The Compact Hilbert Transform

Let $f(\theta)$ be an integrable real function on $[-\pi,\pi]$, with Fourier coefficients as given in Equation (4). As was shown in [#!CAoRFIII!#] the real function $g(\theta)=v(1,\theta)$ which is the Fourier-conjugate function to $f(\theta)=u(1,\theta)$ has the same Fourier coefficients, but with the meanings of $\alpha_{k}$ and $\beta_{k}$ interchanged, as shown in Equation (8). The relations in Equations (4) and (8) can be understood as the following collection of integral identities satisfied by all pairs of Fourier-conjugate integrable real functions,

$$\int_{-\pi}^{\pi}d\theta\, \cos(k\theta) f(\theta) = \int_{-\pi}^{\pi}d\theta\, \sin(k\theta) g(\theta),$$

$$\int_{-\pi}^{\pi}d\theta\, \sin(k\theta) f(\theta) = -\, \int_{-\pi}^{\pi}d\theta\, \cos(k\theta) g(\theta),$$ (10)

for $k\in\{1,2,3,\ldots,\infty\}$. It is well known that this replacement of $\cos(k\theta)$ with $\sin(k\theta)$ and of $\sin(k\theta)$ with $-\cos(k\theta)$ can be effected by the use of the Hilbert transform. However, that transform was originally introduced by Hilbert for real functions defined on the whole real line, rather that on the unit circle as is our case here. Therefore, the first thing that we will do here is to define a compact version of the Hilbert transform that applies to real functions defined on the unit circle.

Since the Fourier coefficient $\alpha_{0}$ of $f(\theta)$ has no effect on the definition of the Fourier-conjugate function $g(\theta)$, and in order for this pair of real functions to be related in a unique way, we will assume that $f(\theta)$ is also a zero-average real function,

$$\int_{-\pi}^{\pi}d\theta\, f(\theta) = 0,$$ (11)

thus implying for its $k=0$ Fourier coefficient that $\alpha_{0}=0$. This does not affect, of course, any subsequent arguments about the convergence of the Fourier series. According to the construction presented in [#!CAoRFI!#] and reviewed in Section 2, from the other Fourier coefficients $\alpha_{k}$ and $\beta_{k}$, for $k\in\{1,2,3,\ldots,\infty\}$, we may construct the complex coefficients $c_{k}=\alpha_{k}-\mbox{\boldmath$\imath$}\beta_{k}$, for $k\in\{1,2,3,\ldots,\infty\}$, where we now have $c_{0}=0$, and from these we may construct the corresponding inner analytic function $w(z)$ shown in Equation (7), which is now, in fact, a proper inner analytic function, since $c_{0}=0$ implies that $w(0)=0$.

Since $u(\rho,\theta)$ and $v(\rho,\theta)$ are harmonic conjugate functions to each other, it is now clear that there is a one-to-one correspondence between $u(\rho,\theta)$ and $v(\rho,\theta)$, and in particular between $u(1,\theta)$ and $v(1,\theta)$. Therefore, there is a one-to-one correspondence between $f(\theta)$ and $g(\theta)$, in this case valid almost everywhere on the unit circle, since we have shown in [#!CAoRFI!#] that $f(\theta)=u(1,\theta)$ and that $g(\theta)=v(1,\theta)$, both almost everywhere over the unit circle. Therefore, a transformation must exist that produces $g(\theta)$ from $f(\theta)$ almost everywhere over the unit circle, as well as an inverse transformation that recovers $f(\theta)$ from $g(\theta)$ almost everywhere over the unit circle. In this section we will show that the following definition accomplishes this.

Definition 1   : Compact Hilbert Transform

If $f(\theta)$ is an arbitrarily given zero-average integrable real function defined on the unit circle, then its compact Hilbert transform $g(\theta)$ is the real function defined by

$$g(\theta) = {\cal H}_{\rm c}[f(\theta)]$$

$$= -\, \frac{1}{2\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1... ...ht]} {\sin\!\left[\rule{0em}{2ex}(\theta_{1}-\theta)/2\right]}\, f(\theta_{1}),$$ (12)

where $\mbox{\rm PV}\!\!\,$ stands for the Cauchy principal value, and where ${\cal H}_{\rm c}[f(\theta)]$ is the notation we will use for the compact Hilbert transform applied to the real function $f(\theta)$.

We will now prove the following theorem.

Theorem 1   : The zero-average integrable real functions $f(\theta)$ and $g(\theta)$, which are such that $f(\theta)=u(1,\theta)$ and $g(\theta)=v(1,\theta)$ almost everywhere on the unit circle, are related to each other by this transform, that is, we have that $g(\theta)={\cal H}_{\rm c}[f(\theta)]$ almost everywhere on the unit circle, and that $f(\theta)={\cal H}_{\rm c}^{-1}[g(\theta)]$ almost everywhere on the unit circle, where the inverse transform is simply given by ${\cal H}_{\rm c}^{-1}[g(\theta)]=-{\cal H}_{\rm c}[g(\theta)]$.

