Cayley Transform and Eigenvalues

Question

I have a particular operator, namely $A=-i\frac{d}{dx}$ that I would like to Cayley transform. $A$ is defined on the Hilbert space $L^{2}[0,1]$ and has domain $\mathcal{D}_{\alpha}=\{g:g \in C^{\infty}, \, g(1)=e^{i \alpha}g(0)\}$.

I can find the eigenfunctions and eigenvalues for $A$:

Eigenfunctions: $\, f^{\alpha}_{n}(x)=e^{i(2 \pi n + \alpha)x}, \, n \in \mathbb{Z},$

Eigenvalues: $\, 2 \pi n + \alpha, \, n \in \mathbb{Z}.$

As previously stated, my interest is in the Cayley transform of $A$, namely

$$U(A)=(A-i)(A+i)^{-1}.$$

The problem that I am having is computation of the inverse $(A+i)^{-1}$. I have only been able to write down the trivial steps

$$(A+i)^{-1}=(-i\frac{d}{dx}+i)^{-1}=(-i(\frac{d}{dx}-1))^{-1},$$

and am stuck at this point. To elucidate my motivation for the calculation I am searching for the eigenvalues of the Cayley transform $U(A).$

Answer 1

There are two reasonable methods that I can see.

Method 1: If you want $g=(A-\lambda I)^{-1}f$, then $g$ is the unique solution of the ODE $$ -ig'(x)-\lambda g(x) = f(x),\\ g(1)=e^{i\alpha}g(0). $$ You can solve this equation for all $f$ iff $\lambda\notin\sigma(A)$. The solution is obtained by multiplying by the integrating factor $e^{-i\lambda x}$, integrating both from $0$ to $t$, and solving for the unknown $g(0)$ by imposing the second condition. The method is basic Calculus, but a little messy. However, this method is far more fundamental and basic than the next method.

Method 2: You can see that $e_{n}(x)=e^{(2\pi n+\alpha) ix}$ are eigenfunctions of $A$ with eigenvalues $\lambda_{n}=(2\pi n+\alpha)$ because $e_{n}(1)=e^{i\alpha}e_{n}(0)$. And $\{ e^{(2\pi n+\alpha)ix}\}_{n=-\infty}^{\infty}$ is a complete orthonormal basis of $L^{2}(0,1)$ because the ordinary exponentials are an orthonormal basis. Therefore, $$ (A-\lambda I)^{-1}f = \sum_{n=-\infty}^{\infty}\frac{1}{2\pi n + \alpha-\lambda}\left(\int_{0}^{1}f(t)e^{-(2\pi n+\alpha)it}dt\right)e^{(2\pi n+\alpha)ix}. $$ This method relies on knowing the completeness of the exponentials and the selfadjointness of $A$, and realizing that $(U_{\alpha}f)(t)=e^{i\alpha t}f(t)$ is a unitary map on $L^{1}(0,1)$ that maps the complete orthonormal basis $\{ e^{2\pi in x}\}$ onto another complete orthonormal basis $\{ e^{(2\pi n+\alpha)ix} \}$. (This map also trivially relates $A_{\alpha}$ to $A_{0}$.)

Answer 2

If $$(-i\frac{\partial}{\partial x}+i)=g$$ then

$$(1-\frac{\partial}{\partial x})f=-ig$$

$$f=\sum_{k=0}^{\infty}-i\frac{\partial^n}{\partial x^n}g$$

Now $h(t)=e^{-it\alpha}g(t)$ satisfies $h(0)=h(1)$. Writing $h$ as a Fourier series gives you $g$ as a sum of exponentials for which the expression for $(A+i)^{-1}$ above is easy to compute.