} \le \varepsilon \end{align*} This implies two results. First, since $g = g_m - (g_m - g)$, and both terms on the right are Hölder continuous, $g \in C^{0,\alpha}(\overline{\Omega})$. Second, $[g_m - g]_{C^{0,\alpha}} \to 0$ as $m \to \infty$. Combined with uniform convergence, $\|g_m - g\|_{C^{0,\alpha}} \to 0$. Thus, $C^{0,\alpha}(\overline{\Omega})$ is complete.

**Step 2: Derivative Closure Lemma** We establish that uniform limits of derivatives behave as expected. Let $U \subseteq \mathbb{R}^n$ be open. Suppose $\{u_m\} \subseteq C^1(U)$ satisfies: 1. $u_m \to u$ uniformly on compact subsets of $U$. 2. $\partial_i u_m \to v_i$ uniformly on compact subsets of $U$.

We show that $\partial_i u = v_i$. Fix $x \in U$ and choose $r > 0$ such that $\overline{B(x,r)} \subseteq U$. Let $h \in \mathbb{R} \setminus \{0\}$ with $|h| < r$. By the Fundamental Theorem of Calculus applied along the segment connecting $x$ and $x + h e_i$: \begin{align*} u_m(x + h e_i) - u_m(x) = \int_0^h \partial_i u_m(x + t e_i) \, d\mathcal{L}^1(t) \end{align*} Since $\partial_i u_m$ converges uniformly to $v_i$ on $\overline{B(x,r)}$, we pass to the limit as $m \to \infty$: \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 $h$, we estimate the difference quotient: \begin{align*} \left| \frac{u(x + h e_i) - u(x)}{h} - v_i(x) \right| &= \left| \frac{1}{h} \int_0^h (v_i(x + t e_i) - v_i(x)) \, d\mathcal{L}^1(t) \right| \\ &\le \sup_{t \in [0, h]} |v_i(x + t e_i) - v_i(x)| \end{align*} Since $v_i$ is the uniform limit of continuous functions ($\partial_i u_m$), $v_i$ is continuous on $U$. Therefore, as $h \to 0$, the supremum term approaches $0$. We conclude that $\partial_i u(x)$ exists and equals $v_i(x)$.

**Step 3: Completeness of $C^{k,\alpha}(\overline{\Omega})$** Let $\{f_m\}$ be a Cauchy sequence in $C^{k,\alpha}(\overline{\Omega})$. For every multi-index $\beta$ with $|\beta| \le k$, the definition of the norm implies that $\{D^\beta f_m\}$ is Cauchy in the uniform norm ($L^\infty$). Therefore, there exist continuous functions $g_\beta$ such that: \begin{align*} D^\beta f_m \to g_\beta \quad \text{uniformly on } \overline{\Omega} \end{align*} For the specific case where $|\beta| = k$, the sequence $\{D^\beta f_m\}$ is also Cauchy in the $C^{0,\alpha}$ seminorm. By the completeness established in **Step 1**, we have: \begin{align*} D^\beta f_m \to g_\beta \quad \text{in } C^{0,\alpha}(\overline{\Omega}) \end{align*} In particular, $g_\beta \in C^{0,\alpha}(\overline{\Omega})$ for $|\beta| = k$.

**Step 4: Identification of Limits** We define $f := g_0$. We must show that $D^\beta f = g_\beta$ for all $|\beta| \le k$. We proceed by induction on the order of the derivative. The base case $|\beta|=0$ is the definition $f = g_0$. Assume that for some order $j < k$, $D^\gamma f = g_\gamma$ for all $|\gamma| \le j$. Consider a multi-index $\beta$ with $|\beta| = j$ and any coordinate $i \in \{1, \dots, n\}$. Let $u_m = D^\beta f_m$. From our convergence results: \begin{align*} u_m &\to g_\beta \quad \text{uniformly on } \overline{\Omega} \\ \partial_i u_m = D^{\beta + e_i} f_m &\to g_{\beta + e_i} \quad \text{uniformly on } \overline{\Omega} \end{align*} Applying the result from **Step 2** to the interior $\Omega$, we conclude that $\partial_i g_\beta$ exists and equals $g_{\beta + e_i}$. By induction, $D^\beta f$ exists and equals $g_\beta$ for all $|\beta| \le k$.

**Step 5: Conclusion** We have established that $f \in C^k(\overline{\Omega})$ and that its derivatives $D^\beta f$ match the limits $g_\beta$. Specifically, for $|\beta|=k$, $D^\beta f = g_\beta \in C^{0,\alpha}(\overline{\Omega})$. Thus, $f \in C^{k,\alpha}(\overline{\Omega})$. Finally, the convergence $D^\beta f_m \to D^\beta f$ holds uniformly for $|\beta| < k$ and in the $C^{0,\alpha}$ norm for $|\beta| = k$. Summing these components confirms that $\|f_m - f\|_{C^{k,\alpha}} \to 0$.