The Completeness Relation

Let us now show that the completeness relation of the Fourier basis also follows from the structure of complex analysis. In order to do this, we must first show that the Dirac delta ``function'' can be represented in terms of the analytic structure within the open unit disk. We denote the Dirac delta ``function'' centered at $\theta=\theta_{1}$ on the unit circle by $\delta(\theta-\theta_{1})$. The definition of this mathematical object is that it is a symbolic representation of a limiting process which has the following four properties:

1. $\delta(\theta-\theta_{1})$ tends to zero when one takes the defining limit with $\theta\neq\theta_{1}$;

2. $\delta(\theta-\theta_{1})$ diverges to positive infinity when one takes the defining limit with $\theta=\theta_{1}$;

3. in the defining limit the integral

   
   

   $$\int_{a}^{b}d\theta\, \delta(\theta-\theta_{1}) = 1,$$
   
   

   has the value shown, for any interval $(a,b)$ which contains the point $\theta_{1}$;

4. given any continuous function $g(\theta)$, in the defining limit the integral

   
   

   $$\int_{a}^{b}d\theta\, g(\theta) \delta(\theta-\theta_{1}) = g(\theta_{1}),$$
   
   

   has the value shown, for any interval $(a,b)$ which contains the point $\theta_{1}$.

Of course no real function exists that can have all these properties, which justifies the quotes in which we wrap the word ``function'' when referring to it. In order to construct the Dirac delta ``function'' we must first give an object or set of objects over which the limiting process can be defined, and then define that limiting process. In order to fulfill this program, we consider the complex function given within the open unit disk by

$$w_{\delta}(z) = \frac{1}{2\pi} - \frac{1}{\pi}\, \frac{z}{z-z_{1}},$$

as well as its restrictions to circles of radius $\rho$ centered at the origin, with $0<\rho<1$, where $z=\rho\exp(\mbox{\boldmath$\imath$}\theta)$ and where $z_{1}=\exp(\mbox{\boldmath$\imath$}\theta_{1})$ is a point on the unit circle. This function is analytic within the open unit disk, but it is not an inner analytic function, because $w_{\delta}(0)$ is not zero. However, we may write it in terms of another function $w(z)$ as

$$\begin{eqnarray*} w_{\delta}(z) & = & \frac{1}{2\pi} + w(z), \\ w(z) & = & -\, \frac{1}{\pi}\, \frac{z}{z-z_{1}}. \end{eqnarray*}$$

Strictly speaking, $w(z)$ is not an inner analytic function either, because it does not reduce to a real function over the real axis. However, it does reduce to a real function over the straight line $z=\chi z_{1}$, with real $\chi$, since in this case we have

$$w(z) = -\, \frac{1}{\pi}\, \frac{\chi}{\chi-1}.$$

We see therefore that $w(z)$ is an inner analytic function rotated around the origin by the angle $\theta_{1}$ associated to $z_{1}$. Therefore this is just a simple extension of the structure we defined here. The limiting process to be used for the definition is just the $\rho\to 1$ limit to the unit circle. We will now show that the real part of $w_{\delta}(z)$, taken on the $\rho\to 1$ limit, satisfies all the required properties defining the Dirac delta ``function''. In order to recover the real and imaginary parts of this complex function, we must now rationalize it,

$$\begin{eqnarray*} w_{\delta}(z) & = & \frac{1}{2\pi} - \frac{1}{\pi}\, \fr... ...$}\rho\sin(\Delta\theta)} {\rho^{2}-2\rho\cos(\Delta\theta)+1}, \end{eqnarray*}$$

where $\Delta\theta=\theta-\theta_{1}$. We now examine the real part of this function,

$$\Re[w_{\delta}(z)] = \frac{1}{2\pi} - \frac{1}{\pi}\, ... ...a)\right]} {\left(\rho^{2}+1\right)-2\rho\cos(\Delta\theta)}.$$

If we now take the limit $\rho\to 1$, under the assumption that $\Delta\theta\neq 0$, we get

$$\begin{eqnarray*} \lim_{\rho\to 1}\Re[w_{\delta}(z)] & = & \frac{1}{2\pi} - ... ...} \\ & = & \frac{1}{2\pi} - \frac{1}{2\pi} \\ & = & 0, \end{eqnarray*}$$

which is the correct value for the case of the Dirac delta ``function''. Thus we see that the first property holds.

If, on the other hand, we calculate $\Re[w_{\delta}(z)]$ for $\Delta\theta=0$ and $\rho<1$ we obtain

$$\begin{eqnarray*} \Re[w_{\delta}(z)] & = & \frac{1}{2\pi} - \frac{1}{\pi}\,... ... & = & \frac{1}{2\pi} - \frac{1}{\pi}\, \frac{\rho}{\rho-1}, \end{eqnarray*}$$

which diverges to positive infinity as $\rho\to 1$ from below, as it should in order to represent the singular Dirac delta ``function''. This establishes that the second property holds.

