The spaces $C^k([a,b])$ of $k$ times continuously differentiable functions on $[a,b]$ with
$$||x||_{C^k}=||x||_\infty+||x^{(k)}||_\infty$$
Since it is clear that this is a normed vector space we need only show that it is complete with respect to the sup norm.
Let $(f_n)$ be a Cauchy sequence in $C^k([a,b])$. We have
$$|f_m(x)-f_n(x)| \leq ||f_m-f_n||$$
and hence $(f_n(x))$ converges for each $x \in [a,b]$ since $[a,b]$ is complete. Define $f:[a,b] \to \mathbb{R}$
$$f(x)=\lim_{n \to \infty}f_n(x)$$
Now it seems the problem is to show that this function is in fact $k$ time differentiable? I believe the result follows a.e. but I think I need more than that. I was wondering if we could do something off the limit definition of the derivative?
Use Taylor's formula with integral form of remainder. If $\{f_n\}$ is Cauchy you will see that $f_n (0)+f_n'(0) x /(1!) +...+f_n^{(k-1)} (0) x^{k-1} /(k-1)!$ converges uniformly. Put $x=0$ and conclude that $f_n (0)$ converges. Then remove this term and divide by x to see that $f_n'(0)$ converges, etc. Hence $\{f_n^{(j)}(0)\}$ converges for each $j<k$. Now $|(f_n-f_m) ^{(k-1)} (x) -(f_n-f_m) ^{(k-1)} (y)| \leq |x-y| |(f_n-f_m) ^{k} (t)|$ for some t by MVT. Conclude that $\{f_n ^{(k-1)}\}$ converges uniformly. Lemma: if $\{f_n\}$ and $\{f_n'\}$ converge uniformly to f and g respectively the f is differentiable and $f'=g$. Apply this basic Calculus result k-times to prove that the Cauchy sequence $\{f_n\}$ converges in the norm of your space. [ You will need the formula $g(x)-g(0)= \int _0 ^{x} g'(t)dt$ to see how the limits of various derivatives are related. I hope I have given enough details for you to complete the argument].