Verification: Prove $F : C^1[0,1] \to C^1[0,1]$ is continuous and well-defined.

Question

Verification: Prove $F : C^1[0,1] \to C^1[0,1]$ is continuous and well-defined. Where $F(f(x)) = \sin(x)f(x)$. and $C^1[0,1]$ is equipped with the norm $\lvert\lvert f \rvert\rvert = \sup_{[0,1]} \lvert f \rvert + \sup_{[0,1]} \lvert f' \rvert$.

$F$ is well defined since can evaluate $F(f), \forall f \in C^1[0,1]$ and $F(f) \in C^1[0,1]$.

The norm induces a metric $d(x,y) = \lvert\lvert x-y \rvert\rvert $ so need to show $\forall f, \forall \epsilon >0, \exists \delta >0 \,\,\, d(F(f),F(g)) < \epsilon$ when $d(f,g) < \delta$.

Let $\epsilon >0,$ set $\delta = \frac{\epsilon}{2}$.

Now, $d(F(f),F(g))= \lvert\lvert \sin(x)f(x) - \sin(x)g(x) \rvert\rvert = \\ \sup_{[0,1]} \lvert\sin(x)f(x) - \sin(x)g(x) \rvert + \sup_{[0,1]} \lvert \sin(x)f'(x) + \cos(x)f(x) - \sin(x)g(x)- \cos(x)g'(x)\vert \le 2\sup_{[0,1]} \lvert f(x) - g(x) \rvert + \sup_{[0,1]} \lvert f'(x) - g'(x) \vert< \epsilon$

When $d(f,g) = \lvert\lvert f-g \rvert\rvert = \sup_{[0,1]} \lvert f - g \rvert + \sup_{[0,1]} \lvert f' - g' \vert < \frac{\epsilon}{2} = \delta$.

Just making sure I haven't missed anything. I haven't taken a course on Functional Analysis so I'm not sure if $\epsilon-\delta$ is used to prove continuity on functionals, I don't see why it wouldn't.

Answer 1

You're not wrong, but there should be $g'$ instead of $g$, and the long line of maths being broken up automatically meant that I didn't understand for a moment (i.e. "why is this less than 2..."). (Also continuity in a normed space doesn't need to be written with a metric, but not an issue.)

Another way to proceed is by using the fact that a linear operator (ps its not a functional, those are maps into a field) is continuous iff it is bounded (in operator norm). So you just have to check the slightly simpler assertion that $\| F(f)\|_{C^1} \le C \|f\|_{C^1}$ for some constant $C>0$: $$ \|F(f)\|_{C^1} = \|\sin(\cdot)f(\cdot)\|_\infty + \|\cos(\cdot)f(\cdot) + \sin(\cdot)f'(\cdot)\|_{\infty} \le 2\|f\|_{\infty}+\|f'\|_\infty \le 2 \|f\|_{C^1}.$$