We then calculate the integral of $\Re[w_{\delta}(z)]$ over the circle of radius $\rho<1$, which is given by

$$\begin{eqnarray*} I_{1} & = & \frac{1}{2\pi} \int_{-\pi}^{\pi}d\theta\, \rh... ...left[\left(\rho^{2}+1\right)/(2\rho)\right]-\cos(\Delta\theta)}, \end{eqnarray*}$$

since $d(\Delta\theta)=d\theta$. Note that this is not the integral of an analytic function over a closed contour, but the integral of a real function over the circle of radius $\rho$. This real integral over $\theta$ can be calculated by residues. We introduce an auxiliary complex variable $\xi=\lambda\exp(\mbox{\boldmath$\imath$}\Delta\theta)$, which becomes simply $\exp(\mbox{\boldmath$\imath$}\Delta\theta)$ on the unit circle $\lambda=1$. We have $d\xi=\mbox{\boldmath$\imath$}\xi d(\Delta\theta)$, and so we may write the integral as

$$\begin{eqnarray*} \int_{-\pi}^{\pi}d(\Delta\theta)\, \frac{1}{\left[\left(1+\r... ...frac{1}{1-\left[\left(1+\rho^{2}\right)/\rho\right]\xi+\xi^{2}}, \end{eqnarray*}$$

where the integral is now over the unit circle $C$ in the complex $\xi$ plane. The two roots of the quadratic polynomial on $\xi$ in the denominator are given by

$$\begin{eqnarray*} \xi_{+} & = & 1/\rho, \\ \xi_{-} & = & \rho. \end{eqnarray*}$$

Since $\rho<1$, only the pole corresponding to $\xi_{-}$ lies inside the integration contour, so we get for the integral

$$\begin{eqnarray*} \int_{-\pi}^{\pi}d\theta\, \frac{1}{\left[\left(1+\rho^{2}\r... ...\rho} \\ & = & 4\pi\, \frac{\rho}{\left(1-\rho^{2}\right)}. \end{eqnarray*}$$

It follows that we have for the integral $I_{1}$

$$\begin{eqnarray*} I_{1} & = & \frac{\left(1-\rho^{2}\right)}{4\pi}\, 4\pi\, \frac{\rho}{\left(1-\rho^{2}\right)} \\ & = & \rho, \end{eqnarray*}$$

so that we have $I_{1}=1$ in the $\rho\to 1$ limit. Once we have this result, and since the integrand goes to zero everywhere on the unit circle except at $\Delta\theta=0$, which means that $\theta=\theta_{1}$, the integral can be changed to one over any open interval on the unit circle containing the point $\theta_{1}$, without any change in its limiting value. This establishes that the third property holds.

In order to establish the fourth and last property, we take an essentially arbitrary inner analytic function $\gamma(z)$, with the single additional restriction that it be well-defined at the point $z_{1}$, in the sense that its $\rho\to 1$ limit exists at $z_{1}$. This inner analytic function corresponds to a pair of FC real functions on the unit circle, both of which are well-defined at $z_{1}$. We now consider the following integral over the circle of radius $\rho<1$,

$$\begin{eqnarray*} I_{2} & = & \frac{1}{2\pi} \int_{-\pi}^{\pi}d\theta\, \rh... ...left[\left(\rho^{2}+1\right)/(2\rho)\right]-\cos(\Delta\theta)}, \end{eqnarray*}$$

since $d(\Delta\theta)=d\theta$. Note once more that this is not the integral of an analytic function over a closed contour, but two integrals of real functions, given by the real and imaginary parts of $\gamma(z)$, over the circle of radius $\rho$. These real integrals over $\theta$ can be calculated by residues, exactly like the one which appeared before in the case of $I_{1}$. The calculation is exactly the same except for the extra factor of $\gamma(z)$ to be taken into consideration when calculating the residue, so that we may write directly that

$$\begin{eqnarray*} \int_{-\pi}^{\pi}d(\Delta\theta)\, \frac {\gamma(z)} {\lef... ...frac{\rho}{\left(1-\rho^{2}\right)} \lim_{\xi\to\rho}\gamma(z). \end{eqnarray*}$$

Note now that since $\xi=\lambda\exp(\mbox{\boldmath$\imath$}\Delta\theta)$ and we must take the limit $\xi\to\rho$, we in fact have that in that limit

$$\lambda\,{\rm e}^{\mbox{\boldmath$\imath$}\Delta\theta} = \rho,$$

which implies that $\lambda=\rho$ and that $\Delta\theta=0$. We must therefore write $\gamma(z)$ at the point given by $\rho$ and $\Delta\theta=0$, that is, at the point given by $\rho$ and $\theta_{1}$,

$$\int_{-\pi}^{\pi}d(\Delta\theta)\, \frac {\gamma(z)} {\l... ...rac{\rho}{\left(1-\rho^{2}\right)}\, \gamma(\rho,\theta_{1}).$$

