$ X=C^1([0,1],\mathbb{R})$ and $||f||=\max\{||f||_{\infty}, ||f'||_{\infty}\}$ form a Banach space

Question

Show that the space $X=C^1([0,1],\mathbb{R})$ of the functions in $[0,1]$ to real values with continuous derivative and with the norm $||f||=\max\{||f||_{\infty}, ||f'||_{\infty}\}$ is a Banach space. Is $(X, ||.||_{\infty})$ Banach?

I'm assuming $||f||$ as defined is already a norm. Then I just need to show that every Cauchy sequence is convergent. However, I don't know how to assume a general Cauchy Sequence. A sequence of functions can have any face. So I think another theorem must be used instead. Maybe something that talks about another definition equivalent to this one, that doesn't involve cauchy sequences. Any hints?

Answer 1

Let $\{f_n\}$ be Cauchy. Then there exist continuous functions $f$ and $g$ such that $f_n \to f$ uniformly and $f_n' \to g$ uniformly. Now $f_n(x)=f_n(0)+\int_o^{x} f_n'(t)\, dt$. Let $n \to \infty$ to see that $f(x)=f(0)+\int_0^{x} g(t)\, dt$. This implies that $f$ is differentiable and $f'=g$. Hence $f \in C^{1}[0,1])$ and it is clear that $f_n \to f$ in the given norm.

Answer 2

A Cauchy Sequence with respect to this metric implies uniform convergence and the uniform limit of continuous functions is continuous, so the sequence of functions converges to a continuous function. Their derivatives also converge to a continuous function, so then the sequences all converge in $C^1([0,1];\mathbb{R})$. It remains to show that $\lim f_n'(x)=f'(x)$. Or as suggested by the next answer. Be careful, if you use the proof bellow you need to justify the passage of the limit through the integral.