Proof 1.1   :

In order to derive these facts from our complex-analytic structure, we start from the Cauchy integral formula for the inner analytic function $w(z)$,

$$w(z) = \frac{1}{2\pi\mbox{\boldmath$\imath$}} \oint_{C}dz_{1}\, \frac{w(z_{1})}{z_{1}-z},$$ (13)

where $C$ can be taken as a circle centered at the origin, with radius $\rho_{1}<1$, and where we write $z$ and $z_{1}$ in polar coordinates as $z=\rho\exp(\mbox{\boldmath$\imath$}\theta)$ and $z_{1}=\rho_{1}\exp(\mbox{\boldmath$\imath$}\theta_{1})$. The integral formula in Equation (13) is valid for $\rho<\rho_{1}$, and in fact, by the Cauchy-Goursat theorem, the integral is zero if $\rho>\rho_{1}$, since both $z$ and $z_{1}$ are within the open unit disk, a region where $w(z)$ is analytic. We must now determine what happens if $\rho=\rho_{1}$, that is, if $z$ is on the circle $C$ of radius $\rho_{1}$. Note that in this case we may slightly deform the integration contour $C$ in order to have it pass on one side or the other of the simple pole of the integrand at $z_{1}=z$. If we use a deformed contour $C_{\ominus}$ that excludes the pole from its interior, then we have, instead of Equation (13),

$$0 = \frac{1}{2\pi\mbox{\boldmath$\imath$}} \oint_{C_{\ominus}}dz_{1}\, \frac{w(z_{1})}{z_{1}-z},$$ (14)

due to the Cauchy-Goursat theorem, while if we use a deformed contour $C_{\oplus}$ that includes the pole in its interior, then we have, just as in Equation (13),

$$w(z) = \frac{1}{2\pi\mbox{\boldmath$\imath$}} \oint_{C_{\oplus}}dz_{1}\, \frac{w(z_{1})}{z_{1}-z}.$$ (15)

Since by the Sokhotskii-Plemelj theorem [#!sokhplem!#] the Cauchy principal value of the integral over the circle $C$ is the arithmetic average of these two integrals, in the limit where the deformations vanish, a limit which does not really have to be considered in detail, so long as the deformations do not cross any other singularities,

$$\mbox{\rm PV}\!\!\oint_{C}dz_{1}\, \frac{w(z_{1})}{z_{1}-z... ...c{1}{2} \oint_{C_{\oplus}}dz_{1}\, \frac{w(z_{1})}{z_{1}-z},$$ (16)

adding Equations (14) and (15) we may conclude that

$$w(z) = \frac{1}{\pi\mbox{\boldmath$\imath$}}\, \mbox{\rm PV}\!\!\oint_{C}dz_{1}\, \frac{w(z_{1})}{z_{1}-z},$$ (17)

where we now have $\rho=\rho_{1}$, that is, both $z_{1}$ and $z$ are on the circle $C$ of radius $\rho_{1}$ within the open unit disk. This formula can be understood as a special version of the Cauchy integral formula, and will be used repeatedly in what follows. We may now write all quantities in this equation in terms of the polar coordinates $\rho_{1}$, $\theta_{1}$ and $\theta$,

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

$$= \frac{1}{\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1}\, \... ...,\theta_{1})} {1-\,{\rm e}^{-\mbox{\boldmath\scriptsize$\imath$}\Delta\theta}},$$ (18)

where $\Delta\theta=\theta_{1}-\theta$. 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 expression, we may now take the $\rho_{1}\to 1_{(-)}$ limit of this equation, in which the principal value acquires its usual real meaning over the unit circle, that is, the meaning that the asymptotic limits of the integral on either side of a non-integrable singularity must be taken in the symmetric way. In that limit we have

$$u(1,\theta)+\mbox{\boldmath$\imath$}v(1,\theta) = \frac{1... ...\,{\rm e}^{-\mbox{\boldmath\scriptsize$\imath$}\Delta\theta}}.$$ (19)

In order to identify separately the real and imaginary parts of this equation, we must now rationalize the integrand of the integral shown. We will use the fact that

$$\frac{1}{1-\,{\rm e}^{-\mbox{\boldmath\scriptsize$\imath$}\Delta\theta}} = \frac { 1 - \,{\rm e}^{\mbox{\boldmath\scriptsize$\imath$}\Delta\... ...{2ex} 1 - \,{\rm e}^{\mbox{\boldmath\scriptsize$\imath$}\Delta\theta} \right) }$$

$$= \frac { \left[ \rule{0em}{2ex} 1 - \cos(\Delta\theta) \right] - \mbox{\boldmath$\imath$} \sin(\Delta\theta) } { 2-2\cos(\Delta\theta) }$$

$$= \frac{1}{2} \left[ 1 - \mbox{\boldmath$\imath$}\, \frac { \sin(\Delta\theta) } { 1-\cos(\Delta\theta) } \right].$$ (20)

