The Dirac Delta ``Function''

This is where we begin the discussion of inner analytic functions that have hard singularities with strictly positive degrees of hardness. Let us start by simply introducing a certain particular inner analytic function of $z$. If $z_{1}$ is a point on the unit circle, this function is defined as a very simple rational function of $z$,

$$w_{\delta}(z,z_{1}) = \frac{1}{2\pi} - \frac{1}{\pi}\, \frac{z}{z-z_{1}}.$$ (4)

This inner analytic function has a single point of singularity, which is a simple pole at $z_{1}$. This is a hard singularity with degree of hardness equal to one. Our objective here is to examine the properties of the real part $u_{\delta}(\rho,\theta,\theta_{1})$ of this inner analytic function,

$$w_{\delta}(z,z_{1}) = u_{\delta}(\rho,\theta,\theta_{1}) + \mbox{\boldmath$\imath$} v_{\delta}(\rho,\theta,\theta_{1}).$$ (5)

We will prove that in the $\rho\to 1_{(-)}$ limit $u_{\delta}(\rho,\theta,\theta_{1})$ can be interpreted as a Schwartz distribution [#!DTSchwartz!#,#!DTLighthill!#], namely as the singular object known as the Dirac delta ``function'', which we will denote by $\delta(\theta-\theta_{1})$. This object is also known as a generalized real function, since it is not really a real function in the usual sense of the term. In the Schwartz theory of distributions this object plays the role of an integration kernel for a certain distribution. Note that $w_{\delta}(z,z_{1})$ can, in fact, be written explicitly as a function of $\rho$ and $\theta-\theta_{1}$. Since we have that $z=\rho\exp(\mbox{\boldmath$\imath$}\theta)$ and that $z_{1}=\exp(\mbox{\boldmath$\imath$}\theta_{1})$, we have at once that

$$w_{\delta}(z,z_{1}) = \frac{1}{2\pi} - \frac{1}{\pi}\, ... ...}^{\mbox{\boldmath\scriptsize$\imath$}(\theta-\theta_{1})}-1}.$$ (6)

The definition of the Dirac delta ``function'' is that it is a symbol for a limiting process, which satisfies certain conditions. In our case here the limiting process will be the limit $\rho\to 1_{(-)}$ from the open unit disk to the unit circle. The limit of $u_{\delta}(\rho,\theta,\theta_{1})$ represents the delta ``function'' in the sense that it satisfies the conditions that follow.

1. The defining limit of $\delta(\theta-\theta_{1})$ tends to zero when one takes the $\rho\to 1_{(-)}$ limit while keeping $\theta\neq\theta_{1}$.

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

3. In the $\rho\to 1_{(-)}$ limit the integral

   
   

   $$\int_{a}^{b}d\theta\, \delta(\theta-\theta_{1}) = 1$$ (7)

   
   

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

4. Given any continuous integrable function $g(\theta)$, in the $\rho\to 1_{(-)}$ limit the integral

   
   

   $$\int_{a}^{b}d\theta\, g(\theta) \delta(\theta-\theta_{1}) = g(\theta_{1})$$ (8)

   
   

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

   This is the usual form of this condition, when it is formulated in strictly real terms. However, we will impose a slight additional restriction on the real functions $g(\theta)$, by assuming that the limit to the point $z_{1}$ on the unit circle that corresponds to $\theta_{1}$, of the corresponding inner analytic function $w_{\gamma}(z)$, exists and is finite. This implies that $w_{\gamma}(z)$ may have at $z_{1}$ a soft singularity, but not a hard singularity.

Note that, although it is customary to list both separately, the third condition is in fact just a particular case of the fourth condition. It is also arguable that the second condition is not really necessary, because it is a consequence of the others. We may therefore consider that the only really essential conditions are the first and the fourth ones.

The functions $g(\theta)$ are sometimes named test functions within the Schwartz theory of distributions [#!DTSchwartz!#,#!DTLighthill!#]. The additional part of the fourth condition, that the limit to the point $z_{1}$ of the corresponding inner analytic function $w_{\gamma}(z)$ must exist and be finite, consists of a weak limitation on these test functions, and does not affect the definition of the singular distribution itself. This is certainly the case for our definition here, since we define this object through a definite and unique inner analytic function.

