Let $X=C^1([0,1])$ be the space of all continuous complex-valued functions on $[0,1]$ that have continuous derivatives on $[0,1]$. Define $\lVert x \rVert = \lVert x \rVert_\infty + \Vert x' \rVert_\infty$.
Suppose $x_n$ is a Cauchy sequence in $X$, then prove there are $y,z \in C([0,1])$ such that $\lVert x_n – y\rVert_\infty \to 0$ and $\lVert x_n'-z\rVert_\infty \to 0$. Also prove that
$$y(t)-y(0)=\int_{0}^{t} z(s) ds$$
It is easy to find $y$ since $C([0,1])$ is complete with the sup norm. I want to show that $x_n'$ is Cauchy so that there is $z$ such that $x_n' \to z$.
I want to finally show that $C^1([0,1])$ is Banach.
Best Answer
It is better to start with $z$ and then reconstruct $y$. By definition of the norm on $C^1([0,1])$, the sequence $x_n'$ is Cauchy in $C([0,1])$ so (possibly after passing to a subsequence) we can write $x_n' \rightarrow z$ (in $C([0,1])$). Since $x_n$ is also Cauchy in $C([0,1])$, we can also assume (after passing to another subsequence) that $x_n \rightarrow y$ in $C([0,1])$. Now
$$ y(t) = \lim_{n \to \infty} x_n(t) = \lim_{n \to \infty} \left( x_n(0) + \int_0^t x_n'(s) \, ds \right) = y(0) + \int_0^t \lim_{n \to \infty} x_n'(s) \, ds = y(0) + \int_0^t z(s) \, ds$$
where are allowed to exchange the limit and the integral because $x_n' \rightarrow z$ uniformly over $[0,1]$.
Finally, the equation above together with the Fundamental theorem of calculus shows that $y$ is differentiable and $y'(t) = z(t)$ for all $t \in [0,1]$ so $x_n \rightarrow y$ in $C^1([0,1])$ and so $C^1([0,1])$ is Banach.