I am quite new to Fredholm operators, can anyone help me with the following problem, thank you.
For any given $ g \in C[0,1] $ let $ A: C^{(1)}[0,1] \to C[0,1] $ be an operator such that $$ Af(x) = f'(x) + g(x)f(x) $$ Show that $A$ is a Fredholm operator and find its index.
We compute $\ker A$:
$f(x) \in \ker A \Longleftrightarrow f'(x) + g(x)f(x) = 0; \tag 1$
the unique solution, with $f(0)$ specified, to this last equation is
$f(x) = f(0) e^{-\int_0^x g(s) \; ds}; \tag 2$
since $f(0) \in \Bbb R$ is arbitrary, (2) expresses a $1$-dimensional family of solutions to (1); thus,
$\dim{\ker A} = 1; \tag 2$
we next observe that
$\text{coker} A = \{0\}; \tag 3$
this follows from the fact that for any
$h(x) \in C[0, 1] \tag 4$
the equation
$f'(x) + g(x)f(x) = h(x), f(0) \; \text{arbitrary}, \tag 5$
has the (again, unique) solution
$f(x) = e^{-\int_0^x g(s) \; ds}(f_0 + \displaystyle \int_0^x e^{\int_0^u g(s) \; ds} h(u) \; du), \tag 6$
from which we see that
$\text{Range}(A) = C[0, 1]; \tag 7$
therefore,
$\text{coker} A = C[0, 1]/\text{Range}(A) = \{0\}, \tag 8$
whence
$\dim{\text{coker} A } = 0; \tag 9$
since the dimensions of both the kernel and cokernel of $A$ are finite, we see that $A$ is Fredholm and we thus may conclude with
$\text{index}(A) = \dim{\ker A} - \dim{\text{coker} A } = 1 - 0 = 1, \tag{10}$
as per request.