[proofplan] We first prove completeness for $C^{0,\gamma}(X;\mathbb R^m)$ directly from the completeness of $\mathbb R^m$: a [Cauchy sequence](/page/Cauchy%20Sequence) in the Hölder norm is uniformly Cauchy, hence has a uniform pointwise limit. The Hölder estimates pass to the limit pointwise, and the original Cauchy property then gives convergence in the full Hölder norm. For $C^{k,\gamma}(\overline U;\mathbb R^m)$, we apply the $C^{0,\gamma}$ result to the highest-order derivative arrays and the uniform part of the norm to all lower-order derivatives. The only compatibility issue is to prove that the limiting derivative arrays are derivatives of one limiting function; this follows locally on line segments in $U$ from the integral identity for one-variable restrictions of $C^1$ functions. [/proofplan]

[step:Prove completeness of $C^{0,\gamma}(X;\mathbb R^m)$ from a Cauchy sequence]Let $(f_j)_{j\in\mathbb N}$ be a Cauchy sequence in $C^{0,\gamma}(X;\mathbb R^m)$ with respect to $\|\cdot\|_{C^{0,\gamma}(X)}$. Thus, for every $\varepsilon>0$, there exists $N\in\mathbb N$ such that for all $j,l\ge N$, \begin{align*} \|f_j-f_l\|_{\infty}+[f_j-f_l]_{C^{0,\gamma}(X)}<\varepsilon. \end{align*} For each $x\in X$, the sequence $(f_j(x))_{j\in\mathbb N}$ is Cauchy in $\mathbb R^m$, hence converges. Define the map \begin{align*} f:X&\to\mathbb R^m \end{align*} by declaring $f(x):=\lim_{j\to\infty}f_j(x)$ for each $x\in X$. Fix $\varepsilon>0$ and choose $N\in\mathbb N$ as above. Passing to the limit $l\to\infty$ in the inequality $|f_j(x)-f_l(x)|\le\varepsilon$ gives \begin{align*} |f_j(x)-f(x)|\le\varepsilon \end{align*} for all $j\ge N$ and all $x\in X$. Hence $f_j\to f$ uniformly on $X$, and $f$ is bounded because $f_N$ is bounded and $\|f-f_N\|_\infty\le\varepsilon$. Now let $x,y\in X$ with $x\ne y$. For $j\ge N$ and $l\ge N$, \begin{align*} |f_j(x)-f_l(x)-f_j(y)+f_l(y)|\le\varepsilon |x-y|^\gamma. \end{align*} Letting $l\to\infty$ gives \begin{align*} |f_j(x)-f(x)-f_j(y)+f(y)|\le\varepsilon |x-y|^\gamma. \end{align*} Taking the supremum over all distinct $x,y\in X$ yields \begin{align*} [f_j-f]_{C^{0,\gamma}(X)}\le\varepsilon. \end{align*} Therefore $\|f_j-f\|_{C^{0,\gamma}(X)}\le 2\varepsilon$ for all $j\ge N$. In particular $f_j\to f$ in $C^{0,\gamma}(X;\mathbb R^m)$. It remains only to check that $f\in C^{0,\gamma}(X;\mathbb R^m)$. Choose $j\in\mathbb N$ with $f_j-f\in C^{0,\gamma}(X;\mathbb R^m)$. Since $f=f_j+(f-f_j)$ and both summands are bounded and Hölder continuous with exponent $\gamma$, the function $f$ is bounded and Hölder continuous with exponent $\gamma$. Thus $C^{0,\gamma}(X;\mathbb R^m)$ is complete.[/step]

[guided]We begin with a Cauchy sequence \begin{align*} (f_j)_{j\in\mathbb N} \end{align*} in the norm \begin{align*} \|h\|_{C^{0,\gamma}(X)}=\|h\|_\infty+[h]_{C^{0,\gamma}(X)}. \end{align*} The first term of this norm tells us that the sequence is uniformly Cauchy. More precisely, for every $\varepsilon>0$, there is $N\in\mathbb N$ such that whenever $j,l\ge N$, \begin{align*} \|f_j-f_l\|_\infty+[f_j-f_l]_{C^{0,\gamma}(X)}<\varepsilon. \end{align*} In particular, for every fixed $x\in X$, \begin{align*} |f_j(x)-f_l(x)|<\varepsilon. \end{align*} Since $\mathbb R^m$ is complete, the sequence $(f_j(x))_{j\in\mathbb N}$ has a limit in $\mathbb R^m$. This lets us define a candidate limit map \begin{align*} f:X&\to\mathbb R^m \end{align*} by \begin{align*} f(x):=\lim_{j\to\infty}f_j(x). \end{align*} The pointwise definition alone is not enough; we need convergence in the Hölder norm. Fix $\varepsilon>0$ and choose $N$ as above. If $j\ge N$, then for every $l\ge N$ and every $x\in X$, \begin{align*} |f_j(x)-f_l(x)|\le\varepsilon. \end{align*} Letting $l\to\infty$ and using the definition of $f(x)$ gives \begin{align*} |f_j(x)-f(x)|\le\varepsilon \end{align*} for all $x\in X$. Taking the supremum over $x\in X$ proves \begin{align*} \|f_j-f\|_\infty\le\varepsilon. \end{align*} Now we use the Hölder part of the Cauchy condition. For distinct $x,y\in X$ and $j,l\ge N$, the estimate \begin{align*} [f_j-f_l]_{C^{0,\gamma}(X)}<\varepsilon \end{align*} means exactly that \begin{align*} |f_j(x)-f_l(x)-f_j(y)+f_l(y)|\le\varepsilon |x-y|^\gamma. \end{align*} Again let $l\to\infty$. Since $f_l(x)\to f(x)$ and $f_l(y)\to f(y)$ in $\mathbb R^m$, the Euclidean norm is continuous, so \begin{align*} |f_j(x)-f(x)-f_j(y)+f(y)|\le\varepsilon |x-y|^\gamma. \end{align*} After dividing by $|x-y|^\gamma$ and taking the supremum over all distinct $x,y\in X$, we obtain \begin{align*} [f_j-f]_{C^{0,\gamma}(X)}\le\varepsilon. \end{align*} Together with the uniform estimate, this gives convergence in the full Hölder norm: \begin{align*} \|f_j-f\|_{C^{0,\gamma}(X)}\le 2\varepsilon \end{align*} for all $j\ge N$. Finally, the limit belongs to the same space. Choose one index $j$ for which $f_j-f\in C^{0,\gamma}(X;\mathbb R^m)$, which is possible by the convergence just proved. Since \begin{align*} f=f_j+(f-f_j), \end{align*} and the sum of two bounded Hölder maps with exponent $\gamma$ is again bounded and Hölder with exponent $\gamma$, the map $f$ lies in $C^{0,\gamma}(X;\mathbb R^m)$. Thus every Cauchy sequence converges in the space, so $C^{0,\gamma}(X;\mathbb R^m)$ is Banach.[/guided]

[step:Construct the limiting derivative fields for a Cauchy sequence in $C^{k,\gamma}(\overline U;\mathbb R^m)$] Let $(f_j)_{j\in\mathbb N}$ be Cauchy in $C^{k,\gamma}(\overline U;\mathbb R^m)$. If $k=0$, this is exactly the result already proved with $X=\overline U$. Assume $k\ge1$. For each multi-index $\alpha\in\mathbb N_0^n$ with $|\alpha|\le k$, the sequence $(D^\alpha f_j)_{j\in\mathbb N}$ is Cauchy in the sup norm on $\overline U$. Hence there exists a bounded map \begin{align*} g_\alpha:\overline U&\to\mathbb R^m \end{align*} such that \begin{align*} \|D^\alpha f_j-g_\alpha\|_\infty\to0. \end{align*} Because each $D^\alpha f_j$ is continuous on $\overline U$ and a uniform limit of continuous maps into $\mathbb R^m$ is continuous, each map $g_\alpha:\overline U\to\mathbb R^m$ is continuous. For each multi-index $\alpha$ with $|\alpha|=k$, the same sequence is Cauchy in $C^{0,\gamma}(\overline U;\mathbb R^m)$, so the first part of the proof gives \begin{align*} g_\alpha\in C^{0,\gamma}(\overline U;\mathbb R^m) \end{align*} and \begin{align*} \|D^\alpha f_j-g_\alpha\|_{C^{0,\gamma}(\overline U)}\to0. \end{align*} Set $f:=g_0$, where $0=(0,\dots,0)$ is the zero multi-index. [/step]

