About the orthogonality of the Hamiltonian eigenstates for the the continuous energy spectrum I would like first to describe a strange case that I encountered. $ \ \  - $
I solved the Schrodinger equation with a potential barrier (a potential well limited by a finite height wall which decrease with the distance $r$ from the center of the well). Relevant for me was the continuous spectrum of energies. I selected the set of solutions ${\phi (r, E)}$ regular at $r=0$ - see definition in end of the text. Then, I picked a certain function, $S(r, t)$, which is not an eigenfunction, but is regular at $r=0$. 
I thought that $S(r, t)$ can be fully developed as a superposition of the regular eigenfunctions, i.e. $S(r, t) = \sum _E A(E, t) \ \phi (r; E)$.
But, I discovered that $S(r, t)$ has a non-null projection on the irregular eigenfunctions.

Now, my question: is there some general proof that the eigenfunctions of a Hamiltonian, in the continuous energy spectrum are mutually orthogonal? Could it be that they are not? 

I mention that the spectral theorem doesn't seem helpful for the continuous spectra.  

DEFINITIONS : The Schrodinger equation may have (as in my case), two types of solutions, finite at $r=0$ which we call regular, and infinite at $r=0$ which we call irregular. The regular solutions are physical, while the irregular are non-physical. 
 A: Eigenvectors exist only for the point spectrum of an operator. For any other point of the spectrum one can only find a sequence of vectors for which $(A-\lambda I)u_n\to0$, where $A$ is said operator, and $\lambda$ is a point in the spectrum which is not an isolated point. So in this case there is a sequence of approximate eigenvectors. With a bit of extra details, if a point $\lambda$ comes from the continuous spectrum of the operator, there is no vector $\psi$ such that $A\psi=\lambda\psi$. For any $\epsilon>0$, though, you can choose a vector $\psi_\epsilon$ such that $\Vert(A-\lambda I)\psi_\epsilon\Vert<\epsilon$, hence $\psi_\epsilon$ is just an approximate eigenvector.
This can also be observed from the spectral theory for $A$, where there exists a projection-valued measure $E$ supported by the spectrum of such that
$$A=\int\limits_{\sigma(A)}\lambda\text dE(\lambda).$$
If $\lambda$ is an isolated point, then $E(\{\lambda\})$ is a non-zero projection which projects onto the $\lambda$-eigenspace, but if $\lambda$ is not isolated then such a projection is simply zero.
As for the decomposition of the function $S$, given a othonormal system of functions (say the eigenfunctions for the point spectrum of $A$), this can be completed to an orthonormal basis of the Hilbert space. Any function $S$ can then be decomposed w.r.t to such a basis. The only problem is that there is no clear meaning of the rest of the vectors in the basis for the operator $A$.
A: This doesn't directly answer your question of orthogonality, but may still address your concern.  
I need to point out that you seem to be working on a scattering problem and resonant states. Resonant wave functions DO NOT belong to the Hilbert space.  They are not even eigenfunctions of the Hamiltonian in an usual sense.  You already know that they do not represent stable states, and since they have a life time this should not come as a surprise.  They are obtained from special treatments, refer to specialized books for this topic.
And yes you should have a non-zero overlap with irregular wave functions if you are doing scattering problems. And irregular wave functions, since they contribute to scattering problems, are physical.   
