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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.

Mr. G
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  • This has a "similar" look as a path integral for a circle path in the complex plane $(\int_{\alpha}f(\alpha(t))\alpha'(t)dt $ with $\alpha(t) = re^{it}$, $0\le t \le 2\pi$). Perhaps that will help in looking for a construction of $g(\alpha,\theta)$. – cwa Mar 13 '14 at 16:51

1 Answers1

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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 $g$.

user127096
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  • Is $\hat{K}{\alpha}(n)$ the Fourier transform of $K{\alpha}(t)$? If that's the case, then if I just look up that transform I see $K_{\alpha}(t) = -\sqrt{2\pi} \alpha e^{-\alpha t}$, so I guess that tells you what $C$ is. Also, what's $T$ in the limits of the integral in $(1)$, is it just $[0, 2\pi]$? – Mr. G Mar 13 '14 at 21:34
  • @Mr.G Yes, the Fourier transform. I wrote $\mathbb T$ for $[0,2\pi]$ to emphasize wrap-around in the integral, i.e., $\phi$ and $\theta$ are understood as elements of $\mathbb R \mod 2\pi$. Otherwise $\theta-\phi$ would get out of $[0,2\pi]$, where $K_\alpha$ is defined. – user127096 Mar 13 '14 at 21:38