Suppose that for constant $A_n (n=1,2,3,...)$, we have \begin{equation} A_1\,\sin(\frac{\pi\,x}{L})+ A_2\,\sin(\frac{2\,\pi\,x}{L})+ A_3\,\sin(\frac{3\,\pi\,x}{L})+ ...+ A_n\,\sin(\frac{n\,\pi\,x}{L})+ ... =0 \end{equation} for $x\in (0,L)$. Could we conclude that $A_n=0$, $n=1,2,3,...$? I've tried to multiply both sides by $\sin(\frac{n\,\pi\,x}{L})$ and then integrate from $0$ to $L$, and found $a_n=0$. Is this enough to conclude the desired result?
Any reference, suggestion, idea, or comment is welcome. Thank you!
This is a consequence of Uniqueness of Representation by Trigonometric Series which states that if $$\frac{1}{2} a_0 + \sum_{n=0}^\infty \left(a_n \cos nx + b_n \sin nx\right)$$ converges everywhere to $0$, then all its coefficients are zero. This is a non trivial theorem. See for example the book of Sygmund - Trigonometric series, page 326.
The issue of "your proof" in the original question is that you can't permute the $\sum$ and $\int$ without proper assumptions on the series $A_1\,\sin(\frac{\pi\,x}{L})+ A_2\,\sin(\frac{2\,\pi\,x}{L})+ A_3\,\sin(\frac{3\,\pi\,x}{L})+ ...+ A_n\,\sin(\frac{n\,\pi\,x}{L})+ ...$.