Show that the mapping $(C^1, ||.||_{C^1}) \rightarrow (C, ||.||_\infty) $ is continuous

Question

Assume

$D:(C^1, ||.||_{C^1}) \rightarrow (C, ||.||_{\infty}),$

$$D(f) = f',$$

is a mapping with

$$||f||_{C^1} := ||f||_{\infty} + ||f'||_{\infty},$$

$$||f||_{\infty} := sup_{x \in [a, b]} f(x).$$

I have to show that the mapping is continuous. We already proved that $C([a, b])$ and $C^1([a,b])$ are complete, therefore every Cauchy-sequence converges in both of these rooms.

My own approach looks like this:

Assume $(f_n) \in C^1([a, b])$ is a Cauchy-sequence. Since $C^1([a, b])$ is complete, $(f_n)$ converges (pointwise or uniform, I guess it doesn't matter yet), such that

$\lim_{n\to \infty} (f_n) = f$

for a certain $f \in C^1([a, b]).$ Interestingly enough, this already looks exactly like the premise I would normally use for proving continuity. Following the definition of the mapping, I have to show that

$\lim_{n\to \infty} (f'_n) = f'.$

Remembering our theorems, the conclusion would follow directly when I could assure the following statements:

• $(f_n)$ is continuous and differentiable (true since $(f_n) \in C^1[(a, b)]),$
• $(f_n)$ converges pointwise or uniformly (true since $(f_n)$ is a Cauchy-sequence),
• $(f'_n)$ converges uniformly.

Since $C([a, b])$ is complete, $(f'_n) \in C([a, b])$ is a Cauchy-sequence, and therefore it converges at least pointwise - but what gives me the uniform convergence here?

Answer 1

You should have a look of the various definitions of convergence again. Convergence in the $||.||_\infty $ norm is uniform convergence, and it does matter, because only uniform convergence allows you to conclude that the limit of a sequence of continuous functions is continuous.

Having said that it's in fact almost trivial resp a tautological statement that convergence in $C^1$ implies that differentiation is a continuous map $C^1 \rightarrow C^0$ when these spaces are equipped with the norms you've written down, since this then just states that $f_n^\prime$ (and $f_n$) converge uniformly, which means that the respecitve limit is continous. The only possible pitfall is the question of whether the limit $g$ of the sequence $f_n^\prime$ is, in fact, the derivative of the limit $f$ of $f_n$. This I leave as an exercise to you.