I'm trying to prove that $\|f\|=\sup \{|f(x)|x\in [a,b]\}+ \sup \{|f'(x)|:x\in [a,b\}\}$ defines a norm on $C^1[a,b]$ (the vector spaces of continuously functions on [a,b] with continuous first derivatives), and that this norm make $C^1[a,b]$ a Banach space.
I've got the norm part down. For the banach space, my attempt is to relate this to the normal $\sup$ norm on $C^0[a,b]$. So, I assume $\{f_n\}$ is Cauchy in $C^1$. By construction of the norm, $f_n$ and $f'_n$ are cauchy in $C_0$, so each converge in $C_0$ to $f$ and $f^*$, respectively. My attempt is to show that $f_n$ converges to $f$ in $C_1$ as well, and that $f'=f^*$.
First to show that $f'=f^*$, my attempt is as follows: $$f'(x)=\lim _{h\to 0}\frac {f(x+h)-f(x)}{h}=\lim _{h\to 0} \lim _{n\to \infty}\frac {f(x+h)-f(x)}{h}$$
First question is at that step, can I add that limit, since $n$ doesn't appear it shouldn't change anything? Then, I do the standard add and subtract trick to get $$\lim _{h\to 0}\lim _{n\to \infty }\frac {f(x+h)-f_n(x+h)+f_n(x+h)-f_n(x)+f_n(x)-f(x)} {h}$$
Now, my next question comes up on is breaking up the limits acceptable, since I can show each limit exists (assuming the interchange step is legal), that brings us to
$$\lim _{h\to 0}\lim _{n\to \infty }\frac {f(x+h)-f_n(x+h)}{h} +\lim _{h\to 0} \lim _{n \to \infty} \frac {f_n(x+h)-f_n(x)}{h}+\lim _{h\to 0}\lim _{n\to \infty}\frac {f_n(x)-f(x)}{h} $$
Now, the next question: evaluating the first and third limits, doing the inner limit first, since $f \to f_n$ in $C^0$, I should get the inner limit is $\frac 0 h=0$, so the first and third limits should be 0, correct?
Finally on the middle limit, can I exchange the order of limits to get $$\lim _{h\to 0} \lim _{n \to \infty} \frac {f_n(x+h)-f_n(x)}{h}=\lim _{n\to \infty} \lim _{h \to 0} \frac {f_n(x+h)-f_n(x)}{h}=\lim _{n\to \infty} f'_n(x)=f^*(x)$$
If all of that's valid, then I get $f'=f^*$, and from that I'm pretty sure I can finish the proof that $f_n\to f$ in the norm described.