An Infinite Collection of Identities

In order to obtain a certain infinite collection of identities satisfied by all zero-average integrable real functions and their Fourier-conjugate real functions, which will be very important later, we start by examining the action of the compact Hilbert transform on the products of arbitrarily given integrable real functions and the elements of the Fourier basis. We will prove the following theorem.

Theorem 4   : Given an arbitrary zero-average integrable real function $f(\theta)$ on the unit circle, and the corresponding Fourier-conjugate real function $g(\theta)$, these two real functions satisfy almost everywhere the following infinite collection of identities:

$$f(\theta) = \frac{1}{2\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1}\, ... ...e{0em}{2ex}(k+1/2)\Delta\theta\right] g(\theta_{1}) } { \sin(\Delta\theta/2) },$$

$$g(\theta) = \frac{1}{2\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1}\, ... ...e{0em}{2ex}(k+1/2)\Delta\theta\right] f(\theta_{1}) } { \sin(\Delta\theta/2) },$$ (43)

where $\Delta\theta=\theta_{1}-\theta$ and $k\in\{1,2,3,\ldots,\infty\}$.

Proof 4.1   :

In order to prove this theorem we start from the expression in Equation (17), exchanging $w(z)$ for the product $z^{k}w(z)$, which is also an inner analytic function so long as $z^{k}$ is an arbitrary positive integer power, which it is since we assume that $k\in\{1,2,3,\ldots,\infty\}$. We therefore have

$$z^{k} w(z) = \frac{1}{\pi\mbox{\boldmath$\imath$}}\, \m... ...m PV}\!\!\oint_{C}dz_{1}\, \frac{z_{1}^{k}w(z_{1})}{z_{1}-z},$$ (44)

where both $z_{1}$ and $z$ are on the circle $C$ of radius $\rho_{1}$ within the open unit disk. We may now write all quantities in this equation in terms of $\rho_{1}$, $\theta_{1}$ and $\theta$,

$$\rho_{1}^{k} \,{\rm e}^{\mbox{\boldmath\scriptsize$\imath$}k\theta} w(\rho_{1},\theta) = \frac{1}{\pi\mbox{\boldmath$\imath$}}\, \mbox{\rm PV}\!\!\int_{-\... ...rho_{1}\,{\rm e}^{\mbox{\boldmath\scriptsize$\imath$}\theta}} \;\;\;\Rightarrow$$

$$w(\rho_{1},\theta) = \frac{1}{\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1}\, \... ...k\theta_{1}}} {1-\,{\rm e}^{-\mbox{\boldmath\scriptsize$\imath$}\Delta\theta}},$$ (45)

where $\Delta\theta=\theta_{1}-\theta$ and $k\in\{1,2,3,\ldots,\infty\}$. Note that, since by construction the real and imaginary parts $u(\rho_{1},\theta)$ and $v(\rho_{1},\theta)$ of $w(\rho_{1},\theta)$ for $\rho_{1}=1$ are integrable real functions on the unit circle, and since there are no other dependencies on $\rho_{1}$ in this equation, we may now take the $\rho_{1}\to 1_{(-)}$ limit of this expression, in which the principal value acquires its usual real meaning on the unit circle, thus obtaining

$$u(1,\theta) + \mbox{\boldmath$\imath$} v(1,\theta) = \... ...{\rm e}^{-\mbox{\boldmath\scriptsize$\imath$}\Delta\theta} },$$ (46)

where $\Delta\theta=\theta_{1}-\theta$ and $k\in\{1,2,3,\ldots,\infty\}$. Once more, in order to identify separately the real and imaginary parts of this equation, we must now rationalize the integrand. Using this time the form shown in Equation (22) for the factor to be rationalized, we get

$${ u(1,\theta) + \mbox{\boldmath$\imath$} v(1,\theta) }$$

$$= \frac{1}{2\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1}\, ... ... e}^{\mbox{\boldmath\scriptsize$\imath$}\Delta\theta/2}} {\sin(\Delta\theta/2)}$$

$$= \frac{1}{2\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1}\, ... ...\mbox{\boldmath\scriptsize$\imath$}(k+1/2)\Delta\theta}} {\sin(\Delta\theta/2)}$$

$$= \frac{1}{2\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1}\, ... ...{\boldmath$\imath$} \sin(k_{1}\Delta\theta) \right] } { \sin(\Delta\theta/2) },$$ (47)

where $\Delta\theta=\theta_{1}-\theta$ and $k_{1}=k+1/2$, with $k\in\{1,2,3,\ldots,\infty\}$. Expanding the numerator in the integrand of this integral we have

$${ \left[ \rule{0em}{2ex} v(1,\theta_{1}) - \mbox{\boldmath$\imath... ...{1}\Delta\theta) + \mbox{\boldmath$\imath$} \sin(k_{1}\Delta\theta) \right] }$$

$$= \left[ \rule{0em}{2ex} \sin(k_{1}\Delta\theta) u(1,\theta_{1}) + \cos(k_{1}\Delta\theta) v(1,\theta_{1}) \right] +$$

$$+ \mbox{\boldmath$\imath$} \left[ \rule{0em}{2ex} \sin(k_{1}\Delta\theta) v(1,\theta_{1}) - \cos(k_{1}\Delta\theta) u(1,\theta_{1}) \right],$$ (48)

and therefore we are left with

$${ u(1,\theta) + \mbox{\boldmath$\imath$} v(1,\theta) }$$

$$= \frac{1}{2\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1}\, ... ...,\theta_{1}) + \cos(k_{1}\Delta\theta) v(1,\theta_{1}) } {\sin(\Delta\theta/2)}$$

$$+ \mbox{\boldmath$\imath$}\, \frac{1}{2\pi}\, \mbox{\rm PV}\!\!\i... ...\theta_{1}) - \cos(k_{1}\Delta\theta) u(1,\theta_{1}) } {\sin(\Delta\theta/2)},$$ (49)

where $\Delta\theta=\theta_{1}-\theta$ and $k_{1}=k+1/2$, with $k\in\{1,2,3,\ldots,\infty\}$. Separating the real and imaginary parts we therefore obtain an infinite collection of identities in the form

$${ u(1,\theta) }$$

$$= \frac{1}{2\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1}\, ... ...0em}{2ex}(k+1/2)\Delta\theta\right] v(1,\theta_{1}) } { \sin(\Delta\theta/2) },$$

$${ v(1,\theta) }$$

$$= \frac{1}{2\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1}\, ... ...0em}{2ex}(k+1/2)\Delta\theta\right] u(1,\theta_{1}) } { \sin(\Delta\theta/2) },$$ (50)

where $\Delta\theta=\theta_{1}-\theta$ and $k\in\{1,2,3,\ldots,\infty\}$. Recalling now that $f(\theta)=u(1,\theta)$, and also that $g(\theta)=v(1,\theta)$, almost everywhere over the unit circle, one obtains the results in Equation (43), and therefore this completes the proof of Theorem 4.

Note that since this is an infinite collection of integral identities, satisfied by $f(\theta)$ and $g(\theta)$ for all strictly positive $k$, it follows that the right-hand sides of the equations above do not, in fact, depend on $k$. If one recognizes in the first term of each one of these two equations the well-known result for the $k^{\rm th}$ partial sums of the corresponding Fourier series in terms of Dirichlet integrals [#!FSchurchill!#], then it follows that the other terms must be the corresponding remainders. This provides us with some level of understanding of the nature of this infinite set of identities. In the next section we will prove that one does obtain in fact the partial sums and remainders of the corresponding Fourier series directly from our complex-analytic structure.