# Orthogonal Polynomials and Sturm Liouville operators

Classical Orthogonal polynomials (e.g., Hermite, Legendre) are eigenfunctions of Sturm Liouville operators. For example, define $$L[u]=u''-xu'$$, then the $$n$$-th order Hermite polynomial satisfies $$LH_n (x) = -nH_n (x)$$.

The only proof of this miracle that I know of goes through solving the SL problem by power-series method, and on the other hand obtaining the orthogonal polynomials via Gram-Schmidt algorithm.

My question: I am going to give a talk about this property for graduate students. Is there another, perhaps a more "natural" explanation for this miracle?

(this is related to a previous question of mine, but not entirely the same)

It seems to be most usual to take a spectral theory viewpoint (and it would be odd to avoid it because it is also most natural). One can give a proof of the spectral theorem for compact self-adjoint operators and motivate the introduction of compact operators by talking briefly about Green's functions and their relation to differential operators (one can mention the ubiquitous example of the Laplacian with Dirichlet/Neumann BC's and its Green's function). After proving the spectral theorem one can spend a little time developing the corresponding operators for Sturm-Liouville equations to justify the properties needed in the theorem do in fact hold. Then one has demonstrated existence of an orthonormal basis of eigenfunctions for the differential operators (since they are eigenfunctions for the compact operator). It may take more than one talk to do it properly.

Edit: To show that certain orthogonal polynomials do in fact form an orthonormal basis of eigenfunctions is a bit more involving for many cases (demonstrating completeness for example). But it should be clear by inspection that they are in fact eigenfunctions, which leaves showing they are orthogonal and complete. This might be explained in a few computations for some specific examples. Of course, the above suggestions could be substituted with more computations if you are inclined to talk more about this. These suggestions are heavily motivated by some lecture notes from one of T. Ransford's mini-courses for grad students here. Starting on page 19 compact operators are briefly described, then the spectral theorem, then SL equations.

• This describes why orthogonal polynomials are closely related to the spectral theory of Jacobi difference operators, but the OP is interested in differential operators, and of course polynomials can never be eigenfunctions of these since they fail to lie in $L^2$. – Christian Remling Nov 27 '18 at 18:17
• @ChristianRemling they're not in $L^2 (\mathbb{R})$, but they are in $L^2 ([0,1])$. Isn't that enough? – Amir Sagiv Nov 27 '18 at 18:22
• But you can only obtain very special polynomials in this way (wrt Lebesgue measure, and there are further strong restrictions, from oscillation theory). The general story in connection with the spectral theorem really is that arbitrary orthogonal polynomials are associated with a Jacobi matrix, and they are orthogonal wrt its spectral measure. – Christian Remling Nov 27 '18 at 18:31
• @ChristianRemling Agreed that the reference that was attached does not address the cases that the OP addresses (for instance, for Hermite polynomials for which the corresponding operator is not of the form $-(py')'+qy$). However there are SL problems which have solutions which can be interpreted as eigenfunctions of an operator $y\rightarrow -(py')'+qy$ which form an orthonormal basis for $L^2$. A more general answer incorporating your observations might be useful. – Josiah Park Nov 27 '18 at 18:42
• @AmirSagiv P.S. The reference is sufficient for Legendre polynomials (which certainly do not fail to be in $L^2[-1,1]$). Also, by considering weights one can cover more interesting cases in the same framework. – Josiah Park Nov 27 '18 at 19:27

Let $$Lu=u''-xu'$$, and suppose $$Lu=\lambda u$$. Let $$v$$ be such that $$v'=u$$. By integrating the equation $$Lu=\lambda u$$, you find $$Lv=(\lambda-1)v+C$$, where $$C$$ is a constant. You can make this constant zero by choosing the integration constant in $$v$$ appropriately. Now observe that $$L(1)=0$$ and integrate successively.

• Hi, Thanks for your answer! I followed all the way, but lost you at the end - what happens when we integrate successively? – Amir Sagiv Nov 27 '18 at 18:12
• You start with $L(1)=0$, integrate and find $L(x)=-x$. Integrate again, you find $L(x^2/2)=-x^2+1$, hence $L(x^2/2-1/2)=-2(x^2/2-1/2)$. Continue in this fashion. – Michael Renardy Nov 28 '18 at 0:29