It follows that we have for the integral $I_{2}$

$$\begin{eqnarray*} I_{2} & = & \frac{\left(1-\rho^{2}\right)}{4\pi}\, 4\pi\, ... ...gamma(\rho,\theta_{1}) \\ & = & \rho\gamma(\rho,\theta_{1}). \end{eqnarray*}$$

Finally, we may now take the $\rho\to 1$ limit, since $\gamma(\rho,\theta)$ is well-defined in that limit, and thus obtain

$$\lim_{\rho\to 1} I_{2} = \gamma(1,\theta_{1}).$$

Once we have this result, and since the integrand goes to zero everywhere on the unit circle except at $\Delta\theta=0$, which means that $\theta=\theta_{1}$, the integral can be changed to one over any open interval on the unit circle containing the point $\theta_{1}$, without any change in its value. This establishes that the fourth and last property holds. We may then write symbolically that

$$\delta(\theta-\theta_{1}) = \lim_{\rho\to 1} \Re[w_{\delta}(z)].$$

Note that in order to obtain this result it was not necessary to assume that $\gamma(z)$ is continuous at $z_{1}$ in the direction of $\theta$ along the unit circle. It was necessary to assume only that $\gamma(z)$ is continuous as a function of $\rho$, in the direction perpendicular to the unit circle. One can see therefore that, once more, we get a bit more than we bargained for, because we were able to establish the result with slightly weaker hypotheses than at first expected.

We are now in a position to establish the completeness relation using this representation of the Dirac delta ``function''. If we use once again the Cauchy integral formulas for $w(z)$ we get for the coefficients of the Taylor expansion of $w(z)$

$$\begin{eqnarray*} a_{k} & = & \frac{1}{2\pi\mbox{\boldmath$\imath$}} \oint_{... ...math$}} \oint_{C}dz\, \frac{1}{z^{k}}\, \frac{(-1)}{z-z_{1}}, \end{eqnarray*}$$

for $k\geq 1$, since $w(z)$ is a rotated inner analytic function. We now observe that the second ratio in the integrand can be understood as the sum of a geometric series, which is convergent so long as $\rho<1$,

$$\frac{(-1)}{z-z_{1}} = \frac{1}{z_{1}} \sum_{n=0}^{\infty} \left(\frac{z}{z_{1}}\right)^{n},$$

so that we may now write

$$\begin{eqnarray*} a_{k} & = & \frac{1}{2\pi^{2}\mbox{\boldmath$\imath$}} \oi... ...m_{n=0}^{\infty} \frac{1}{z_{1}^{n+1}} \oint_{C}dz\, z^{n-k}, \end{eqnarray*}$$

since a convergent power series can always be integrated term-by-term. As we have already discussed before, in the previous section, the remaining integral is zero except if $n=k-1$, in which case it has the value $2\pi\mbox{\boldmath$\imath$}$. Note that this condition relating $n$ and $k$ can always be satisfied since $k\geq 1$. We therefore get for the coefficients

$$\begin{eqnarray*} a_{k} & = & \frac{1}{2\pi^{2}\mbox{\boldmath$\imath$}} \su... ...h$\imath$}\,\delta_{n,k-1} \\ & = & \frac{1}{\pi z_{1}^{k}}. \end{eqnarray*}$$

As a result, we get for the Taylor expansion of $w(z)$

$$w(z) = \frac{1}{\pi} \sum_{k=1}^{\infty} \left(\frac{z}{z_{1}}\right)^{k}.$$

We now write both $z$ and $z_{1}$ in polar coordinates, to obtain

$$\begin{eqnarray*} w(z) & = & \frac{1}{\pi} \sum_{k=1}^{\infty} \rho^{k} \,... ...k\theta) \sin(k\theta_{1}) \right] \rule{0em}{3ex} \right\}. \end{eqnarray*}$$

If we now write the real part of $w_{\delta}(z)$ we get

$$\Re[w_{\delta}(z)] = \frac{1}{2\pi} + \frac{1}{\pi} \s... ...os(k\theta_{1}) + \sin(k\theta) \sin(k\theta_{1}) \right],$$

and, if we then take the $\rho\to 1$ limit, we get the expression

$$\delta(\theta-\theta_{1}) = \frac{1}{2\pi} + \frac{1}{\... ...os(k\theta_{1}) + \sin(k\theta) \sin(k\theta_{1}) \right],$$

which is the completeness relation in its usual form, a bilinear form on the Fourier basis functions, at two separate points along the unit circle. Note that the constant function, which is an element of the complete Fourier basis, is included in the first term. Note also that this time it was necessary to take the $\rho\to 1$ limit, and that the completeness of the Fourier basis is valid only on the unit circle. This is to be expected, of course, since the unit circle is where the corresponding space of real functions, which is generated by the basis, is defined.