Completeness of the Dirichlet sine system on an interval

[Orthonormality] For $n,m\ge 1$, [F3] gives $2\sin (n\pi x\mathrm{/}L)\sin (m\pi x\mathrm{/}L)=\cos ((n-m)\pi x\mathrm{/}L)-\cos ((n+m)\pi x\mathrm{/}L)$. If $n\mathrm{\ne }m$, then $n-m\mathrm{\ne }0$ and $n+m\mathrm{\ne }0$, and by [F3] ${\int }_0^L\cos \text{\hspace{0.17em}⁣}\left(\frac{k\pi x}{L}\right)dx=\frac{L}{k\pi }(\sin (k\pi )-\sin 0)=0(k\mathrm{\ne }0),$ so $⟨e_n,e_m⟩=0$. If $n=m$, the first cosine is $1$ and ${\int }_0^L(1-\cos (2n\pi x\mathrm{/}L))dx=L$, so $⟨e_n,e_n⟩=(2\mathrm{/}L)\cdot (L\mathrm{/}2)=1$. Hence $(e_n)$ is orthonormal.

[Reduction to real-valued $f$] Let $f\in L^2(0,L;\mathbb{C})$ satisfy $⟨f,e_n⟩=0$ for all $n\ge 1$. By [F6] write $f=f_1+i{f}_2$ with real-valued $f_1,f_2\in L^2$; since each $e_n$ is real-valued, $0=⟨f,e_n⟩=⟨f_1,e_n⟩+i⟨f_2,e_n⟩$, so both real parts have vanishing sine coefficients. It suffices to prove $f_1=f_2=0$; hence assume from now on that $f$ is real-valued with ${\int }_0^{L}f(x)\sin (n\pi x\mathrm{/}L)dx=0$ for all $n\ge 1$.

[Odd extension] Define $g:(-L,L)\to \mathbb{R}$ by $g(x)=f(x)$ for $x\in (0,L)$, $g(x)=-f(-x)$ for $x\in (-L,0)$, and $g(0)=0$. Then $g$ is measurable, odd, and ${\int }_{-L}^L\mathrm{\mid }g{\mathrm{\mid }}^2=2{\int }_0^L\mathrm{\mid }f{\mathrm{\mid }}^2<\mathrm{\infty }$, so $g\in L^2(-L,L)$ with exactly the same class as the odd extension. For every $n\in \mathbb{Z}$, ${\int }_{-L}^{L}g(x)e^{-i\pi nx\mathrm{/}L}dx=0.$ Indeed, writing $e^{-i\pi nx\mathrm{/}L}=\cos (n\pi x\mathrm{/}L)-i\sin (n\pi x\mathrm{/}L)$, the product $g\cos (n\pi x\mathrm{/}L)$ is odd, so its integral vanishes, while $g\sin (n\pi x\mathrm{/}L)$ is even, so its integral equals $2{\int }_0^{L}f(x)\sin (n\pi x\mathrm{/}L)dx$, which is $0$ for $n\ge 1$ by hypothesis and also for $n\le -1$ by oddness of sine; for $n=0$ the sine factor vanishes.

[Transport to the torus] By [F5], choose a Borel representative of the real-valued class $g$ and assign any values at the endpoints $\pm L$, which does not change its $L^2$ class. Define $G:\mathbb{T}\to \mathbb{C}$ using the unique representative $t\in [0,1)$ of each torus point by $G([t]):=g(2Lt-L)$. The representative map $[t]\mapsto t\in [0,1)$ is Borel: the quotient map restricts to a homeomorphism from $(0,1)$ onto $\mathbb{T}\setminus \{[0]\}$, and the remaining singleton is Borel; hence $G$ is measurable. The affine integral identity [F4] gives [step 2.1, F1, F4, F5, algebra] ${\int }_{\mathbb{T}}\mathrm{\mid }G{\mathrm{\mid }}^{2}d{m}_{\mathbb{T}}=\frac{1}{2L}{\int }_{-L}^L\mathrm{\mid }g(x){\mathrm{\mid }}^{2}dx<\mathrm{\infty }\mathrm{.}$ For $k\in \mathbb{Z}$, the substitution $x=2Lt-L$ gives $e^{-2\pi ikt}=e^{-i\pi k}{e}^{-i\pi kx\mathrm{/}L}=(-1{)}^k{e}^{-i\pi kx\mathrm{/}L}$; hence $\widehat{G}(k)=\frac{(-1{)}^k}{2L}{\int }_{-L}^{L}g(x)e^{-i\pi kx\mathrm{/}L}dx=0$ by step 2.1. The harmless phase factor is retained explicitly.

[Conclusion] By [F1] the characters form an orthonormal basis of $L^2(\mathbb{T})$, so by [F2] a vector with all Fourier coefficients zero is zero: $G=0$ in $L^2(\mathbb{T};\mathbb{C})$. Since $\phi (t)=2Lt-L$ transports the fundamental interval onto $[-L,L)$ and the norm identity in step 3.1 is exact, $g=0$ almost everywhere on $(-L,L)$, hence $f=0$ almost everywhere on $(0,L)$. Therefore no nonzero class is orthogonal to all $e_n$, and [F2] makes $(e_n{)}_{n\ge 1}$ complete. With step 1.1 this proves the statement.