$C^k$ Space

Let $\Omega \subset \mathbb{R}^d$ be an open set and $C^k(\bar{\Omega})$ be the set of all bounded functions $u : \Omega \to \mathbb{R}$ whose partial derivatives $D^a u$ for $0 \leq |a| \leq k$ are continuous on $\Omega$ and can be continuously extended to $\bar{\Omega}$ in a bounded way. (Here $D^a u$ represents partial derivative where $a=(a_1,a_2,\cdots,a_d)$ is a multi-index). Define a norm on $C^k(\bar{\Omega})$ by $\|u\|_{C^k} = \underset{0 \leq |a| \leq k}{\max} \underset{x \in \bar{\Omega}} \sup|D^au(x)|$. Then $(C^k(\bar{\Omega}),\|\cdot\|_{C^k})$ is a Banach Space.

Note that $\Omega$ can be unbounded here. The norm is still meaningful as the functions themselves are bounded. The following Mean-Value theorem in higher dimensions is used in the proof. Let $f : \Omega \to \mathbb{R}$ be differentiable (e.g. all partial derivatives of $f$ are continuous). Take two points $x,y\in \Omega$ and assume that the line joining the two points is also contained in $\Omega$. Define $g : [0,1] \to \mathbb{R}$ by $g(t) = f((1-t)x+ty)$. Now, clearly $g$ is differentiable and hence by the Mean Value Theorem, $g(1) - g(0) = g^\prime(c)$ where $c\in (0,1)$. Hence, $f(y) - f(x) = \langle \nabla f((1-c)x+cy),(y-x)\rangle$.

Assume now that $\{u_n\}$ is a cauchy-sequence and let $0 \leq |a| < k$. Also, let $\lim_{n\to \infty} u_n = v^0$. This limit exists and belongs to $C(\bar{\Omega})$. Similarly, let $\lim_{n\to \infty} D^au_n = v^a$. We need to show that $D^av^0 = v^a$.

As $\{u_n\}$ is cauchy, for a given $\epsilon > 0$, there exists $N$ such that $\underset{x \in \bar{\Omega}} \sup |\frac{\partial}{\partial x_i} D^a u_n (x) - \frac{\partial}{\partial x_i} D^a u_m (x)| < \epsilon$ for all $m,n \geq N$. Choose $f = D^au_n - D^au_m : \Omega \to \mathbb{R}$ in the previous paragraph. Then \begin{array} a\frac{\|(D^au_n(y) - D^au_m(y)) - (D^au_n(x) - D^au_m(x))\|}{\|y - x\|} &\leq \|\nabla f((1-c)x+cy)\| \\ &\leq \epsilon \sqrt{d} \end{array}

Now let $x\in \Omega$ and $\phi_n^i : \Omega_i \to \mathbb{R}$ given by $\phi_n^i(y) = \frac{\|D^au_n(x_1,x_2,\cdots,x_{i-1},y,x_{i+1},\cdots,x_d)- D^au_n(x_1,x_2,\cdots,x_{i-1},x_i,x_{i+1},\cdots,x_d)\|} {|y - x_i|}$. Here $\Omega_i \subset \mathbb{R}$ is an open ball around $x_i$ minus the point $x_i$ From the above we get that $\phi_n^i$ is uniformly-cauchy and hence converges uniformly in $\Omega_i$. Let $\lim_{n\to\infty} \phi_n^i(y)= \phi^i(y)$ in $\Omega_i$. Then, $\lim_{y \to x_i}\lim_{n \to \infty} \phi_n^i(y)= \lim_{n \to \infty} \lim_{y \to x_i} \phi_n^i(y)$ (i.e. the limits can be interchanged because of uniform convergence, see Rudin's PrinMathAnalysis Theorem 7.11) if $\lim_{y \to x_i} \phi_n^i(y)$ exists. In this case, this is true. Let $b=(a_1,a_2,\cdots,a_i+1,\cdots,a_n)$, then we get using recursion, \[D^bu(x)= \lim_{n\to \infty} D^{b}u_n(x)\]