Action on the Fourier Basis

We will now determine the action of the compact Hilbert transform on the elements of the Fourier basis of functions. The case of the constant function, which constitutes the $k=0$ element of the basis, that is the single member of the basis which is not a zero-average function, must be examined in separate. We will now prove the following simple theorem.

Theorem 2   : Given any constant real function $f(\theta)=R$, for any real constant $R$, its compact Hilbert transform is zero, that is, ${\cal H}_{\rm c}[R]=0$.

Proof 2.1   :

We start from the expression in Equation (17) for the very simple case $w(z)=1$,

$$1 = \frac{1}{\pi\mbox{\boldmath$\imath$}}\, \mbox{\rm PV}\!\!\oint_{C}dz_{1}\, \frac{1}{z_{1}-z},$$ (29)

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

$$1 = \frac{1}{\pi\mbox{\boldmath$\imath$}}\, \mbox{\rm PV}\!\!\int_{-\... ...ath$}\theta_{1}}-\rho_{1}\,{\rm e}^{\mbox{\boldmath\scriptsize$\imath$}\theta}}$$

$$= \frac{1}{\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1}\, \frac{1}{1-\,{\rm e}^{-\mbox{\boldmath\scriptsize$\imath$}\Delta\theta}},$$ (30)

where $\Delta\theta=\theta_{1}-\theta$. Note that, since there are no remaining dependencies on $\rho_{1}$, 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. Just as in the previous section, in order to identify separately the real and imaginary parts of this equation, we must now rationalize the integrand. Using the result in Equation (21) we obtain

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

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

$$= 1 -\, \frac{\mbox{\boldmath$\imath$}}{2\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1}\, \frac{\cos(\Delta\theta/2)}{\sin(\Delta\theta/2)}.$$ (31)

It follows therefore that we have

$$-\, \frac{1}{2\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\... ...1}\, \frac{\cos(\Delta\theta/2)}{\sin(\Delta\theta/2)} = 0,$$ (32)

which is the statement that ${\cal H}_{\rm c}[1]=0$. Note that, since $\Delta\theta=\theta_{1}-\theta$, which in the context of this integral implies that $d\theta_{1}=d(\Delta\theta)$, by means of a trivial transformation of variables this integral can also be shown to be zero by simple parity arguments. Given the linearity of the compact Hilbert transform, it is equally true that, for any real constant $R$, we have that ${\cal H}_{\rm c}[R]=0$, so that all constant functions are mapped to the null function. This completes the proof of Theorem 2.

Let us now consider all the remaining elements of the Fourier basis of functions. We will prove the following theorem.

Theorem 3   : Given the elements of the Fourier basis of functions, $\cos(k\theta)$ and $\sin(k\theta)$, for $k\in\{1,2,3,\ldots,\infty\}$, the following relations between them hold:

$$\cos(k\theta) = - {\cal H}_{\rm c}\!\left[\rule{0em}{2ex}\sin(k\theta)\right],$$

$$\sin(k\theta) = {\cal H}_{\rm c}\!\left[\rule{0em}{2ex}\cos(k\theta)\right].$$ (33)

Proof 3.1   :

In order to prove this theorem we start from the expression in Equation (17) for the case $w(z)=z^{k}$, where $k\in\{1,2,3,\ldots,\infty\}$, that is, for a strictly positive power of $z$, which is therefore an inner analytic function. Note that these are all the elements of the complex Taylor basis of functions, with the exception of the constant function. We have therefore

$$z^{k} = \frac{1}{\pi\mbox{\boldmath$\imath$}}\, \mbox{\rm PV}\!\!\oint_{C}dz_{1}\, \frac{z_{1}^{k}}{z_{1}-z},$$ (34)

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

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

$$\,{\rm e}^{\mbox{\boldmath\scriptsize$\imath$}k\theta} = \frac{1}{\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1}\, \... ...k\theta_{1}}} {1-\,{\rm e}^{-\mbox{\boldmath\scriptsize$\imath$}\Delta\theta}},$$ (35)

where $\Delta\theta=\theta_{1}-\theta$. Note that, since there are no remaining dependencies on $\rho_{1}$, 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. In the limit we have

$$\cos(k\theta) + \mbox{\boldmath$\imath$} \sin(k\theta) ... ...\,{\rm e}^{-\mbox{\boldmath\scriptsize$\imath$}\Delta\theta}}.$$ (36)

Just as in the previous cases, in order to identify separately the real and imaginary parts of this equation, we must now rationalize the integrand. Using the result in Equation (21) we obtain

$${ \cos(k\theta) + \mbox{\boldmath$\imath$} \sin(k\theta) }$$

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

$$= \frac{1}{2\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1}\, ... ...}{2ex} \cos(k\theta_{1}) + \mbox{\boldmath$\imath$} \sin(k\theta_{1}) \right] +$$

$$+ \frac{1}{2\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1}\... ...m}{2ex} \sin(k\theta_{1}) - \mbox{\boldmath$\imath$} \cos(k\theta_{1}) \right],$$ (37)

where $\Delta\theta=\theta_{1}-\theta$ and $k\in\{1,2,3,\ldots,\infty\}$. The first two integrals in the last form of the equation above are zero for all $k>0$ because they are integrals of cosines and sines over integer multiples of their periods, so that we may now separate the real and imaginary parts of the remaining terms and thus get

