# Do the Fourier sine and cosine series individually a complete basis by Sturm Liouville theorem?

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.

• 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/…
– yoyo
Jul 7, 2023 at 2:29

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 $$f \in L^2[0,\pi]$$.
• +1. Should it be $\pi$ instead of $2\pi$ ? Jul 8, 2023 at 2:56
• @Balajisb : Thank you for the correction. Yes it should be $\pi$ instead of $2\pi$. I changed my answer. Jul 8, 2023 at 18:02