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The eigenfunctions of distinct eigenvalues for a Hermitian operator are proved to be orthogonal. Why does the same not apply to Legendre polynomials and functions that have different eigenvalues ?

http://en.wikipedia.org/wiki/Legendre_polynomials

Just for clarity, I am talking of orthogonality of non-polynomial solution and polynomial solution here.

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  • $\begingroup$ It says in the link "An important property of the Legendre polynomials is that they are orthogonal with respect to the $L_2$ inner product on the interval $−1 \le x \le 1$" $\endgroup$ Sep 7, 2014 at 20:49
  • $\begingroup$ @MarkBennet : please look at the edited post. $\endgroup$
    – cleanplay
    Sep 7, 2014 at 20:54

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An eigenfunction/eigenvalue problem involves boundary conditions, not just the ODE. The proof of orthogonality of eigenfunctions for different eigenvalues essentially relies on the boundary conditions.

Legendre's differential equation being singular at $\pm 1$, the boundary conditions are somewhat implicit: they amount to the statement that the solution $y$ must stay bounded at both endpoints. This could be recast as the Dirichlet boundary condition for the function $(1-x^2)y(x)$.

Legendre functions do not obey the aforementioned boundary condition: they blow up at the endpoints, as the linked article shows.

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  • $\begingroup$ The proof of orthogonality of eigenfunctions for different eigenvalues essentially relies on the boundary conditions. Can you explain further ? Thank you! $\endgroup$
    – cleanplay
    Sep 8, 2014 at 6:17
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    $\begingroup$ It involves integration by parts. And for that one needs to know what happens on the boundary. $\endgroup$
    – user147263
    Sep 8, 2014 at 10:35

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