The orthogonality interval of a Sturm-Liouville problem Consider the Sturm-Liouville problem $(p(x)y)'+(q(x)+λ r(x))y =0$.

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*How to specify the interval (limits of integration) of orthogonality of eigenfunctions $y_m$ with respect to weight function $r$ (in the two cases: when the boundary conditions are given and the case in which there are no boundary conditions)?
Do the limits of integration depend on $p$? If yes, how? For example, why are eigenfunctions orthogonal only on interval $(-\infty, \infty)$ for the Hermite equation but they are orthogonal on interval $(0, \infty)$ only for the Laguerre equation?


*Is there a relation between the interval of convergence and the interval on which orthogonality holds? If we know one of them can we deduce the other?
 A: The interval over which the problem is considered is usually dictated by the underlying Physical problem. However, in order to study the general problem, you look at an interval $(a,b)$ which is finite, semi-infinite, or infinite. It is standard to consider intervals over which $p > 0$.
For example, the Legendre eigenvalue problem is
$$
                     -((1-x^{2})f')' = \lambda f,
$$
which is limited to $(-1,1)$. Here $p=1-x^{2}$, $q=0$, $r\equiv 1$. So the space is $L^2_r(-1,1) = L^2(-1,1)$ because $r\equiv 1$.
The problem $-(f')'=\lambda f$ can be considered on any finite, semi-infinite, or infinite interval $I$ because $p\equiv 1$. And $r\equiv 1$ means that the space used is $L^2(I)$.
The Bessel equation is
$$
                  x^{2}f''+xf'+\lambda x^{2}f =0,
$$
which is put into selfadjoint form by dividing by $x$:
$$
                  -(xf')' = \lambda xf.
$$
This equation has to do with the radial equation coming from separation of variables for the Laplacian in cylindrical coordinates. Problems inside a cylinder would correspond to $(0,r_0]$; problems outside would correspond to $[r_0,\infty)$ and annular regions would be $[r_0,r_1]$. The space considered here is $L^2_x(I)$, meaning that the inner product is $(f,g) =\int_{I}f\,\overline{g}xdx$. The weight function $x$ corresponds naturally to the radial weight in the cylindrical coordinate system: $dV = rdr d\phi dz$ ($x=r$ for Bessel's equation.)
To get a truly selfadjoint problem the operator $L$ in question must be restricted to a domain $\mathcal{D}(L)$ where the functions satisfy endpoint conditions at the ends of the interval. Some problems do not require endpoint conditions at one or both of the endpoints. Others require one condition. And, there are periodic conditions that are allowed as well. The number of conditions has to do with whether the equation is in the limit-point case or the limit-circle case at the endpoint; this language comes out of an old theorem proved by H. Weyl concerning the dimension of space of eigenfunctions with non-real eigenvalues that are in $L^2$ near an endpoint.
All of the regular problems where $p \ge \epsilon > 0$ on a finite interval $[a,b]$ require conditions such as $Af(a)+Bf'(a)=0$ or $Cf(b)+Df'(b)=0$. The Legendre equation is singular, the conditions have to do numbers in asymptotic expansions of functions near $\pm 1$. Every $f$ in the unrestricted domain has an asymptotic because near $x=1$ given by $f \approx \alpha + \beta\ln(1-x)$. The condition in this case has the form $A\alpha +B\beta = 0$. One typically requires $\beta = 0$, which is equivalent to requiring the functions in the domain to be bounded near the endpoints.
Once all of the require conditions have been imposed to obtain a selfadjoint problem, then
$$
                (Lf,g)_r = (f,Lg)_r, \;\;\; f,g\in \mathcal{D}(L),
$$
where the domain consists of functions $f$ satisfying the endpoint conditions. Once you imposed real conditions, then the symmetry implies that eigenfunctions corresponding to different eigenvalues are orthogonal, just as it does for matrices, because
$$
    Lf = \lambda f, \;\; Lg = \mu g \implies (\lambda-\mu)(f,g)_r = (Lf,g)_r-(f,Lg)_r = 0,
$$
which either implies that $\lambda=\mu$ or $(f,g)_r= 0$. Historically, symmetry arguments of this type came out of studying such ODEs, and worked its way down to matrices.
As a final note, the Hermite equation on $(-\infty,\infty)$ requires no endpoint conditions at either endpoint because the equation is in the limit point case at both endpoints. No conditions are possible. The assumption that $f \in L^2$ and $Lf = -f''+x^2f \in L^2$ is enough to guarantee that the integration by parts evaluation terms must be $0$, giving $(Lf,g)=(f,Lg)$ for all $f,g \in \mathcal{D}(L)$. This problem is selfadjoint without endpoint conditions.
