The space $C^1[a,b]$ is complete. 
Denote $C^1[a,b]$ to be the set of all $C^1$ functions $f: [a,b] \to \Bbb R$. Prove $C^1[a,b]$ is complete under $\| f \| = \max |f| + \max |f'|$.

So my idea is this
(1) We know $\| f_n - f_m \| < \epsilon$ for $n,m > N$ for some $N$.
(2) So then the quantities, $\max |f_n - f_m|$ and $\max |f_n' - f'_m|$ can be made as small as possible (since their sum is, and they are both nonnegative).
(3) This also means then the quantities $|f_n - f_m|$ and $|f_n' - f_m'|$ can be made as small as we possible as well.
(4) The sequence in (3) are real numbers for each choice of $x \in [a,b]$, so this means they converge (because Cauchy sequence converges in $\Bbb R$). 
Is this off or not?
 A: Let $\{f_n\}$ be a Cauchy sequence in $C^1([a,b])$.  Define $f(x):= \lim_{n\rightarrow \infty} f_n(x)$.  Since for each $x$ we have $$|f_m(x) - f_n(x)| \leq \max_{a\leq x \leq b}|f_m(x) - f_n(x)| \leq \|f_m - f_n\|$$ (and this last expression is Cauchy in $\mathbb{R}$), the limit exists and is well-defined.
Considering the second inequality, $\max_{a\leq x \leq b}|f_m(x) - f_n(x)| \leq \|f_m - f_n\|$, we can clearly choose $n$ and $m$ large enough that the two functions differ by at most $\epsilon$ on the interval.  Since $\epsilon>0$ is arbitrary, the convergence $f_n \rightarrow f$ is uniform.  In the same way, for $$g(x) := \lim_{n\rightarrow \infty}f'_n(x),$$ the derivatives converge uniformly: $f'_n \rightarrow g$.
Now, since the $f_n$ and the $f'_n$ are continuous functions, uniform convergence implies that their limits, $f$ and $g$, are also continuous functions.  If it happens that $f' = g$, then $f$ will be a continuous function on $[a,b]$ with a continuous derivative, i.e. $f \in C^1([a,b])$.
The most elementary way (that I can think of) to show that $g = f'$ is to interchange the following limits:
$$ f'(x) = \lim_{t \rightarrow 0} {f(x+t) - f(x) \over t} = \lim_{t\rightarrow 0} \lim_{n\rightarrow \infty} {f_n(x+t) - f_n(x) \over t}$$ to get
$$ \lim_{n\rightarrow \infty} \lim_{t\rightarrow 0} {f_n(x+t) - f_n(x) \over t} = \lim_{n\rightarrow \infty} f_n'(x) = g(x). $$
Since for any $n$ the limit $\lim_{t\rightarrow 0} {f_n(x+t) - f_n(x) \over t}$ exists, and for any $t>0$, the limit $\lim_{n\rightarrow \infty} {f_n(x+t) - f_n(x) \over t}$ exists, the two limits are interchangeable.  This proves that $g = f'$ and so $f \in C^1([0,1])$.
A: I'm assuming you know that $C[a,b]$ is complete. Therefore, if $\{ f_{n} \}_{n=1}^{\infty}$ is a Cauchy sequence in $C^{1}[a,b]$, then $\{ f_{n} \}$ and $\{ f_{n}' \}$ converge in $C[a,b]$ to continuous functions $f$ and $g$, respectively. By the fundamental theorem of Calculus,
$$
       f_{n}(x)=f_{n}(a)+\int_{a}^{x}f_{n}'(t)\,dt,\;\;\; n \ge 1, \;\; a\le x\le b.
$$
Because the convergence of $\{ f_{n} \}$ and $\{ f_{n}'\}$ is uniform, then
$$
                  f(x) = f(a)+\int_{a}^{x}g(t)\,dt,\;\;\; a\le x\le b.
$$
Therefore, $f \in C^{1}[a,b]$ with $f'=g$, which shows that
$$
     \lim_{n}\|f_{n}-f\|_{C^{1}[a,b]}=\lim_{n}(\|f_{n}-f\|_{C}+\|f_{n}'-f'\|_{C})=0.
$$