In this section we will prove the following theorem.

Theorem 1   : The $\rho\to 1_{(-)}$ limit of the real part of the inner analytic function $w_{\delta}(z,z_{1})$ converges to the generalized function $\delta(\theta-\theta_{1})$.

Before we attempt to prove this theorem, our first task is to write explicitly the real and imaginary parts of $w_{\delta}(z,z_{1})$. In order to do this we must now rationalize it,

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

$$= \frac{1}{2\pi} - \frac{1}{\pi}\, \frac {\rho^{2}-\rho\cos(\Delta\... ...\boldmath$\imath$}\rho\sin(\Delta\theta)} {\rho^{2}-2\rho\cos(\Delta\theta)+1},$$ (9)

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

$$u_{\delta}(\rho,\theta,\theta_{1}) = \frac{1}{2\pi} - \... ...a)\right]} {\left(\rho^{2}+1\right)-2\rho\cos(\Delta\theta)}.$$ (10)

We are now ready to prove the theorem, which we will do by simply verifying all the properties of the Dirac delta ``function''.

Proof 1.1   :

If we take the limit $\rho\to 1_{(-)}$, under the assumption that $\Delta\theta\neq 0$, we get

$$\lim_{\rho\to 1_{(-)}}u_{\delta}(\rho,\theta,\theta_{1}) = \frac{1}{2\pi} - \frac{1}{\pi}\, \frac {1-\cos(\Delta\theta)} {2-2\cos(\Delta\theta)}$$

$$= 0,$$ (11)

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

If, on the other hand, we calculate $u_{\delta}(\rho,\theta,\theta_{1})$ for $\Delta\theta=0$ and $\rho<1$ we obtain

$$u_{\delta}(\rho,\theta_{1},\theta_{1}) = \frac{1}{2\pi} - \frac{1}{\pi}\, \frac {\rho(\rho-1)} {(\rho-1)^{2}}$$

$$= \frac{1}{2\pi} - \frac{1}{\pi}\, \frac{\rho}{\rho-1},$$ (12)

which diverges to positive infinity as $\rho\to 1_{(-)}$, as it should in order to represent the singular Dirac delta ``function''. This establishes that the second condition is satisfied.

We now calculate the real integral of $u_{\delta}(\rho,\theta,\theta_{1})$ over the circle of radius $\rho<1$, which is given by

$$\int_{-\pi}^{\pi}d\theta\, \rho\, u_{\delta}(\rho,\theta,\theta_{1}) = \frac{1}{2\pi} \int_{-\pi}^{\pi}d\theta\, \rho \left\{ 1 - \frac ... ...Delta\theta)\right]} {\left(\rho^{2}+1\right)-2\rho\cos(\Delta\theta)} \right\}$$

$$= \frac{\rho}{2\pi} \int_{-\pi}^{\pi}d(\Delta\theta)\, \frac {\left(1-\rho^{2}\right)} {\left(\rho^{2}+1\right)-2\rho\cos(\Delta\theta)}$$

$$= \frac{\left(1-\rho^{2}\right)}{4\pi} \int_{-\pi}^{\pi}d(\Delta\th... ...\, \frac {1} {\left[\left(\rho^{2}+1\right)/(2\rho)\right]-\cos(\Delta\theta)},$$ (13)

since $d(\Delta\theta)=d\theta$. This real integral over $\Delta\theta$ 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 on the right-hand side as

$$\int_{-\pi}^{\pi}d(\Delta\theta)\, \frac{1}{\left[\left(1+\rho^{2}\right)/(2\rho)\right]-\cos(\Delta\theta)} = \oint_{C}d\xi\, \frac{1}{\mbox{\boldmath$\imath$}\xi}\, \frac{2}{\left[\left(1+\rho^{2}\right)/\rho\right]-\xi-1/\xi}$$

$$= 2\mbox{\boldmath$\imath$} \oint_{C}d\xi\, \frac{1}{1-\left[\left(1+\rho^{2}\right)/\rho\right]\xi+\xi^{2}},$$ (14)