[step:Verify that the limiting fields are the derivatives of the limiting function]We prove that, for every multi-index $\alpha$ with $|\alpha|\le k$, the restriction $g_\alpha|_U$ is equal to $D^\alpha f$ on $U$. It is enough to prove the following compatibility statement: if $|\alpha|\le k-1$ and $e_i\in\mathbb N_0^n$ denotes the multi-index with $1$ in the $i$th coordinate and $0$ elsewhere, then \begin{align*} \partial_{x_i} g_\alpha=g_{\alpha+e_i} \end{align*} on $U$. Fix such an $\alpha$, fix $i\in\{1,\dots,n\}$, and fix $a\in U$. Since $U$ is open, choose $r>0$ such that $B(a,r)\subset U$. Let $h\in\mathbb R$ satisfy $0<|h|<r$. The line segment from $a$ to $a+he_i$ is contained in $B(a,r)\subset U$. For each $j\in\mathbb N$, the function \begin{align*} \phi_j:[0,1]&\to\mathbb R^m \end{align*} defined by \begin{align*} \phi_j(t):=D^\alpha f_j(a+the_i) \end{align*} is continuously differentiable, and its derivative is \begin{align*} \phi_j'(t)=hD^{\alpha+e_i}f_j(a+the_i). \end{align*} Let $\mathcal L^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $[0,1]$. By the [fundamental theorem of calculus](/theorems/632) for continuously differentiable functions on $[0,1]$, \begin{align*} D^\alpha f_j(a+he_i)-D^\alpha f_j(a)=h\int_0^1 D^{\alpha+e_i}f_j(a+the_i)\,d\mathcal L^1(t). \end{align*} Passing to the limit $j\to\infty$ in the two endpoint terms uses [uniform convergence](/page/Uniform%20Convergence) of $D^\alpha f_j$ to $g_\alpha$. For the integral term, the estimate \begin{align*} \left|\int_0^1 \left(D^{\alpha+e_i}f_j(a+the_i)-g_{\alpha+e_i}(a+the_i)\right)\,d\mathcal L^1(t)\right|\le \|D^{\alpha+e_i}f_j-g_{\alpha+e_i}\|_\infty \end{align*} shows that the integrals converge to the integral of $g_{\alpha+e_i}$ with respect to $\mathcal L^1$ by uniform convergence on $[0,1]$. Therefore \begin{align*} g_\alpha(a+he_i)-g_\alpha(a)=h\int_0^1 g_{\alpha+e_i}(a+the_i)\,d\mathcal L^1(t). \end{align*} Since $g_{\alpha+e_i}$ is continuous on $\overline U$, it is continuous at $a$. Dividing by $h$ and letting $h\to0$ yields \begin{align*} \lim_{h\to0}\frac{g_\alpha(a+he_i)-g_\alpha(a)}{h}=g_{\alpha+e_i}(a). \end{align*} Thus $\partial_{x_i}g_\alpha(a)=g_{\alpha+e_i}(a)$. Since $a\in U$ was arbitrary, the compatibility statement holds on $U$. Starting from $g_0=f$, repeated application of this compatibility statement over the entries of $\alpha$ gives \begin{align*} D^\alpha f=g_\alpha \end{align*} on $U$ for every $|\alpha|\le k$. Therefore $f\in C^k(U;\mathbb R^m)$, and every derivative $D^\alpha f$ with $|\alpha|\le k$ extends continuously to $\overline U$ by $g_\alpha$.[/step]

[guided]The possible obstruction is that the uniform limits $g_\alpha$ might be unrelated functions rather than the derivatives of a single function. We rule this out by proving the first-order compatibility relation \begin{align*} \partial_{x_i}g_\alpha=g_{\alpha+e_i} \end{align*} whenever $|\alpha|\le k-1$. Fix a multi-index $\alpha$ with $|\alpha|\le k-1$, fix a coordinate direction $i\in\{1,\dots,n\}$, and let $e_i$ denote the multi-index with $1$ in the $i$th coordinate and $0$ elsewhere. Choose a point $a\in U$. Because $U$ is open, there exists $r>0$ such that \begin{align*} B(a,r)\subset U. \end{align*} If $h\in\mathbb R$ and $0<|h|<r$, then the whole segment \begin{align*} \{a+the_i:0\le