Eigenvectors not in the Underlying Hilbert Space Eigenvectors for the momentum operator: $P=-i\dfrac{d}{dx}$ are functions $f_p(x):=e^{ipx}$ which are not in $L^2(\mathbb{R})$. The notion of orthogonality can be made rigorous by use of Fourier transform which allows us to represent a general $L^2$ function as a superposition of functions $e^{ipx}$.
quoting wikipedia:
https://en.wikipedia.org/wiki/Self-adjoint_operator#Spectral_theorem
"Other versions of the spectral theorem are similarly intended to capture the idea that a self-adjoint operator can have "eigenvectors" that are not actually in the Hilbert space in question."
Can anyone provide other examples of eigenvectors for self-adjoint operators that are not in the underlying Hilbert space of the operator and the strategies to prove that they are, in fact, eigenvectors?
(i am most interested in examples involving trigonometric functions.)

$e_{\lambda}(x) = \frac{1}{\sqrt{2\pi}}e^{i\lambda x}$ is an eigenfunction of $L=\frac{1}{i}\frac{d}{dx}$, and $Le_{\lambda}(x)=\lambda e_{\lambda}$, but it is not an eigenvector in $L^2(\mathbb{R})$. However, the following is in $L^2(\mathbb{R})$ and is an approximate eigenvector:
\begin{align}
   e_{\lambda,\epsilon}(x)&=\int_{\lambda-\epsilon}^{\lambda+\epsilon} e_{\lambda'}(x)d\lambda' \\
     &=\frac{1}{\sqrt{2\pi}}\int_{\lambda-\epsilon}^{\lambda+\epsilon}e^{i\lambda' x}d\lambda' \\
     &= \frac{1}{\sqrt{2\pi}ix}(e^{i(\lambda+\epsilon)x}-e^{i(\lambda-\epsilon)x}) \\
     &=\sqrt{\frac{2}{\pi}}e^{i\lambda x}\frac{\sin(\epsilon x)}{x}.
\end{align}
The $L^2$ norm of $e_{\lambda,\epsilon}$ can be determined from the Plancherel identity:
$$
      \|e_{\lambda,\epsilon}\|_{L^2}^{2}=\int_{\lambda-\epsilon}^{\lambda+\epsilon}1d\lambda'=2\epsilon.
$$
Note that
$$
    \|Le_{\lambda,\epsilon}-\lambda e_{\lambda,\epsilon}\|^2=\left\|\frac{1}{\sqrt{2\pi}}\int_{\lambda-\epsilon}^{\lambda+\epsilon}(\lambda'-\lambda)e^{i\lambda' x}d\lambda'\right\|^2 \\
       = \int_{\lambda-\epsilon}^{\lambda+\epsilon}(\lambda'-\lambda)^2d\lambda'=\frac{2\epsilon^3}{3}
$$
Therefore
$$
              \|(L-\lambda I)e_{\lambda,\epsilon}\|^2=\frac{\epsilon^2}{3}\|e_{\lambda,\epsilon}\|^2.
$$
