Looking for a proof of the completeness of $C^{n}[0,1]$. Can someone refer me to a verification of the completeness of $C^{n}[0,1]$ under the norm $\|f\|_{C^{n}} = \sum_{k=0}^{n}\|f^{(k)}\|_{\infty}$?
I tried to follow the same approach as the standard proof the $C[0,1]$ is complete under the supremum norm, but I run into the problem that the limits I end up with do not necessarily define a function in $C^n[0,1]$.
 A: Let me do it for $C^1([0,1])$ then you can adapt the argument for $C^n ([0,1])$.
We want to show that $C^1([0,1])$ is complete with respect to $\|f \| := \| f \|_\infty + \| f^\prime \|_\infty$. So let $f_n$ be a Cauchy sequence. Then
(i) $f_n$ converges pointwise to some $f: [0,1] \to \mathbb{R}$
For this you observe that $f_n(x_0)$ is a Cauchy sequence in $\mathbb{R}$ for all $x_0$ in $[0,1]$. $\mathbb{R}$ is complete so $f(x_0)$ is also in $\mathbb{R}$. By the same argument $f_n^\prime$ converges pointwise to some $g(x) := \lim_{n \to \infty} f_n(x)$.
(ii) Next you want to show that the pointwise limit function $f$ is in $C^1([0,1])$. Here you want to show two things: that $f$ is continuous and that it's differentiable. 
Let's do continuous first: Let $\varepsilon > 0$ and $x_0 \in [0,1]$. Then $|f(x_0) - f(x)| \leq |f(x_0) - f_n(x_0)| + |f_n(x_0) - f_n(x)| + |f(x) - f_n(x)|$
Now you can argue that each term can be made less than $\frac{\varepsilon}{3}$. For $n$ big enough you get $|f(x_0) - f_n(x_0)| < \frac{\varepsilon}{3}$ and $|f(x) - f_n(x)| < \frac{\varepsilon}{3}$ and because $f_n$ is continuous you can find a $\delta$ such that $|x-x_0| < \delta$ implies $|f_n(x_0) - f_n(x)| < \frac{\varepsilon}{3}$. 
Use the same argument to show that $g = \lim_{n \to \infty} f^\prime_n$ is continuous.
Next you want to show that $f$ is differentiable. In particular you claim that $f^\prime = \lim_{n \to \infty} f^\prime_n = g$. To do this you can use the fundamental theorem of calculus to rewrite $f_n$ as $f_n(x) = f_n(0) + \int_0^x f^\prime_n (t) dt$. 
Also using the fundamental theorem your claim translates into $f(x) = f(0) + \int_0^x \lim_{n \to \infty} f_n^\prime(t) dt$. To show this you want to  show $|f(x) - f(0) - \int_0^x \lim_{n \to \infty} f_n^\prime(t) dt| < \varepsilon$ as follows:
Choose an $n$ such that $\| f_n - f\| < \varepsilon$ then you have $\| f_n - f\|_\infty < \varepsilon$ and $\| f_n^\prime - g\|_\infty < \varepsilon$. Then use the triangle inequality:
$$\begin{align} 
&|f(x) - f(0) - \int_0^x \lim_{n \to \infty} f_n^\prime(t) dt| \\
&\leq |f(x) - f_n(x)| + |f(0) - f_n(0)| + |f_n(x) - f_n(0) - \int_0^x f_n^\prime(t) dt | + | \int_0^x f_n^\prime(t) - \lim_{n \to \infty} f_n^\prime(t) dt| 
\end{align}$$
The first two terms can be made less than $\frac{\varepsilon}{4}$ for $n$ large enough, the third term is $0$ and for last term we can choose $n$ so large that $| f_n^\prime(t) - \lim_{n \to \infty} f_n^\prime(t)| < \frac{\varepsilon}{4}$ and hence
$$ \left | \int_0^x f_n^\prime(t) - \lim_{n \to \infty} f_n^\prime(t) dt \right |  \leq 
\int_0^x | f_n^\prime(t) - \lim_{n \to \infty} f_n^\prime(t)| dt \leq
\int_0^x \frac{\varepsilon}{4} dt = x \frac{\varepsilon}{4} \leq  \frac{\varepsilon}{4}$$
(iii) The last thing you need to show is that $\| f - f_n \| \xrightarrow{n \to \infty} 0$:
$$ \| f - f_n \|  = \| f  - f_n \|_\infty + \| f^\prime - f_n^\prime \|_\infty \leq 2 \max ( \| f  - f_n \|_\infty, \| f^\prime - f_n^\prime \|_\infty )$$
So to finish the proof you want to show that $f_n$ converges in the sup norm (and by the same argument that $f^\prime_n$ does, too). Let's do it for $f_n$:
$| f(x) - f_n(x) | \leq |f(x) - f_N(x)| + |f_N(x) - f_n(x)|$
Now choose $N$ such that for $n,m\geq N$ you have $|f_N(x) - f_n(x)| < \frac{\varepsilon}{2}$ and $|f_m(x) - f_N(x)| < \frac{\varepsilon}{2}$. Then $\lim_{m \to \infty} |f_m(x) - f_N(x)| \leq \frac{\varepsilon}{2}$ for all $x$ in $[0,1]$. Hence $\| f  - f_n \|_\infty \xrightarrow{n \to \infty} 0$.
You can find a proof of this for the norm $\|f\| := \max(\|f\|_\infty, \|f^\prime\|_\infty)$ on $C^1([0,1])$ and for the norm $\|f\| := \max_{\operatorname{deg}{(\partial_\alpha)} \leq k} \|\partial_\alpha f\|_\infty$ on $C^n_b(\Omega)$ where $\Omega$ is an open set in $\mathbb{R}^d$ here starting on page 35.
Hope this helps.