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

$$-\, \frac{\mbox{\boldmath$\imath$}}{2\pi}\, \mbox{\rm PV}\!\!\int... ...ta_{1}\, \frac{\cos(\Delta\theta/2)}{\sin(\Delta\theta/2)}\, \cos(k\theta_{1}),$$ (38)

where $\Delta\theta=\theta_{1}-\theta$ and $k\in\{1,2,3,\ldots,\infty\}$. We therefore obtain the action of the compact Hilbert transform on the elements of the Fourier basis,

$$\cos(k\theta) = - {\cal H}_{\rm c}\!\left[\rule{0em}{2ex}\sin(k\theta)\right]$$

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

$$\sin(k\theta) = {\cal H}_{\rm c}\!\left[\rule{0em}{2ex}\cos(k\theta)\right]$$

$$= -\, \frac{1}{2\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1}\, \frac{\cos(\Delta\theta/2)}{\sin(\Delta\theta/2)}\, \cos(k\theta_{1}),$$ (39)

where $\Delta\theta=\theta_{1}-\theta$ and $k\in\{1,2,3,\ldots,\infty\}$. The second equation above is the transform applied to the cosines and the first equation is the inverse transform applied to the sines. As one can see, the transform does indeed have the property of replacing cosines with sines and sines with minus cosines, as expected. This completes the proof of Theorem 3.

One can now see that the application of the compact Hilbert transform to the Fourier series of an arbitrarily given zero-average integrable real function $f(\theta)$ on the unit circle will produce the Fourier series of its Fourier-conjugate real function $g(\theta)$. Given the linearity of the transform, if we apply it to the Fourier series of $f(\theta)$ we get

$${\cal H}_{\rm c}\!\left\{ \sum_{k=1}^{\infty} \left[ \rule{0em}{2ex} \alpha_{k}\cos(k\theta) + \beta_{k}\sin(k\theta) \right] \right\} = \sum_{k=1}^{\infty} \left\{ \rule{0em}{2.5ex} \alpha_{k} {\cal H}... ... \beta_{k} {\cal H}_{\rm c}\!\left[\rule{0em}{2ex}\sin(k\theta)\right] \right\}$$

$$= \sum_{k=1}^{\infty} \left[ \rule{0em}{2ex} \alpha_{k} \sin(k\theta) - \beta_{k} \cos(k\theta) \right],$$ (40)

where this last one is the Fourier series of the real function $g(\theta)$, which is the Fourier conjugate of $f(\theta)$. Hence, if $S(\rho,\theta)$ is the complex power series given in Equation (6), if $S^{F,f}(\theta)=\Re[S(1,\theta)]$ is the Fourier series of $f(\theta)$ and $S^{F,g}(\theta)=\Im[S(1,\theta)]$ is the Fourier series of $g(\theta)$, then we have that

$$S^{F,g}(\theta) = {\cal H}_{\rm c}\!\left[\rule{0em}{2.2ex}S^{F,f}(\theta)\right].$$ (41)

Note that the same is true for the corresponding partial sums, as well as for the corresponding remainders, so long as the latter exist at all. If $S_{N}(\rho,\theta)$ is the $N^{\rm th}$ partial sum and $R_{N}(\rho,\theta)$ is the $N^{\rm th}$ remainder of the complex power series given in Equation (6), and if $S_{N}^{F,f}(\theta)=\Re[S_{N}(1,\theta)]$ is the $N^{\rm th}$ partial sum of the real Fourier series of $f(\theta)$, if $S_{N}^{F,g}(\theta)=\Im[S_{N}(1,\theta)]$ is the $N^{\rm th}$ partial sum of the Fourier series of $g(\theta)$, if $R_{N}^{F,f}(\theta)=\Re[R_{N}(1,\theta)]$ is the $N^{\rm th}$ remainder of the Fourier series of $f(\theta)$ and if $R_{N}^{F,g}(\theta)=\Im[R_{N}(1,\theta)]$ is the $N^{\rm th}$ remainder of the Fourier series of $g(\theta)$, then we have

$$S_{N}^{F,g}(\theta) = {\cal H}_{\rm c}\!\left[\rule{0em}{2ex}S_{N}^{F,f}(\theta)\right],$$

$$R_{N}^{F,g}(\theta) = {\cal H}_{\rm c}\!\left[\rule{0em}{2ex}R_{N}^{F,f}(\theta)\right].$$ (42)

Of course, in each one of these cases the inverse mapping holds as well, using the inverse transform to take us from the quantities related to $g(\theta)$ back to the corresponding quantities related to $f(\theta)$.