The sine and cosine systems are complete orthogonal systems of L^2 on an interval and on a rectangle

[One-interval inner products] Let $m,m^{\mathrm{\prime }}\ge 0$ and $p,p^{\mathrm{\prime }}\ge 1$ be integers. By [F6], $\cos \frac{m\pi x}{L}\cos \frac{m^{\mathrm{\prime }}\pi x}{L}=\frac{1}{2}[\cos \frac{(m-m^{\mathrm{\prime }})\pi x}{L}+\cos \frac{(m+m^{\mathrm{\prime }})\pi x}{L}]$ and $\sin \frac{p\pi x}{L}\sin \frac{p^{\mathrm{\prime }}\pi x}{L}=\frac{1}{2}[\cos \frac{(p-p^{\mathrm{\prime }})\pi x}{L}-\cos \frac{(p+p^{\mathrm{\prime }})\pi x}{L}]$; for every nonzero integer $k$ the function $x\mapsto \frac{L}{k\pi }\sin \frac{k\pi x}{L}$ has derivative $\cos \frac{k\pi x}{L}$ on $[0,L]$ by [F5], so ${\int }_0^L\cos \frac{k\pi x}{L}dx=\frac{L}{k\pi }[\sin (k\pi )-\sin 0]=0$ by [F5], [F6] (the case $k=0$, i.e. $m=m^{\mathrm{\prime }}$ or $p=p^{\mathrm{\prime }}$, leaves the constant $1$ or the difference $0$). The integrands are continuous, hence Riemann integrable [F5], and their Riemann and Lebesgue integrals agree [F5]. Therefore $⟨c_m,c_{m^{\mathrm{\prime }}}⟩=0$ for $m\mathrm{\ne }{m}^{\mathrm{\prime }}$, $⟨s_p,s_{p^{\mathrm{\prime }}}⟩=0$ for $p\mathrm{\ne }{p}^{\mathrm{\prime }}$, while $⟨c_m,c_m⟩=\frac{L}{2}(1+{\delta }_{m0})$ and $⟨s_p,s_p⟩=\frac{L}{2}$ are nonzero; each displayed member is nonzero because it equals $1$ at some point of $[0,L]$. Hence both families of clause (a) are orthogonal in the sense of [F1].

[Even extensions have vanishing torus coefficients] Let $f\in L^2(0,L)$ satisfy $⟨f,c_m⟩=0$ for every $m\ge 0$. Define $F(x):=f(\mathrm{\mid }x\mathrm{\mid })$ on $(-L,L)$ and let $G$ be the $2L$-periodic extension of $F$ to $\mathbb{R}$; define $g([t]):=G(2Lt)$ on $\mathbb{T}$. Then $g$ is a real-valued element of $L^2(\mathbb{T};\mathbb{C})$: by [F1], [F3] the change of variables $x\mapsto 2Lt$ and the periodicity give ${\int }_{\mathbb{T}}\mathrm{\mid }g{\mathrm{\mid }}^{2}d{m}_{\mathbb{T}}={\int }_0^1\mathrm{\mid }G(2Lt){\mathrm{\mid }}^{2}dt=\frac{1}{2L}{\int }_0^{2L}\mathrm{\mid }G{\mathrm{\mid }}^2=\frac{1}{2L}{\int }_{-L}^L\mathrm{\mid }F{\mathrm{\mid }}^2=\frac{1}{L}\mathrm{\parallel }f{\mathrm{\parallel }}_2^2<\mathrm{\infty }$. Splitting the integral at $0$, substituting $x\mapsto -x$ on $(-L,0)$ by [F3] and using $F(-x)=F(x)$, its character coefficients are $\widehat{g}(k)={\int }_0^{1}G(2Lt)e^{-2\pi ikt}dt=\frac{1}{2L}{\int }_{-L}^{L}F(x)e^{-i\pi kx\mathrm{/}L}dx=\frac{1}{2L}{\int }_0^{L}f(x)(e^{-i\pi kx\mathrm{/}L}+e^{i\pi kx\mathrm{/}L})dx=\frac{1}{L}{\int }_0^{L}f(x)\cos \frac{k\pi x}{L}dx=\frac{1}{L}⟨f,c_{\mathrm{\mid }k\mathrm{\mid }}⟩=0$ for every $k\in \mathbb{Z}$, using the affine substitutions, the periodicity of $G$, the cartesian form of [F2] and the parity of [F6] for the identification with $\cos \frac{k\pi x}{L}=\cos \frac{\mathrm{\mid }k\mathrm{\mid }\pi x}{L}$.

[Trivial orthogonal complement in $L^2$] We record the Hilbert-space step used below. Let $H$ be a real or complex Hilbert space, let $V\subseteq H$ be the closed linear span of a family $(v_i{)}_{i\in I}$, and suppose the only $w\in H$ with $⟨w,v_i⟩=0$ for every $i$ is $w=0$. Then $V=H$: for $u\in H$ put $d:={\inf }_{v\in V}\mathrm{\parallel }u-v\mathrm{\parallel }$; under [F2] (countable choice) choose $v_n\in V$ with $\mathrm{\parallel }u-v_n\mathrm{\parallel }\to d$, and the parallelogram identity $\mathrm{\parallel }{v}_n-v_m{\mathrm{\parallel }}^2=2\mathrm{\parallel }u-v_n{\mathrm{\parallel }}^2+2\mathrm{\parallel }u-v_m{\mathrm{\parallel }}^2-4\mathrm{\parallel }u-\frac{v_n+v_m}{2}{\mathrm{\parallel }}^2\le 2\mathrm{\parallel }u-v_n{\mathrm{\parallel }}^2+2\mathrm{\parallel }u-v_m{\mathrm{\parallel }}^2-4d^2$ of [F1] makes $(v_n)$ Cauchy, so $v_n\to v_{\ast }\in V$ by closedness and completeness; if $d>0$ then $z:=u-v_{\ast }\mathrm{\ne }0$ and for every $w\in V$ and real $\epsilon$ one has $d^2\le \mathrm{\parallel }z-\epsilon w{\mathrm{\parallel }}^2=d^2-2\epsilon \mathrm{Re}⟨z,w⟩+{\epsilon }^2\mathrm{\parallel }w{\mathrm{\parallel }}^2$, so taking $\epsilon =\mathrm{Re}⟨z,w⟩\mathrm{/}\mathrm{\parallel }w{\mathrm{\parallel }}^2$ for $w\mathrm{\ne }0$ forces $\mathrm{Re}⟨z,w⟩=0$; when $H$ is complex, applying that conclusion to $iw$ in place of $w$ gives $\mathrm{Im}⟨z,w⟩=0$ as well, so $⟨z,w⟩=0$ in either case; in particular $⟨z,v_i⟩=0$ for all $i$, contradicting the hypothesis, so $d=0$ and $u\in V$.

[Torus elements with all coefficients zero vanish] If $h\in L^2(\mathbb{T};\mathbb{C})$ has $\widehat{h}(k)=0$ for every $k$, then $h=0$: by [F2] the characters are complete orthonormal, so Cauchy–Schwarz [F1] makes $⟨h,\phi ⟩=0$ for every $\phi$ in their closed span, which is all of $L^2(\mathbb{T};\mathbb{C})$, and $⟨h,h⟩=0$ gives $h=0$.

[Odd extensions have vanishing torus coefficients] Let $f\in L^2(0,L)$ satisfy $⟨f,s_p⟩=0$ for every $p\ge 1$. Define $F(x):=\mathrm{sgn}(x)f(\mathrm{\mid }x\mathrm{\mid })$ (so $F$ is odd and $F=f$ on $(0,L)$) and define $G,g$ from $F$ as in step 1.2. The same change of variables and periodicity give $\mathrm{\parallel }g{\mathrm{\parallel }}^2=\frac{1}{L}\mathrm{\parallel }f{\mathrm{\parallel }}_2^2<\mathrm{\infty }$, and now the reflection substitution yields $\widehat{g}(k)=\frac{1}{2L}{\int }_{-L}^{L}F(x)e^{-i\pi kx\mathrm{/}L}dx=\frac{1}{2L}{\int }_0^{L}f(x)(e^{-i\pi kx\mathrm{/}L}-e^{i\pi kx\mathrm{/}L})dx=-\frac{i}{L}{\int }_0^{L}f(x)\sin \frac{k\pi x}{L}dx$; by the parity of sine this is $-\frac{i\mathrm{sgn}(k)}{L}⟨f,s_{\mathrm{\mid }k\mathrm{\mid }}⟩=0$ for $k\mathrm{\ne }0$ and $0$ for $k=0$.

[Interval completeness, cosine case] Let $f\in L^2(0,L)$ be orthogonal to every $c_m$. Step 1.2 produces $g\in L^2(\mathbb{T};\mathbb{C})$ whose character coefficients all vanish, so $g=0$ by step 1.4, and hence $G(2Lt)=0$ for almost every $t\in [0,1)$. By [F3] (the linear map $t\mapsto 2Lt$ scales Lebesgue measure by $2L$) the set of $x\in (0,2L)$ with $G(x)\mathrm{\ne }0$ is null, so $F$, hence $f$, vanishes almost everywhere on $(0,L)$. Thus the only element of $L^2(0,L)$ orthogonal to the cosine family is $0$; by step 1.3 with the family $(c_m)$ the closed span of the cosine family is $L^2(0,L)$, and step 1.1 gives orthogonality.

[Rectangle orthogonality] Let $c$ denote either of the two $x_1$-families and $d$ either of the two $x_2$-families of clause (b); all members are continuous, hence bounded and measurable on the closed rectangle. For members $c_m\otimes d_n$ and $c_{m^{\mathrm{\prime }}}\otimes d_{n^{\mathrm{\prime }}}$, the product $(c_m{c}_{m^{\mathrm{\prime }}})(d_n{d}_{n^{\mathrm{\prime }}})$ is bounded and measurable on the finite-measure rectangle, hence in $L^1$, and Fubini [F4] factors the inner product as $⟨c_m\otimes d_n,c_{m^{\mathrm{\prime }}}\otimes d_{n^{\mathrm{\prime }}}⟩=⟨c_m,c_{m^{\mathrm{\prime }}}⟩⟨d_n,d_{n^{\mathrm{\prime }}}⟩$; this vanishes whenever $(m,n)\mathrm{\ne }(m^{\mathrm{\prime }},n^{\mathrm{\prime }})$ by step 1.1 (a member with a matching index has nonzero squared norm $\frac{L_1}{2}(1+{\delta }_{m0})$ or $\frac{L_1}{2}$, or the analogous $x_2$-norm). Hence each of the four product families is orthogonal in $L^2(R)$.

[Interval completeness, sine case] The same argument with step 2.1 in place of step 1.2 shows that the only element of $L^2(0,L)$ orthogonal to the sine family is $0$, so its closed span is $L^2(0,L)$ by step 1.3, and it is orthogonal by step 1.1; this proves clause (a) together with step 2.2.

[Rectangle completeness] Fix one of the four product families $c\otimes d$ in $L^2(R)$ and let $f\in L^2(R)$ be orthogonal to every $c_m\otimes d_n$. By Fubini applied to the integrable function $\mathrm{\mid }f{\mathrm{\mid }}^2$ [F4], the section $f(\cdot ,x_2)$ lies in $L^2(0,L_1)$ for almost every $x_2$, and the identities of [F4] applied to the integrable function $f{c}_m$ define measurable functions ${\phi }_m(x_2):={\int }_0^{L_1}f(x_1,x_2)c_m(x_1)d{x}_1$ almost everywhere, with ${\int }_0^{L_2}\mathrm{\mid }{\phi }_m{\mathrm{\mid }}^2\le \mathrm{\parallel }{c}_m{\mathrm{\parallel }}_2^2\mathrm{\parallel }f{\mathrm{\parallel }}_2^2<\mathrm{\infty }$ by Cauchy–Schwarz [F7], so ${\phi }_m\in L^2(0,L_2)$; and Fubini applied to $f{c}_m{d}_n$ gives $⟨{\phi }_m,d_n⟩=⟨f,c_m\otimes d_n⟩=0$ for every $n$. Since the $d$-family is complete in $L^2(0,L_2)$ by step 2.2 or step 3.1, Cauchy–Schwarz [F1] makes ${\phi }_m$ orthogonal to the whole closed span, so ${\phi }_m=0$ almost everywhere. There are countably many $m$, so [F2] (countable choice) selects the countably many exceptional null sets whose union $N$ is null, and for every $x_2\mathrm{\notin }N$ the section $f(\cdot ,x_2)\in L^2(0,L_1)$ is orthogonal to every $c_m$, hence vanishes almost everywhere in $x_1$ by step 2.2 or step 3.1. Fubini applied to $\mathrm{\mid }f{\mathrm{\mid }}^2$ [F4] now gives $\mathrm{\parallel }f{\mathrm{\parallel }}_2^2=0$, so $f=0$. By step 1.3 the closed span of the product family is $L^2(R)$; with step 2.3 this proves the first claim of clause (b) for all four parity choices, and the identification of [F9] gives the claim for the Neumann family $\{h_{mn}\}$ and the Dirichlet family $\{e_{mn}\}$.

[Expansions] For each system $(u_i)$ of clauses (a) and (b), put $v_i:=u_i\mathrm{/}\mathrm{\parallel }{u}_i{\mathrm{\parallel }}_2$; the norms are nonzero and $(v_i)$ is a complete orthonormal system of [F1]. By [F8], every $f$ in the space is the norm limit of the finite-subset net ${\sum }_{i\in F}⟨f,v_i⟩v_i$. Since $⟨f,v_i⟩v_i=(⟨f,u_i⟩\mathrm{/}\mathrm{\parallel }{u}_i{\mathrm{\parallel }}_2^2)u_i$, the corresponding coefficients in the original orthogonal system are $⟨f,u_i⟩\mathrm{/}\mathrm{\parallel }{u}_i{\mathrm{\parallel }}_2^2$, obtained by dividing the normalized coefficients by $\mathrm{\parallel }{u}_i{\mathrm{\parallel }}_2$.