Showing that $(Ax)(t) = x'(t)$ is a Fredholm operator Can you help me to prove that the operator $A \colon C^1[0,1] \rightarrow C[0,1]$,
$$(Ax)(t) = x'(t)$$
is a Fredholm operator and find $\alpha(A) = \dim(\operatorname{Ker} A)$ and $\beta(A) = \operatorname{codim} (\operatorname{Im} A)$? 
I have no idea.
 A: A linear operator $A$ is Fredholm, by definition, if an only if both $\ker A$ and $\text{coker} A$ are of finite dimension, and the range of $A$ is closed.
I assume both $C([0, 1])$ and $C^1([0, 1])$ are understood in the usual sense, that is, as the spaces of continuous and once-continuously differentiable complex-valued functions on $[0, 1]$, respectively with the usual norms; that is, for 
$f(x) \in C([0,1]) \tag 1$
we have
$\Vert f(x) \Vert_0 = \sup_{x \in [0,1]} \vert f(x) \vert, \tag 2$
whilst for
$f(x) \in C^1([0,1]), \tag 3$
$\Vert f(x) \Vert_1 = \sup_{x \in [0,1]} (\vert f(x) \vert + \vert f'(x) \vert); \tag 4$
it is then clear that the operator
$A:C^1([0, 1]) \to C([0, 1]), \; A(f(x)) = f'(x) \tag 5$
is bounded, since
$\Vert A(f(x)) \Vert_0 = \Vert f'(x) \Vert_0 = \sup_{x \in [0, 1]}\vert f'(x) \vert$
$\le \sup_{x \in [0, 1]}(\vert f(x) \vert + \vert f'(x) \vert) = \Vert f(x) \Vert_1 \tag 6$
which shows that
$\Vert A \Vert \le 1. \tag7$
Now if
$A(f(x)) = f'(x) = 0, \tag 8$
then
$f(x) = c,\; \text{a constant}, \tag 9$
and since the derivative of a constant function vanishes, it follows that
$\ker A = \{c, \; c \in \Bbb C \}, \tag{10}$
so clearly,
$\alpha(A) = \dim \ker A = 1. \tag{11}$
Now if
$g(x) \in C([0, 1]), \tag{12}$
we may define
$f(x) = \displaystyle \int_0^x g(s) \; ds \in C^1([0, 1]); \tag{13}$
then clearly
$A(f(x)) = f'(x) = g(x); \tag{13}$
thus $A$ is surjective:
$\text{Range}(A) = C([0, 1]), \tag{14}$
from which we see that
$\text{coker}(A) = C([0, 1]) / \text{Range}(A) = C([0, 1]) / C([0, 1]) = \{0\}, \tag{15}$
whence
$\beta(A) = \dim \text{coker}(A) = 0. \tag{16}$
By (14), $\text{Range}(A)$ is evidently closed.
We have thus verified that $A$ is indeed a Fredholm operator.