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

$$\xi_{(+)} = 1/\rho,$$

$$\xi_{(-)} = \rho.$$ (15)

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

$$\int_{-\pi}^{\pi}d\theta\, \frac{1}{\left[\left(1+\rho^{2}\right)/(2\rho)\right]-\cos(\Delta\theta)} = 2\mbox{\boldmath$\imath$}(2\pi\mbox{\boldmath$\imath$}) \lim_{\xi\to\rho} \frac{1}{\xi-1/\rho}$$

$$= 4\pi\, \frac{\rho}{\left(1-\rho^{2}\right)}.$$ (16)

It follows that we have for the real integral in Equation (13)

$$\int_{-\pi}^{\pi}d\theta\, \rho\, u_{\delta}(\rho,\theta,\theta_{1}) = \frac{\left(1-\rho^{2}\right)}{4\pi}\, 4\pi\, \frac{\rho}{\left(1-\rho^{2}\right)}$$

$$= \rho,$$ (17)

and thus we have that the integral is equal to $1$ in the $\rho\to 1_{(-)}$ limit. Once we have this result, and since according to the first condition the integrand goes to zero everywhere on the unit circle except at $\Delta\theta=0$, which is the same as $\theta=\theta_{1}$, the integral can be changed to one over any open interval $(a,b)$ on the unit circle containing the point $\theta_{1}$, without any change in its limiting value. This establishes that the third condition is satisfied.

In order to establish the validity of the fourth and last condition, we consider an essentially arbitrary integrable real function $g(\theta)$, with the additional restriction that it be continuous at the point $z_{1}$. As was established in [#!CAoRFI!#], it corresponds to an inner analytic function

$$w_{\gamma}(z) = u_{\gamma}(\rho,\theta) + \mbox{\boldmath$\imath$} v_{\gamma}(\rho,\theta),$$ (18)

where we also assume that $g(\theta)$ is such that $w_{\gamma}(z)$ may have at $z_{1}$ a soft singularity, but not a hard singularity, so that its limit to $z_{1}$ exists. We now consider the following real integral² over the circle of radius $\rho<1$,

$$\int_{-\pi}^{\pi}d\theta\, \rho\, u_{\gamma}(\rho,\theta) u_{\delta}(\rho,\theta,\theta_{1}) = \frac{1}{2\pi} \int_{-\pi}^{\pi}d\theta\, \rho\, u_{\gamma}(\rho,... ...Delta\theta)\right]} {\left(\rho^{2}+1\right)-2\rho\cos(\Delta\theta)} \right\}$$

$$= \frac{\displaystyle \rho}{\displaystyle 2\pi} \int_{-\pi}^{\pi}d(... ...rac {\left(1-\rho^{2}\right)} {\left(\rho^{2}+1\right)-2\rho\cos(\Delta\theta)}$$

$$= \frac{\left(1-\rho^{2}\right)}{4\pi} \int_{-\pi}^{\pi}d(\Delta\th... ...rho,\theta)} {\left[\left(\rho^{2}+1\right)/(2\rho)\right]-\cos(\Delta\theta)},$$ (19)

since $d(\Delta\theta)=d\theta$. This real integral over $\Delta\theta$ can be calculated by residues, exactly like the one in Equation (13) which appeared before in the case of the third condition. The calculation is exactly the same except for the extra factor of $u_{\gamma}(\rho,\theta)$ to be taken into consideration when calculating the residue, so that we may write directly that

$$\int_{-\pi}^{\pi}d(\Delta\theta)\, \frac {u_{\gamma}(\rho,\theta)} {\left[\left(\rho^{2}+1\right)/(2\rho)\right]-\cos(\Delta\theta)} = 2\mbox{\boldmath$\imath$}(2\pi\mbox{\boldmath$\imath$}) \lim_{\xi\to\rho} \frac{u_{\gamma}(\rho,\theta)}{\xi-1/\rho}$$

$$= 4\pi\, \frac{\rho}{\left(1-\rho^{2}\right)} \lim_{\xi\to\rho}u_{\gamma}(\rho,\theta).$$ (20)

