I have the following:
Show that the function $$\phi:(V, \|\cdot\|_{C^1}) \rightarrow (W, \|\cdot\|_{\infty}),~~f \rightarrow f'$$ is continuous. With
$$\|\cdot\|_\infty = \sup\{|f(x)| ~ \big| ~x\in [0,1]\}$$
and
$$\|\cdot\|_{C^1}= \sup\{|f(x)| + |f'(x)| \}$$
Together with $a,b \in \mathbb{R},~ a < b,~V = C^1[a,b];$ the vector space of all once-continouously differentiable functions and $W = C^0[a,b]$ the vector soace of continuous functions from $[a,b] \rightarrow \mathbb{R}$.
I want to prove that saying the following:
A map $\phi:(V, \|\cdot\|_{C^1}) \rightarrow (W, \|\cdot\|_{\infty})$ between two metric spaces is continuous in $x \in (V, \|\cdot\|_{C^1})$, if
$$\lim_{n \rightarrow \infty} \phi(x_n) = \phi(x) \in (W, \|\cdot\|_{\infty})$$
is true for a sequence $(x_n)_{n\in\mathbb{N}}$ that approaches $x$.
$\phi$ is continuous, if it is continouous in all points $x$.
Now let $f_n \in (V, \|\cdot\|_{C^1})$ be a sequence that approaches $f$. Thus we have
$$\|f_n - f\| = \sup\{|f_n - f| + |f'_n - f'|\} < \epsilon ~~~~\text{for large enough n}$$
$$\Rightarrow |f_n - f| + |f'_n - f'| < \epsilon$$ $$\Rightarrow |f_n - f| < \frac{\epsilon}{2} ~~~~\text{and} ~~~~|f'_n - f'| < \frac{\epsilon}{2}$$
So that means that for the function itself one gets:
$$\|\phi(f_n) - \phi(f)\|_\infty = \sup\{|\phi(f_n) - \phi(f)|\} = \sup\{|f'_n - f'|\} < \frac{\epsilon}{2}$$
So I have just hopefully proven that
$$\text{if }~~ \lim_{n \rightarrow \infty} f_n = f ~~\text{ then }~~ \lim_{n \rightarrow \infty} \phi(f_n) = \phi(f)$$
Is that correct? Especially because $\|\cdot\|_\infty$ is only defined for $[0,1]$ this makes me doubtful. Do I have to make any restrictions for $f_n$?
Thank you very much for the help.
FunkyPeanut