t\le1\} \end{align*} lies in $B(a,r)$, so all derivatives below are evaluated inside $U$. For each $j\in\mathbb N$, define the one-variable map \begin{align*} \phi_j:[0,1]&\to\mathbb R^m \end{align*} by \begin{align*} \phi_j(t):=D^\alpha f_j(a+the_i). \end{align*} Since $f_j\in C^{k,\gamma}(\overline U;\mathbb R^m)$ and $|\alpha|\le k-1$, the derivative $D^\alpha f_j$ is continuously differentiable along this segment. By the chain rule applied to the affine map $t\mapsto a+the_i$, we have \begin{align*} \phi_j'(t)=hD^{\alpha+e_i}f_j(a+the_i). \end{align*} Let $\mathcal L^1$ denote one-dimensional Lebesgue measure on $[0,1]$. The fundamental theorem of calculus for continuously differentiable functions then gives \begin{align*} D^\alpha f_j(a+he_i)-D^\alpha f_j(a)=h\int_0^1 D^{\alpha+e_i}f_j(a+the_i)\,d\mathcal L^1(t). \end{align*} Now pass to the limit in this identity. The left-hand side converges to \begin{align*} g_\alpha(a+he_i)-g_\alpha(a) \end{align*} because $D^\alpha f_j\to g_\alpha$ uniformly on $\overline U$. The integrand on the right-hand side converges uniformly on $[0,1]$ to the map \begin{align*} t\mapsto g_{\alpha+e_i}(a+the_i), \end{align*} because $D^{\alpha+e_i}f_j\to g_{\alpha+e_i}$ uniformly on $\overline U$. Uniform convergence justifies passing the limit through the integral with respect to $\mathcal L^1$, so \begin{align*} g_\alpha(a+he_i)-g_\alpha(a)=h\int_0^1 g_{\alpha+e_i}(a+the_i)\,d\mathcal L^1(t). \end{align*} Divide by $h$: \begin{align*} \frac{g_\alpha(a+he_i)-g_\alpha(a)}{h}=\int_0^1 g_{\alpha+e_i}(a+the_i)\,d\mathcal L^1(t). \end{align*} Because $g_{\alpha+e_i}$ is continuous at $a$, the integrand converges uniformly in $t\in[0,1]$ to $g_{\alpha+e_i}(a)$ as $h\to0$. Hence the right-hand side converges to \begin{align*} \int_0^1 g_{\alpha+e_i}(a)\,d\mathcal L^1(t)=g_{\alpha+e_i}(a). \end{align*} Therefore \begin{align*} \partial_{x_i}g_\alpha(a)=g_{\alpha+e_i}(a). \end{align*} Since $a\in U$ was arbitrary, this holds throughout $U$. Starting from $g_0=f$, repeated use of this identity shows that every $g_\alpha$ is the corresponding derivative $D^\alpha f$ on $U$. Thus the limiting data are compatible, and $f$ is a $C^k$ map on $U$ whose derivatives of order at most $k$ extend continuously to $\overline U$ by the functions $g_\alpha$.[/guided]

[step:Conclude convergence in the full $C^{k,\gamma}$ norm] For every multi-index $\alpha$ with $|\alpha|\le k$, we have already shown \begin{align*} \|D^\alpha f_j-D^\alpha f\|_\infty=\|D^\alpha f_j-g_\alpha\|_\infty\to0. \end{align*} For every multi-index $\alpha$ with $|\alpha|=k$, the first part of the proof applied on $X=\overline U$ gives \begin{align*} [D^\alpha f_j-D^\alpha f]_{C^{0,\gamma}(\overline U)}=[D^\alpha f_j-g_\alpha]_{C^{0,\gamma}(\overline U)}\to0. \end{align*} Therefore \begin{align*} \|f_j-f\|_{C^{k,\gamma}(\overline U)}\to0. \end{align*} Also, each $D^\alpha f$ with $|\alpha|=k$ is Hölder continuous on $\overline U$, because $g_\alpha\in C^{0,\gamma}(\overline U;\mathbb R^m)$. Hence $f\in C^{k,\gamma}(\overline U;\mathbb R^m)$. Every Cauchy sequence in $C^{k,\gamma}(\overline U;\mathbb R^m)$ converges in that norm, so $C^{k,\gamma}(\overline U;\mathbb R^m)$ is a [Banach space](/page/Banach%20Space). [/step]