Solve: $\int_{0}^{2\pi}g \psi e^{i n \theta}\,\text{d}\theta = n/(n-i\alpha) \int_{0}^{2\pi}\psi e^{i n \theta}\,\text{d}\theta$ For $\alpha>0$, I want to find a $g(\alpha, \theta)$ such that
$$
\int_{0}^{2\pi}g(\alpha, \theta)\psi(\theta)e^{i n \theta}\,\text{d}\theta = \frac{n}{n-i\alpha} \int_{0}^{2\pi}\psi(\theta)e^{i n \theta}\,\text{d}\theta
$$
for some $\psi(\theta)$. Note that neither $g$ nor $\psi$ are functions of $n$. I don't even know if this is possible for a general $\psi$, but I would like to know if anyone has some ideas.
 A: For general $\psi$ this is impossible, but for "most" $\psi$ you can do it. You have   a bounded operator $T$ on $L^2(\mathbb T)$ defined by multiplication on the Fourier side: 
$$\widehat{T\psi}(n) = \frac{n}{n-i\alpha}\widehat{\psi}(n),\quad n\in\mathbb Z$$
The multiplier is of the form $1+O(1/n)$. We can write $T$ as the sum of identity operator and a convolution with the square-integrable kernel $K_\alpha$ determined by $\widehat{K_\alpha}(n)=i\alpha/(n-i\alpha)$: 
$$ {T\psi}(\theta) = \psi(\theta)+\int_{\mathbb T} \psi(\phi) K_\alpha(\theta-\phi)\,d\phi$$
This sort of tells us what $g$ is: 
$$g(\alpha,\theta ) = 1 + \frac{1}{ \psi(\theta)}\int_{\mathbb T} \psi(\phi) K_\alpha(\theta-\phi)\,d\phi \tag{1}$$
However, it is possible for $\psi$ to vanish on some set of positive measure, and for the convolution $\psi*K_\alpha$ to be nonzero on that set. In this case there is no way to express $\psi+\psi*K_\alpha$ as a product in which $\psi$ is one of the factors. 
If $\psi\ne 0$ a.e., then (1) defines $g$.  You can write down $K_\alpha$ explicitly: it is of the form $K_\alpha(t)=C e^{-\alpha t}$, $t\in [0,2\pi]$, where $C$ is a constant. If $\psi$ is such that the integral (1) can be evaluated, then you have an explicit $g$.
