2
$\begingroup$

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.

$\endgroup$

2 Answers 2

3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

Another proof works like this:

The map $C^1[0,1]\to C[0,1]\oplus \mathbb C$, $f\mapsto (f',f(0))$ is clearly an isomorphism of vector spaces. So the norm on $C[0,1]\oplus\Bbb C$ induces a norm $f\mapsto |f(0)|+\|f'\|_\infty$ on $C^1[0,1]$ wrt to which $C^1[0,1]$ is complete.

Call the norm we started with $\|\cdot\|_A$ and this second norm $\|\cdot\|_B$. One has: $$\|f\|_A≤\|f\|_B≤2\|f\|_A$$ for all $f\in C^1[0,1]$ so the norms are equivalent and $C^1[0,1]$ is a Banach space. Here the first inequality is trivial and the second follows from: $$\|f\|_\infty=\sup_{x\in[0,1]}\left|f(0)+\int_0^xf'(t)\,dt\right|≤|f(0)|+\|f'\|_\infty$$ So $\|f\|_\infty+\|f'\|_\infty≤2\|f\|_B$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .