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Fourier orthogonality isolates a coefficient. Multiplication by $\sin (mx)$ and integration over $[0,2\pi ]$ annihilates the constant, all cosine terms and all sine terms except the one with $n=m$, whose squared integral is $\pi$. Thus the corresponding coefficients on the two sides must agree for each positive integer $m$.

Multiplication by $\cos (mx)$ works in the same way, while integration against the constant mode extracts $a_0$. This is the uniqueness of the Fourier coefficients; it does not follow from comparing a few point values of the series.

Assume $L>0$. For any nonzero integer $k$,For $m\ne n$, the product-to-sum identity expresses the integrand as one half the sum of cosines with nonzero frequencies $m-n$ and $m+n$, so the integral is zero. For $m=n>0$, use ${\cos }^{2}u=(1+\cos 2u)/2$ to get $L$. Finally, for $m=n=0$ the integrand is one. The complete Fourier orthogonality formula isThe constant mode has twice the squared norm of a positive-frequency cosine mode.

With the standard real normalization, the Fourier series of a period-$2L$ function iswhereFor the partial sum $S_N$, Fourier orthogonality eliminates all cross terms. The constant term contributes $2L(a_0/2{)}^2$, while every squared sine or cosine integrates to $L$, givingFor a square-integrable function, completeness of the trigonometric orthogonal basis in the Hilbert space $L^2[-L,L]$ makes $S_N\to f$ in norm. Passing to the limit proves Parseval's identity:Square integrability is the needed hypothesis here; periodicity alone is not an integrability or convergence condition.

With the usual real normalization, the Fourier series iswith Fourier coefficientsThese formulas follow from Fourier orthogonality. Integrability is needed to define the coefficients, and periodicity alone does not guarantee pointwise equality with the series. For a piecewise smooth periodic function, the Dirichlet-Jordan convergence theorem gives the average of the one-sided limits at a jump and the function value at a continuity point.

Use the product-to-sum formula and the double-angle trigonometric identities:By Fourier orthogonality, these are already its only nonzero Fourier series coefficients:

Assume $f\in L^2[-L,L]$, so the squared-error integral is finite. WriteAt a minimum, differentiating with respect to each coefficient givesBy Fourier orthogonality, the constant has squared norm $2L$, each nonconstant cosine has squared norm $L$, and all distinct modes are orthogonal. ThusThese are exactly the Fourier coefficients of the Fourier cosine series of the even function. To check that the stationary choice is a unique global minimum, perturb the coefficients by $\delta a_n$. The residual is orthogonal to every fitted mode, so the cross terms vanish andwith equality only when every perturbation is zero. The truncated series is the orthogonal projection onto a finite-dimensional subspace spanned by the constant and the first $N$ cosine modes.

Take the inner product of the Fourier series with $\sin (n\pi x/L)$. Fourier orthogonality eliminates the constant, all cosine terms, and all sine terms except the $n$th. The remaining squared norm issoFor a square-integrable function, the calculation is justified by taking inner products of finite Fourier partial sums and passing to their mean-square limit.

Multiply by the $(m,n)$ sine product and integrate over the square. Fourier orthogonality gives squared norm ${\ell }^2/4$, and thusThereforeThis double Fourier sine series equals one at interior points. At boundary points every summand vanishes, consistent with the endpoint value of the sine extension.

The Fourier orthogonality relation on this angular interval isProjecting the outer boundary condition onto each angular eigenfunction givesThe solution is unique in the class of bounded harmonic functions with the prescribed boundary trace. For continuous boundary data, the difference of two solutions is zero on the boundary and bounded at the vertex, so the maximum principle for harmonic functions makes it zero throughout the sector. Equivalently all its Fourier sine series coefficients vanish. The usual convergence interpretation applies if the data are only piecewise smooth.

The periodic boundary condition requires $e^{ikM}=1$, so there are exactly $M$ independent wavevectors in a Brillouin zone:The inverse discrete Fourier transform, with the sign convention of the paper, isFourier orthogonality gives the tight binding dispersion relation ${ϵ}_k=-2t\cos k$. For the interaction, first use the canonical anticommutation relations at distinct sites to write $n_{\sigma ,j}{n}_{\sigma ,j+1}=c_{\sigma ,j}^{†}{c}_{\sigma ,j+1}^{†}{c}_{\sigma ,j+1}{c}_{\sigma ,j}$. Summing its discrete Fourier transform over $j$ enforces crystal momentum conservation, givingEvery wavevector is reduced modulo $2\pi$. The phase $e^{iq}$ comes from the displacement of the second site by one lattice spacing. Equivalently replace $e^{iq}$ by $\cos q$: exchanging the two incoming and the two outgoing fermions changes $q$ to $-q$ with two cancelling fermionic signs. Thus the usual interaction vertex is $2V\cos q$ accompanied by the factor $1/(2M)$.

Set $\omega =2\pi /T$ and $q=m+n$. If $q=0$, the integrand is one and its integral is $T$. If $q$ is a nonzero integer,ThusThis is Fourier orthogonality in its unconjugated form. With a complex conjugate on the second exponential, the nonzero case would instead be $m=n$.

Multiply the complex Fourier series by $e^{-i\omega rt}$ and integrate over one period. Fourier orthogonality removes every term except the one with index $r$, givingFor general square-integrable functions, this is an orthogonal projection onto the corresponding Fourier mode. It remains valid when the expansion converges only in L2 space: integrate finite Fourier partial sums first, then pass to the limit using the Cauchy-Schwarz inequality. No unjustified pointwise exchange of an infinite series and an integral is needed.

At $t=0$, the displacement expansion is $f(x)={\sum }_n{A}_n\sin (n\pi x/L)$. Fourier orthogonality givesDifferentiating the modal solution at zero gives $g(x)={\sum }_n{\omega }_n{B}_n\sin (n\pi x/L)$, soThe factor $1/{\omega }_n$ is required because the derivative of $\sin ({\omega }_{n}t)$ at zero is ${\omega }_n$. The dimensions agree: $A_n$ and $B_n$ are displacements, while the sine coefficients of $g$ are velocities. These projections also establish uniqueness within the finite-energy solution class.

For the unitary group, the diagonal maximal torus has coordinates $z_i=e^{i{\theta }_i}$ and Weyl group $S_n$. The Weyl integration formula weights independent uniform angles by $n{!}^{-1}{\prod }_{i<j}|z_i-z_j{|}^2$. The Vandermonde determinant $\det (z_i^{n-j})$ has squared torus integral $n!$, by Fourier orthogonality. Multiplying an irreducible character of highest weight $\lambda$ by this determinant gives an alternating Laurent polynomial. Its leading alternant has coefficient one; character orthogonality for compact groups forces every other alternant coefficient to vanish. This yields the quotient $\det (z_i^{{\lambda }_j+n-j})/\det (z_i^{n-j})$.