Completeness of space of $k$-times differentiable functions from $\Bbb R^n$ to $\Bbb R$ I'm having trouble solving exercise 7.5 from Otto Forsters Analysis 2 book:
Let $\displaystyle U \subset \mathbb{R}^n$ be an open subset and $C^k_b(U)$ the set of all $k$-times continuously differentiable functions $f: U \to \mathbb{R}$ with bounded differentials $D^\alpha f \; \forall\; \alpha \in \mathbb{N}^n$ with $\vert \alpha \vert = \sum_{i=1}^{n} \alpha_i = k $.
And $\vert\vert \cdot \vert\vert_{k} : C^k_b(U) \to \mathbb{R} $
with 
$\vert\vert f \vert\vert_{k} := \sum_{\vert \alpha \vert = k} 
\frac{1}{\alpha!} \sup\{ \vert D^\alpha f(x) \vert, x \in U\} $.
Note that $\alpha! := \prod_{i=1}^n \alpha_i!$ and
and $D^\alpha f := 
\frac{\partial^{\vert\alpha \vert}f}
{
\partial x_1^{\alpha_1} \cdots 
\partial x_n^{\alpha_n}
}$
a) I have shown that $\vert\vert \cdot \vert\vert_{k} : C^k_b(U) \to \mathbb{R} $ is a norm
b) Show for $f, g, \in C^k_b(U)$: $\vert\vert fg \vert\vert_{k} \leq
\vert\vert f \vert\vert_{k} \vert\vert g \vert\vert_{k}$
c) Show that $( C^k_b(U), \vert \vert \cdot \vert \vert_{k})$ is complete.
I'd be happy about any hints to solutions for b) and c).

Somewhere in the book you find the product rule:
$$D^\alpha(fg)=\sum_{\beta\le\alpha}\frac{\alpha!}{\beta!(\alpha-\beta)!}D^\beta fD^{\alpha-\beta}g.$$
Let's write $\sup|\phi|=||\phi||_0$. It's clear that $||\phi\psi||_0\le||\phi||_0||\psi||_0$, hence 
$$||D^\alpha(fg)||_0\le\sum_{\beta\le\alpha}\frac{\alpha!}{\beta!(\alpha-\beta)!}||D^\beta f||_0||D^{\alpha-\beta}g||_0.$$Insert that inequality into the definition of $||fg||_k$. Also write out the product $||f||_k||g||_k$. Something magic happens, and you get (b).
For (c), say $(f_n)$ is Cauchy in $C^k$. Then $(D^\alpha f_n)$ is uniformly Cauchy for all $\alpha$ with $|\alpha|\le k$. So for all such $\alpha$ there exists $g_\alpha$ so that $D^\alpha f_n\to g_\alpha$ uniformly. Let $f=g_0$. Now some theorem from advanced calculus shows that $g_\alpha=D^\alpha f$ and you're done... (why?)
