Orthogonality Relations

The Fourier series of an integrable real function can be understood as the expansion of that real function in the Fourier basis of functions, which consists of the set of functions

$$\left\{ \rule{0em}{2.5ex} 1, \cos(k\theta), \sin(k\theta), k\in\{1,2,3,\ldots,\infty\} \right\}.$$ (13)

Let us now show that this is an orthogonal basis. Of course this can be done using the standard form of the scalar product between two real functions on the unit circle, by simply calculating a set of easy integrals by elementary means. However, what we want to do here is to show that both the form of the scalar product and the relations of orthogonality and norm are contained within the structure of the inner analytic functions, and can be derived from that structure. In fact, we will show that these elements can be obtained from a particular set of functions, the powers $z^{k}$, with $k\geq 0$, and their multiplicative inverses $z^{-k}$. We start by noting that, if $C$ is any circle centered at the origin, including the unit circle, then from the residues theorem we have that

$$\frac{1}{2\pi\mbox{\boldmath$\imath$}} \oint_{C}dz\, z^{p-1} = \delta_{p,0},$$ (14)

where $p$ is an arbitrary integer, and where $\delta_{p,0}$ is the Kronecker delta. This is so because the integral can be calculated by residues, and a function which is a simple power, either positive or negative, is its own Laurent series, which has only one term. Therefore, its residue at $z=0$ is zero unless $p=0$, in which case it is equal to one. Using this result for the case $p=k-k'$, where the integers $k$ and $k'$ are in the set $\{0,1,2,3,\ldots,\infty\}$, we have

$$\frac{1}{2\pi\mbox{\boldmath$\imath$}} \oint_{C}dz\, z^{k-k'-1} = \delta_{k,k'},$$ (15)

while using the same expression for $p=k+k'$, with the limitation that $k+k'>0$, which means that $k$ and $k'$ cannot both be zero, we have

$$\frac{1}{2\pi\mbox{\boldmath$\imath$}} \oint_{C}dz\, z^{k+k'-1} = 0.$$ (16)

This is also a consequence of the Cauchy-Goursat theorem, since in this case the integrand is analytic within the unit disk. Note that the power $z^{k}$ with $k\geq 0$ is itself an inner analytic function. Writing these two relations in terms of the integration variable $\theta$ we have

$$\frac{1}{2\pi}\, \rho^{k-k'} \int_{-\pi}^{\pi}d\theta\, \,{\rm e}... ...tsize$\imath$}k\theta} \,{\rm e}^{-\mbox{\boldmath\scriptsize$\imath$}k'\theta} = \delta_{k,k'},$$

$$\frac{1}{2\pi}\, \rho^{k+k'} \int_{-\pi}^{\pi}d\theta\, \,{\rm e}... ...ptsize$\imath$}k\theta} \,{\rm e}^{\mbox{\boldmath\scriptsize$\imath$}k'\theta} = 0,$$ (17)

since $z=\rho\exp(\mbox{\boldmath$\imath$}\theta)$, where in the second equation we must have $k+k'>0$. So long as $\rho\neq 0$ the powers of $\rho$ can be eliminated from the second equation, and since the right-hand term of the first equation is zero unless $k=k'$, they can also be eliminated from the first equation, so that we have

$$\frac{1}{2\pi} \int_{-\pi}^{\pi}d\theta\, \,{\rm e}^{\mbox{\boldm... ...tsize$\imath$}k\theta} \,{\rm e}^{-\mbox{\boldmath\scriptsize$\imath$}k'\theta} = \delta_{k,k'},$$

$$\frac{1}{2\pi} \int_{-\pi}^{\pi}d\theta\, \,{\rm e}^{\mbox{\boldm... ...ptsize$\imath$}k\theta} \,{\rm e}^{\mbox{\boldmath\scriptsize$\imath$}k'\theta} = 0,$$ (18)

where in the second equation we must have $k+k'>0$. Note that this is valid not only on the unit circle, but for all values of $\rho$ so long as $\rho\neq 0$. Expanding the complex exponentials, with the use of the Euler formula, and collecting real and imaginary parts, we have

$$\frac{1}{\pi} \int_{-\pi}^{\pi}d\theta\, \left[ \cos(k\theta)\cos(k'\theta) + \sin(k\theta)\sin(k'\theta) \right] \hspace{2em}$$

$$+ \mbox{\boldmath$\imath$}\, \frac{1}{\pi} \int_{-\pi}^{\pi}d\theta\, \left[ \sin(k\theta)\cos(k'\theta) - \cos(k\theta)\sin(k'\theta) \right] = 2\delta_{k,k'},$$

$$\frac{1}{\pi} \int_{-\pi}^{\pi}d\theta\, \left[ \cos(k\theta)\cos(k'\theta) - \sin(k\theta)\sin(k'\theta) \right] \hspace{2em}$$

$$+ \mbox{\boldmath$\imath$}\, \frac{1}{\pi} \int_{-\pi}^{\pi}d\theta\, \left[ \sin(k\theta)\cos(k'\theta) + \cos(k\theta)\sin(k'\theta) \right] = 0,$$ (19)

