I came across the following problem:
For an open set $U$ in $\mathbb{R^n}$ we define the set of all k-times continuously differentiable functions $f\colon U\rightarrow \mathbb{R}$ for which $D^\alpha f$ is uniformly bounded in $U$ for all $\alpha \in \mathbb{N_0^n}$ with $|\alpha| \leqslant k$. Define for $f\in C_b^k(U)$ :
$$\|f\|_k := \sum\limits_{|\alpha| \leqslant k} \frac{1}{\alpha!} \sup \{|D^{\alpha }f(x)| : x \in U\}.$$
I proved that this is a norm on $C_b^k$ and now I want to prove that the space ($C_b^k(U), \|.\|_k$) is complete which means each Cauchy sequence converges in this space but since this is a more abstract space than I am used to i find it hard to see how to continue.
I would appreciate some guidelines. Thank you
Sketch of proof. Suppose you have a Cauchy sequence $\{f_n\}\subset C_b^k(U)$, with respect to $\|\cdot\|_k$. Then, for every $\lvert\alpha\rvert\le k$, the sequence $\{D^\alpha f_n\}$ is uniformly Cauchy in $U$ and hence uniformly convergent to a continuous and bounded function, say $f^\alpha\in C_b(U)$ - All the $f^\alpha$'s are for the moment only continuous and bounded in $U$.
Next, we need to show that $f^0\in C_b^k(U)$ and $D^\alpha f=f^\alpha$.
This is done inductively on $\alpha$, starting in the following way:
$$ f_n\to f\quad\text{and}\quad \partial_{x_1} f_n\to f^(1,0,\ldots,0), $$ implies that $f$ is differentiable, with respect to $x_1$ and $\partial_{x_1}=f^(1,0,\ldots,0)$.