Self-adjoint operators, distributions and spectral theorem Suppose we have a self-adjoint operator, $L$,  which maps elements $u \in C^2 \to C^2$ on $(0,\infty)$ using the standard inner product.
Furthermore, we have a trigonometric system over an infinite set of real numbers, $\theta_i$, which converges to $L$ as a distribution- when integrated against a test function.
What i mean by this is that suppose, $ Lu=v$, then we have, for example, a cosine series over all $\theta_i$, which takes in $u$ and outputs a distribution which converges to $v$. Perhaps something like
$$\sum_{i=1}^{\infty}\omega_i\cos(\theta_i) \approx{Lu}$$
$$ \omega_i=\langle\cos(\theta_i),u\rangle$$
where $\approx$ is only achieved after integrating the LHS against a test function.
Here are my questions:

*

*is it possible to use the spectral theorem to draw conclusions about the set of numbers, $\theta_i$? Specifically, can we say that the elements in the set are unique? while the cosine series is orthogonal, i am unsure if we can consider them eigenfunctions as the convergence is only as a distribution and is only achieved after we integrate against a test function.


*is it ever possible to have a self-adjoint operator which sends distributions to distributions? my instinct says no, because we can not take inner product of two distributions.
EDIT (based off of comments from paul garrett):
Let's assume that $C^2$ is the space of all twice-continuously-differentiable functions and the 'standard inner product' is $\langle f,g\rangle=\int_{0}^{\infty}f(x)g(x)dx$. Furthermore, $L$ is not necessarily bounded.
paul garrett also pointed out that the distinct consines in the example are problematic because they are not in $L^2$. The essence of my confusion is whether or not these $\theta$'s are unique, or if a collection of functions can be regarded as eigenfunctions if they only converge as a distribution- eg. the cosines above will not converge to $Lu$, but we can prove that they will as a distribution.
 A: The questions are significant, but ambiguous, in fact. There are several issues here, and it would be hard to do justice to all of them, but perhaps I can dispel some ambiguities/confusions.
First, yes, there are some bounded operators on Hilbert spaces that have no eigenvectors in the space, but do have some sort of generalized eigenvectors lying in a larger space. One primordial version of this is H. Weyl's notion of "generalized eigenvector", corresponding to $\lambda$ in the spectrum which is not an eigenvalue (that is, is not discrete spectrum). This is a sequence of elements of the Hilbert space that better-and-better approach the $\lambda$-eigenvalue condition. But the sequence will not converge in the Hilbert space...
Second, with potentially unbounded (but natural) operators, such as (restrictions of) Laplacians, the happy spectral theory of bounded operators is not available, creating more than one complication. But, yes, these complications are natural, not fake, not pathologies.
Perhaps of interest, avoiding some of the (nevertheless interesting) complications: already the multiply-by operator $f(x) \to x\cdot f(x)$ on $L^2[0,1]$ has no discrete spectrum, but, in an obvious way, has "generalized eigenvectors" consisting of Dirac deltas at $x_o$ in $[0,1]$.
Somewhat more complicatedly, but also naturally, Fourier inversion on $L^2)\mathbb R)$ expresses an $L^2$ function as a "superposition" of exponentials, even though the exponentials are not in $L^2$. This does express an $L^2$ function as a superposition of eigenfunctions for the Laplacian. But not a (countable) sum.
In fact, various functional-analytic gymnastics show that if there is a countable collection of (generalized) eigenfunctions that span a Hilbert space, they must actually be in the space.
And so on... Probably a new question, taking such things into account, would be more useful to others than radical revision of this literal question.
