Eigenfunctions corresponding to different eigenvalues of the Sturm-Liouville problem below are orthogonal w.r.t the inner product $\langle u,v\rangle=\int_{a}^{b} u\, \overline{v}\,r dx $. The problem I am referring to is: $$ ( p(x)v'(x) )' + q(x)v(x) + \lambda r(x)v(x) =0, \, \, \, v(a)=v(b)=0. $$ It is assumed that $p,q,r$ are real functions, and that $p,r$ are positive.
My question is effectively the following:
I know how to prove that if $u,v$ are eigenfunctions corresponding to different eigenvalues then $\int_{a}^{b} u vr dx=0 $ but don't know how to deduce that $\int_{a}^{b} u \overline{v}r dx =0$.
Details: The proof of this usually starts (see e.g. here example) by proving the Sturm-Liouville operator $Lv=(pv')'+qv$ is self-adjoint with respect to the inner product $\int_{a}^b fg dx$, i.e. $\int_{a}^b vLu dx = \int_a^b uLv dx$. Then, the argument is as follows, if $\lambda,\mu$ are eigenvalues of $v,u$ respectively, then $$ vLu=-\lambda urv $$ $$ uLv=-\mu urv $$ which implies that $$ (\mu-\lambda)\int_a^b urv dx =0 \Rightarrow \int_a^b uvr dx =0. $$ Unfortunately, I cannot understand how does this imply that also $\int_a^b u\overline{v}r dx =0$.
I keep finding the same argument everywhere, and it seems like I am missing something really trivial that allows one to move from $\int_a^b uvr dx =0$ to $\int_a^b u\overline{v}r dx =0$. Will be happy to receive some help in this one
If $v$ is an eigenfunction with eigenvalue $\lambda$, then $\overline{v}$ is an eigenfunction with eigenvalue $\overline{\lambda}$. You can see this by conjugating the equation and the endpoint conditions. You have shown that, if $u,v$ are eigenfunctions with different eigenvalues, then $$ \int_a^b uvrdx=0 $$ Keep in mind that, if $v$ is an eigenfunction with eigenvalue $\lambda$, then $\overline{v}$ is an eigenfunction with eigenvalue $\overline{\lambda}$. This is true because the coefficient functions in your equation are all real (all you have to do is conjugate your eigenfunction equation and endpoint conditions.) So you can replace $v$ by $\overline{v}$ in order to instead conclude that $\int_a^b u\overline{v}rdx=0$.