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I am confused by the Sturm-Liouville theorem implication on the Fourier sine and cosine series.

Consider the simple ODE $$y''(x)=-k^2 y(x).$$ It is in Sturm-Liouville form. Let’s impose a boundary condition of $y(0)=0$ and $y(\pi)=0$. The eigen solutions will be $\sin(k x)$.

From the wiki page on the Sturm-Liouville theorem, it is said that “The normalized eigenfunctions form an orthonormal basis under the $w$-weighted inner product in the Hilbert space”.

My first question is: does that mean I can represent any smooth function, in the $(-\pi,\pi)$ region with convergence except at simple discontinuity points?

My second question may seem silly. Consider the function $\sin^2(x)$, it satisfies the boundary conditions above, but it seems cannot be represented by the Fourier sine series. It is obviously represented by a Fourier cosine $1/2-\cos(2x)/2$!

Push that further, if the Fourier sine series is already a complete basis supported by the Sturm-Liouville theorem. How do we represent an even function with a Fourier sine series? The coefficients in tue Fourier sine series evaluated are all zero.

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    $\begingroup$ There are no "even functions" on $(0,\pi)$. Extend a function on $(0, \pi)$ to an odd function on $(-\pi, \pi)$ and consider what you already know about Fourier series to see that $\sin(kx)$ is a basis. This is on the wiki(pedia) page I assume you're referencing: en.wikipedia.org/wiki/… $\endgroup$
    – yoyo
    Jul 7, 2023 at 2:29

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You are discussing conditions at $0$, $\pi$, which means you are looking at convergence in the interval $[0,\pi]$. Sturm-Liouville requires that.

For real $\alpha,\beta$, if you impose general linear endpoint conditions on solutions $f$ of $-f''=\lambda f$ of the form $$ \cos\alpha f(0)+\sin\alpha f'(0) = 0,\\ \cos\beta f(\pi)+\sin\beta f'(\pi)= 0, $$ then the resulting eigenfunctions form an orthogonal basis that can be used to expand a function $f \in L^2[0,\pi]$.

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    $\begingroup$ +1. Should it be $\pi$ instead of $2\pi$ ? $\endgroup$
    – Balaji sb
    Jul 8, 2023 at 2:56
  • $\begingroup$ @Balajisb : Thank you for the correction. Yes it should be $\pi$ instead of $2\pi$. I changed my answer. $\endgroup$ Jul 8, 2023 at 18:02
  • $\begingroup$ I believe your comment is for Wong (the one who posted the question ?) $\endgroup$
    – Balaji sb
    Jul 14, 2023 at 20:17
  • $\begingroup$ @Balajisb : I apologize for that. $\endgroup$ Jul 14, 2023 at 20:44

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