Index of a derivative operator on a circle Let $D: C^{1}(S^{1}) \rightarrow C(S^{1})$ be an operator defined as $D(f)=f'$. 
I would like to find its index (on the road proving that it's a Fredholm operator).
First, if $f \in ker(D)$, then $f$ is constant almost everywhere on a circle. 
What goes about cokernel, its dimension equal the dimension of a kernel of an adjoint operator, by Riez-Markov-Kakutani theorem, the adjoint is $T(f(x)) = \int_{S^{1}}{f'(x) \mu(dx))}$, $\mu$ is some Radon measure.
The claim is that the index equals 0, the both the $ker(D)$ and $ker(D^{*})$ have the same dimension. It's clear that $ker(D) \subset ker(D^{*})$, but also it is true that there is no inverse inclusion -- but still this does not affect the dimension property.
Probably, the idea is to use the Fourier expansion on the cirlce, $$f(x) = \sum_{n \in \mathbb{Z}}{c_{n} e^{inz}}$$ then $$f'(x) = \sum_{n \in \mathbb{Z}}{in c_{n} e^{inz}}$$ but this no seems to be very benefitial so far.
What are the possible approaches to pose the problem?
Any sort of help would be much appreciated.
 A: Actually, one can avoid Fourier theory by identifying the spaces $$C(S^1)=\left\{f\in C([0,1],\mathbb C) \ | \ f(0)=f(1)\right\}$$ and $$C^1(S^1)=\left\{f\in C^1([0,1],\mathbb C) \ | \ f(0)=f(1) \text{ and } f'(0)=f'(1)\right\}.$$
In this way, you get that $f\in \ker(D)$ if and only if $f$ is constant.
Moreover, one can easily notice that $$C(S^1)=Im(D)\oplus \mathbb C$$ where $\mathbb C$ represents the complex vector space of constant functions on $S^1$.
Indeed,


*

*If $f\in Im(D)\cap \mathbb C$ then there exists $g\in C^1(S^1)$ such that $f=g'$ and $f\equiv C$ thus $g(t)=Ct+D$. However $g$ must satisfy $g(0)=g(1)$ and $g'(0)=g'(1)$ hence $g$ is the constant function and $f\equiv 0$.

*Let $f\in C(S^1)$, one can define the function $g:[0,1]\rightarrow \mathbb C$ by $$g(t)=\int_0^tf(s)ds-t\int_0^1f(s)ds.$$
$g$ is continuously differentiable and $$\left\{\begin{array}{l}g(0)=0=g(1) \\ g'(0)=f(0)-\displaystyle  \int_0^1f(s)ds \\ g'(1)=f(1)-\displaystyle  \int_0^1f(s)ds\end{array}\right.$$
thus $g$ belongs to $C^1(S^1)$ since $f(0)=f(1)$ and $$\forall t\in [0,1], \ f=\underbrace{g'}_{\in Im(D)}+\underbrace{\int_0^1f(s)ds}_{\in \mathbb C}.$$


As a consequence, $D$ has a finite dimensional cokernel isomorphic to $\mathbb C$ which means that $D$ is a fredholm operator with index $$ind(D)=\dim \ker(D)-\dim \text{coker}(D)=0.$$