Using the half-angle trigonometric identities we may also write this result as

$$\frac{1}{1-\,{\rm e}^{-\mbox{\boldmath\scriptsize$\imath$}\Delta\theta}} = \frac{1}{2} \left[ 1 - \mbox{\boldmath$\imath$}\, \frac{\cos(\Delta\theta/2)}{\sin(\Delta\theta/2)} \right]$$ (21)

$$= -\, \frac{\mbox{\boldmath$\imath$}}{2}\, \frac {\,{\rm e}^{\mbox{\boldmath\scriptsize$\imath$}\Delta\theta/2}} {\sin(\Delta\theta/2)}.$$ (22)

Using the result shown in Equation (21) back in Equation (19) we obtain

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

$$= \frac{1}{2\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1}\, ... ...{0em}{2ex} u(1,\theta_{1}) + \mbox{\boldmath$\imath$} v(1,\theta_{1}) \right] +$$

$$+ \frac{1}{2\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1}\... ...e{0em}{2ex} v(1,\theta_{1}) - \mbox{\boldmath$\imath$} u(1,\theta_{1}) \right].$$ (23)

Since both $u(1,\theta_{1})$ and $v(1,\theta_{1})$ are zero-average real functions on the unit circle, the first two integrals in the last form of the equation above are zero, so that separating the real and imaginary parts within this expression we are left with

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

$$-\, \frac{\mbox{\boldmath$\imath$}}{2\pi}\, \mbox{\rm PV}\!\!\int... ...heta_{1}\, \frac{\cos(\Delta\theta/2)}{\sin(\Delta\theta/2)}\, u(1,\theta_{1}),$$ (24)

where $\Delta\theta=\theta_{1}-\theta$. Separating the real and imaginary parts of this equation we may now write that

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

$$v(1,\theta) = -\, \frac{1}{2\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1}\, \frac{\cos(\Delta\theta/2)}{\sin(\Delta\theta/2)}\, u(1,\theta_{1}),$$ (25)

where $\Delta\theta=\theta_{1}-\theta$. Recalling that $f(\theta)=u(1,\theta)$ and $g(\theta)=v(1,\theta)$, almost everywhere on the unit circle, we have

$$f(\theta) = {\cal H}_{\rm c}^{-1}[g(\theta)]$$

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

$$g(\theta) = {\cal H}_{\rm c}[f(\theta)]$$

$$= -\, \frac{1}{2\pi}\, \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1}\, \frac{\cos(\Delta\theta/2)}{\sin(\Delta\theta/2)}\, f(\theta_{1}),$$ (26)

two equations which are thus valid almost everywhere as well. These are the transformations relating the pair of Fourier-conjugate functions $f(\theta)$ and $g(\theta)$. The second expression defines the compact Hilbert transformation of $f(\theta)$ into $g(\theta)$, and the first one defines the inverse transformation, which recovers $f(\theta)$ from $g(\theta)$. Note that in this notation the transform is defined with an explicit minus sign, and that its inverse is simply minus the transform itself, ${\cal H}_{\rm c}^{-1}[g(\theta)]=-{\cal H}_{\rm c}[g(\theta)]$. This completes the proof of Theorem 1.

It is interesting to observe that this transform can be interpreted as a linear integral operator acting on the space of zero-average integrable real functions defined on the unit circle. The integration kernel of the integral operator depends only on the difference $\theta-\theta_{1}$, and is given by

$$K_{{\cal H}_{\rm c}}(\theta-\theta_{1}) = -\, \frac{1}{2... ...]} {\sin\!\left[\rule{0em}{2ex}(\theta_{1}-\theta)/2\right]},$$ (27)

so that the action of the operator on an arbitrarily given zero-average real integrable function $f(\theta)$ on the unit circle can be written as

$$g(\theta) = {\cal H}_{\rm c}[f(\theta)]$$

$$= \mbox{\rm PV}\!\!\int_{-\pi}^{\pi}d\theta_{1}\, K_{{\cal H}_{\rm c}}(\theta-\theta_{1}) f(\theta_{1}).$$ (28)

The operator is linear, invertible, and the composition of the operator with itself results in the operation of multiplication by $-1$. Note that, since by hypothesis $f(\theta)$ is integrable on the unit circle, the Cauchy principal value refers only to the explicit non-integrable singularity of the integration kernel at the position $\theta_{1}=\theta$.