Are differentiation and integration continuous functions? Is differentiation a continuous function from $C^1[a,b] \to C[a,b]$?
I think it is but I can't prove it... Would it be possible to prove it using theory about closed sets in $C[a,b]$ and their preimage? My problem here would be to check all closed sets and the closeness of their preimage so I'm feeling I'm on the wrong track!
Also : is integration a continuous function from $C[a,b] \to C[a,b]$? Somehow I feel it's not... Then it would be enough to show that some closed subset of $C[a,b]$ has a non-closed preimage? Wouldn't it? But which one?
I'm not sure this isn't all misleading... Can you help me? Thanks a lot!
PS: I'm considering the sup metric on $C[a,b]$!
 A: I'll assume you consider the topologies induced by the $\mathcal{C}^1$-norm:
$$\|f\|_{\mathcal{C}^1([a,b])}=\sup_{[a,b]}|f| + \sup_{[a,b]}|f'|$$
and by the sup norm.
If you consider the topology of the uniform convergence on both spaces, then the answer is no, as pointed out in other answers.
The operator $f\mapsto D(f)=f'$ is linear from $\mathcal{C}^1([a,b])$ to $\mathcal{C}^0([a,b])$ and
$$\|D(f)\|_{\mathcal{C}^0([a,b])}=\sup_{[a,b]}|D(f)|=\sup_{[a,b]}|f'|\leq \sup_{[a,b]}|f|+\sup_{[a,b]}|f'|=\|f\|_{\mathcal{C}^1([a,b])}$$
i.e. $\|D\|\leq 1$ (operator norm). So it is continuous: a linear operator is continuous iff it is bounded.
As for the operator
$$f(x)\mapsto I_a(f)(x)=\int_{a}^x f(t)dt$$
you just do the same. If your target space is again $\mathcal{C}^0([a,b])$, then
$$\|I_a(f)\|_{\mathcal{C}^0([a,b])}=\sup_{[a,b]}|I_a(f)|\leq (b-a)\sup_{[a,b]}|f|=(b-a)\|f\|_{\mathcal{C}^0([a,b])}$$
so $\|I_a\|\leq b-a$ (as an operator). So it is again continuous; this can be also proved by hand: if $f_n\to f$ uniformly on $[a,b]$, 
$$\sup_{[a,b]}|I_a(f_n)-I_a(f)|\leq (b-a)\sup_{[a,b]}|f_n-f|\to 0\;.$$
You can also show that $I_a$ is continuous from $\mathcal{C}^0([a,b])$ to $\mathcal{C}^1([a,b])$, with $\|I_a\|\leq (b-a)+1$ (operator norm again, just between these two other spaces).
A: No it is not. Consider the sequence $(f_n)\in C^1[0,\pi]$, given by
$$f_n(x)=\frac{\sin nx}{n}$$
Then, $\lim_{n\to\infty}f_n=0$, but $Df_n=\cos nx$ does not converge to zero, as $n\to\infty$.
A: These questions are much easier to approach using the sequential definition of continuity: for metric spaces $X$ and $Y$, a map $L : X \to Y$ is continuous if and only if $L(x_n) \to L(x)$ for every convergent sequence $x_n \to x$ in $X$. 
The answers to your questions of course depend on what topology/metric you use - the counterexample posted by Mateus and MO92 works if you are using the uniform norm on both spaces; but if you are using the $C^1$ norm on $C^1[a,b]$ then differentiation is continuous pretty much by definition: the  $C^1$ norm is 
$$ \Vert f \Vert_{C^1} = \Vert f \Vert_{C^0} + \Vert f' \Vert_{C^0} $$
and thus whenever $\Vert f_n - f \Vert_{C^1} \to 0$ we have $\Vert f_n' - f' \Vert_{C^0} \to 0$.
Regarding the indefinite integral, the answer should also be yes: if two functions differ by at most $\epsilon$ then their integrals should differ by at most $\epsilon(b-a)$.