where in the second equation we must have $k+k'>0$. Since the right-hand sides are real, we have the four real equations

$$\frac{1}{\pi} \int_{-\pi}^{\pi}d\theta\, \left[ \cos(k\theta)\cos(k'\theta) + \sin(k\theta)\sin(k'\theta) \right] = 2\delta_{k,k'},$$

$$\frac{1}{\pi} \int_{-\pi}^{\pi}d\theta\, \left[ \sin(k\theta)\cos(k'\theta) - \cos(k\theta)\sin(k'\theta) \right] = 0,$$

$$\frac{1}{\pi} \int_{-\pi}^{\pi}d\theta\, \left[ \cos(k\theta)\cos(k'\theta) - \sin(k\theta)\sin(k'\theta) \right] = 0,$$

$$\frac{1}{\pi} \int_{-\pi}^{\pi}d\theta\, \left[ \sin(k\theta)\cos(k'\theta) + \cos(k\theta)\sin(k'\theta) \right] = 0,$$ (20)

where we must have $k+k'>0$ in the last two equations. In the case $k+k'=0$, which implies that $k=0$ and $k'=0$, we obtain from the first equation the identity

$$\frac{1}{\pi} \int_{-\pi}^{\pi}d\theta\, \cos(0)\cos(0) = 2,$$ (21)

which is a part of the relations of orthogonality and norm of the Fourier basis, namely the one giving the squared norm of the constant function which is equal to one for all $\theta$. The second equation is just a trivial identity when we have $k=0$ and $k'=0$, which we may therefore ignore. We may now assume that we have $k+k'>0$ for all the four equations. Adding and subtracting the first and third equations we get

$$\frac{1}{\pi} \int_{-\pi}^{\pi}d\theta\, \cos(k\theta)\cos(k'\theta) = \delta_{k,k'},$$

$$\frac{1}{\pi} \int_{-\pi}^{\pi}d\theta\, \sin(k\theta)\sin(k'\theta) = \delta_{k,k'},$$ (22)

for $k+k'>0$, while adding and subtracting the other two equations we get

$$\frac{1}{\pi} \int_{-\pi}^{\pi}d\theta\, \sin(k\theta)\cos(k'\theta) = 0,$$

$$\frac{1}{\pi} \int_{-\pi}^{\pi}d\theta\, \cos(k\theta)\sin(k'\theta) = 0,$$ (23)

for $k+k'>0$, which are just two copies of the same relation. We have therefore the complete set of orthogonality relations, which also includes those relations giving the norms of the basis functions,

$$\frac{1}{\pi} \int_{-\pi}^{\pi}d\theta\, \cos(k\theta)\cos(k'\theta) = \delta_{k,k'},$$

$$\frac{1}{\pi} \int_{-\pi}^{\pi}d\theta\, \sin(k\theta)\sin(k'\theta) = \delta_{k,k'},$$

$$\frac{1}{\pi} \int_{-\pi}^{\pi}d\theta\, \sin(k\theta)\cos(k'\theta) = 0,$$ (24)

where $k+k'>0$, which includes all the relevant cases, that is, all the relevant pairs of elements of the basis in Equation (13), except for the single case for $k=0$ and $k'=0$, which we examined separately before, leading to Equation (21). Note that this derivation included the determination of the form of the scalar product for the basis elements. Given two integrable real functions $f(\theta)$ and $g(\theta)$, their scalar product is given by

$$(f\vert g) = \int_{-\pi}^{\pi}d\theta\, f(\theta)g(\theta),$$ (25)

which induces a positive-definite norm in the space of all integrable real functions defined on the periodic interval, which is thus seen to constitute a Hilbert space. We may therefore conclude that the whole structure of orthogonality and norm of the Fourier basis is contained in the structure of the inner analytic function within the unit disk of the complex plane.

Note that, since all possible inner analytic functions are given by convergent power series within the open unit disk, and since these power series can be understood as infinite linear combinations of the particular set of inner analytic functions given by the non-negative powers $\left\{\rule{0em}{2ex}z^{k},k\in\{0,1,2,3,\ldots,\infty\}\right\}$, we may think that this set of functions forms a basis of the space of inner analytic functions, which we may call the Taylor basis. Since the orthogonality of the Fourier basis was obtained above from the properties of this set of non-negative powers, it becomes clear that the orthogonality of the Fourier basis is a consequence of similar properties that must hold for the Taylor basis. In fact, it is possible to define a complex scalar product within the space of inner analytic functions, according to which this Taylor basis is orthogonal. Since this constitutes a considerable detour from our main line of reasoning here, it will be presented as an appendix. As one can see in Appendix A, this complex scalar product induces in the space of inner analytic functions a positive-definite norm. As was observed in [#!CAoRFI!#], this space forms a vector space over the field of complex numbers, and we thus see that it constitutes in fact a complex Hilbert space.