spectral projector on infinite strips and half spaces In my studies I am coming across spectral theorem, spectral measure and spectral projector. In all books you can find the spectral measure and the spectral projector of the Laplacian on $\mathbb{R}$ and on the interval $[0,1]$, but I would like to have some examples. 
So, I asked myself what about the spectral projector of the Laplacian on the infinite strip $[0,2\pi]\times\mathbb{R}$ and on the half space $\mathbb{R}_+=\{x\in\mathbb{R}\;\vert x\geq 0\}$? 
Let $\Delta_S$ be the Laplacian on the infnite strip with, say Dirichlet boundary conditions, then the eigenvalues are $k^2+\lambda^2\in [0,\infty)$ for $k\in\mathbb{N}$ and $\lambda\in\mathbb{R}$.  While for the Laplacian on the half space $\Delta_H$ is spectrum is simply $[0,\infty)$ and we also here impose Dirichlet b.c. 
Let now $ P_S[0,b]$ and $P_H[0,b]$ be the spectral projector of $\Delta_S$ and $\Delta_H$ respectively. My claim is 
$$ P_S[0,b] f(x,y)=\sum_{k^2\leq b} \left(\int_{\{\lambda\in\mathbb{R}\;\vert\; \lambda^2\leq b- k^2\}} \hat{f}(k,y) e^{i\lambda y}\; dy \right) \sin(kx)$$
and 
$$P_H[0,b] f(x) =\int_{\{\lambda\in\mathbb{R}\;\vert\;\lambda^2\leq b\}} \mathcal{L}(f)(s) e^{isx} \; ds,$$
where $\hat{f}(k,y)$ is the Fourier Transform of $f$ in $x\in [0,2\pi]$ and $\mathcal{L}(f)$ is the Laplace transform of $f$. 
Of course $f$ leaves in the domain of the operator. 
However, I cannot prove this rigorously. Does anybody have an idea or a reference? 
 A: Start with the operator $A=-\frac{d^2}{dx^2}$ on the domain $\mathcal{D}(A)$ consisting of all twice absolutely continuous functions $f\in L^2[0,\infty)$ for which $f(0)=0$, $f' \in L^2[0,\infty)$, and $f'' \in L^2[0,\infty)$. The operator $A$ is a closed, densely-defined, symmetric linear operator on its domain, and $A$ has no selfadjoint extensions. The adjoint $A^*$ has the same domain, except for the condition that $f(0)=0$. So $\mathcal{D}(A)$ is of co-dimension $1$ in $\mathcal{D}(A^*)$, and that's what prevents $A$ from having a selfadjoint extension. This corresponds to the fact that one of the deficiency spaces $\mathcal{D}(A^*+iI)$, $\mathcal{D}(A^*-iI)$ is one-dimensional, while the other is zero-dimensional.
The Laplacian $\Delta$ spectral representation on $[0,2\pi]\times\mathbb{R}$ with Dirichlet conditions can be obtained using separation of variables. A function $f \in L^2([0,2\pi]\times\mathbb{R})$ can be represented using a mixed expansion
$$
      f(x,y) 
% = \frac{1}{2\pi}\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}f(x,y')e^{-iy's}dy'\right)e^{isy}ds \\
%   = \frac{1}{4\pi^2}\int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}\left(\sum_{n=-\infty}^{\infty}\int_{0}^{2\pi}f(x',y')e^{-inx'}dx' e^{inx}\right)e^{-iy's}dy'\right)e^{isy}ds \\
   = \int_{-\infty}^{\infty}\sum_{n=-\infty}^{\infty}C(s,n)\sin(n\pi x/2)e^{isy}ds.
$$
The functions $\sin(n\pi x/2)e^{isy}$ are classical eigenfunctions of the Laplacian with classical eigenvalues $n^2\pi^2 s^2/4$. The coefficient functions $C(s,n)$ are given by classical Fourier analysis:
$$
        C(s,n) = \frac{1}{2\pi^2}\int_{0}^{2\pi}\int_{-\infty}^{\infty}f(x',y')e^{-isy'}\sin(n\pi x'/2)dy'dx'
$$
Then, for $f\in \mathcal{D}(\Delta)$,
\begin{align}   f(x,y) 
   & = \int_{-\infty}^{\infty}\sum_{n=-\infty}^{\infty}C(s,n)\sin(n\pi x/2)e^{isy}ds. \\
     -\Delta f & = \int_{-\infty}^{\infty}\sum_{n=1}^{\infty}\left(\frac{n^2\pi^2}{4}+s^2\right)C(s,n)\sin(n\pi x/2)e^{isy}ds
\end{align}
The spectrum of $-\Delta$ consists of all values of $\frac{n^2}{\pi^2}{4}+s^2$. If you sum over all possible ways that these values lie in $[0,a]$, then you get $P[0,a]f$, where $P$ is the spectral measure for $-\Delta$. Such a "sum" is an integral in the continuous component and a sum in the discrete.
To prove everything rigorously, show that $P[0,a]$ as described above is a projection whose range lies in the domain of the Laplacian, and satisfies properties of a spectral measure with regard to the operator $-\Delta$.
