the problem
I have to show that a function $D: C^1([a, b]) \mapsto C^0([a, b]): f \mapsto f'$ is continuous given a metric $\| \cdot \|_{C^1([a, b])}$.
The metric $\| \cdot \|_{C^0([a, b])}$ is defined as:
$ \| f \|_{C^0([a, b])} = \max_{x \in [a, b]} |f(x)| $
And the metric $\| \cdot \|_{C^1([a, b])}$ is defined as:
$ \| f \|_{C^1([a, b])} = \| f \|_{C^0([a, b])} + \| f' \|_{C^0([a, b])} $
my attempts
This is what I have myself so far:
$\epsilon$-$\delta$
That criterion basically says:
\forall p \in C^1([a, b]): \forall \epsilon > 0: \exists \delta > 0: \forall x: \| x-p \|_{C^1([a, b])} < \delta \Rightarrow \| f(x) - f(p) \|_{C^1([a, b])} < \epsilon
After a little simplification I ended up with:
\| x-p \|_0 < \delta \Rightarrow \| x'' - p'' \|_0 < \epsilon
I do not see how I could choose $\delta$ to make that work.
sequences
The other approach is via a limit of a sequence:
\forall \epsilon > 0: \exists n_0 \in \mathbb N: \forall n > n_0: \| D(x_n) - D(x) \|_{C^1([a, b])} < \epsilon
I used the definition of $D$ and the metric and ended up with:
\| x_n' - x' \|_0 + \| x_n'' - x'' \|_0 < \epsilon
Given that $x_n$ converges to $x$, “it just has to be that way”, but I am not sure how to show that.
The problem also states that with the plain $\| \cdot \|_0$ metric, the function $D$ is not continuous. But that would mean that \| x_n' - x' \|_0 < \epsilon does not hold. With the $\| \cdot \|_1$ metric, the left side of the inequality just gets larger, so I think it would be even stricter and not work if the first one did not even work.
How can I show that this $D$ is continuous?
So with the correct metric, I see the answer:
\forall p \in C^1([a, b]): \forall \epsilon > 0: \exists \delta > 0: \forall x: \| x-p \|_{C^1([a, b])} < \delta \Rightarrow \| D(x) - D(p) \|_{C^0([a, b])} < \epsilon
\begin{align*} \| x-p \|_{C^1([a, b])} &< \delta &\Rightarrow& &\| D(x) - D(p) \|_{C^0([a, b])} &< \epsilon \\ \| x-p \|_{C^0([a, b])} + \| x'-p' \|_{C^0([a, b])} &< \delta &\Rightarrow& &\| x' - p' \|_{C^0([a, b])} &< \epsilon \end{align*}
Since $\| x-p \|_{C^0([a, b])} \geq 0$ can I choose $\delta := \epsilon$, to fulfill the $\Rightarrow$ statement.