vector space of continuously differentiable functions is complete regarding a specific norm Consider $C^1[a, b] = \{f: [a, b] \to \mathbb{C}\mid f\text{ continuously differentiable}\}$.
Consider the following norm:
$$\|f\|_{C^1} = \|f\|_\infty + \|f'\|_\infty$$
Now, it needs to be shown that $C^1[a, b]$ is complete with regard to the $\|\cdot\|_{C^1}$-Norm (using the definition for complete that a metric space $X$ is complete $\Longleftrightarrow$ every Cauchy-series in $X$ also converges in $X$).
I know that $C^1[a, b]$ is not complete in respect to the $\infty$-norm. I guess these special Cauchy-sequences of functions that do not converge in $C^1[a, b]$ using the $\infty$-norm simply aren't Cauchy-sequences in respect to the $C^1$-norm. But how can this be proven? Thanks in advance.
Edit: The linked thread only asks for why $||f||_{C^1}$ is a norm, and doesn't deal with whether or not $C^1[a, b]$ is complete regarding this norm at all, which is the specific topic of this thread.
 A: A Cauchy sequence $(f_k)$ with respect to the norm $\|\cdot\|_{C^1}$ obviously is a Cauchy sequence with respect to $\|\cdot\|_\infty$, and the sequence $(f_k^\prime)$ as well is a Cauchy sequence with respect to that latter norm. Therefore we have a uniformly convergent sequence of differentiable functions such that the derivatives are uniformly convergent as well. Under these conditions one knows by a theorem attributed to Dini (?) that the uniform limit $f$ of the sequence $(f_k)$ is differentiable  and
$f^\prime=\lim\limits_{k\rightarrow\infty}f_k^\prime$
the limit on the right hand side being the uniform one. In particular this limit is continous.
A: Consider the fact that if $(f_n)$ is a Cauchy-series in $C^1[a, b]$ with respect to the $||.||_{C^1}$-norm, then it also a Cauchy-series with respect to $||.||_\infty$, and the series of functions $(f_n')$ is a Cauchy-series with respect to the supremum norm.
Now, consider the space $C[a, b] = \{f: [a, b] \to \mathbb{C}, f$ continuous$\}$. It's easy to proof that $C^1[a, b]$ is a subvector space of $C[a, b]$. I'll use the fact that $C[a, b]$ is complete regarding the supremum norm.
Because $C^1[a, b]$ is a subvector space, $(f_n)$ is also a Cauchy series in $C[a, b]$ with respect to the supremum norm. But because $C[a, b]$ is complete, there exists an $f \in C[a, b]$ with $f_n \to f$ regarding the supremum norm. f is continuous per definition of $C[a, b]$, so it only needs to be shown that f has a continuous derivative.
Now consider the series of continuous functions $(f_n')$, which exists because of the definition of $C^1[a, b]$. As such, it can be considered an element of $C[a, b]$ aswell and therefore converges against a continuous $f' \in C[a, b]$. Now using the fundamental theorem of calculus, f' must be derivative of f and therefore f is a continuous function with a contininuous derivative, and therefore $f \in C[a, b]$.
Is this proof correct as it stands? Thanks for any feedback.