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

$$\lambda\,{\rm e}^{\mbox{\boldmath\scriptsize$\imath$}\Delta\theta} = \rho,$$ (21)

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

$$\int_{-\pi}^{\pi}d(\Delta\theta)\, \frac {u_{\gamma}(\rho... ...\rho}{\left(1-\rho^{2}\right)}\, u_{\gamma}(\rho,\theta_{1}).$$ (22)

It follows that we have for the real integral in Equation (19)

$$\int_{-\pi}^{\pi}d\theta\, \rho\, u_{\gamma}(\rho,\theta) u_{\delta}(\rho,\theta,\theta_{1}) = \frac{\left(1-\rho^{2}\right)}{4\pi}\, 4\pi\, \frac{\rho}{\left(1-\rho^{2}\right)}\, u_{\gamma}(\rho,\theta_{1})$$

$$= \rho u_{\gamma}(\rho,\theta_{1}).$$ (23)

Finally, we may now take the $\rho\to 1_{(-)}$ limit, since $w_{\gamma}(z)$ and thus $u_{\gamma}(\rho,\theta)$ are well defined at $z_{1}$ in that limit, and thus obtain

$$\lim_{\rho\to 1_{(-)}} \int_{-\pi}^{\pi}d\theta\, \rho\, u_{\gamma}(\rho,\theta) u_{\delta}(\rho,\theta,\theta_{1}) = u_{\gamma}(1,\theta_{1}) \;\;\;\Rightarrow$$

$$\int_{-\pi}^{\pi}d\theta\, u_{\gamma}(1,\theta) \left[ \lim_{\rho\to 1_{(-)}} u_{\delta}(\rho,\theta,\theta_{1}) \right] = u_{\gamma}(1,\theta_{1}) \;\;\;\Rightarrow$$

$$\int_{-\pi}^{\pi}d\theta\, g(\theta) \left[ \lim_{\rho\to 1_{(-)}} u_{\delta}(\rho,\theta,\theta_{1}) \right] = g(\theta_{1}),$$ (24)

since $u_{\gamma}(\rho,\theta)$ converges to $g(\theta)$, in the $\rho\to 1_{(-)}$ limit, almost everywhere on the unit circle. Just as before, once we have this result, and since according to the first condition the integrand goes to zero everywhere on the unit circle except at $\Delta\theta=0$, which is the same as $\theta=\theta_{1}$, 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 condition is satisfied.

Having established all the properties, we may now write symbolically that

$$\delta(\theta-\theta_{1}) = \lim_{\rho\to 1_{(-)}} u_{\delta}(\rho,\theta,\theta_{1}).$$ (25)

This concludes the proof of Theorem 1.

It is important to note that, when we adopt as the definition of the Dirac delta ``function'' the $\rho\to 1_{(-)}$ limit of the real part of the inner analytic function $w_{\delta}(z,z_{1})$, the limitations imposed on the test functions $g(\theta)$ and on the corresponding inner analytic functions $w_{\gamma}(z)$ become irrelevant. In fact, this definitions stands by itself, and is independent of any set of test functions. Given any integrable real function $f(\theta)$ and the corresponding inner analytic function $w(z)$ with real part $u(\rho,\theta)$, we may always assemble the real integral over a circle of radius $\rho<1$

$$\int_{-\pi}^{\pi}d\theta\, \rho\, u(\rho,\theta) u_{\delta}(\rho,\theta,\theta_{1}),$$ (26)

which is always well defined within the open unit disk. It then remains to be verified only whether or not the $\rho\to 1_{(-)}$ limit of this integral exists, in order to define the corresponding integral

$$\int_{-\pi}^{\pi}d\theta\, f(\theta) \delta(\theta-\theta_{1}).$$ (27)

This limit may exist for functions that do not satisfy the conditions imposed on the test functions. In fact, one can do this for the real part of any inner analytic function, regardless of whether or not it corresponds to an integrable inner analytic function, so long as the $\rho\to 1_{(-)}$ limit of $u(\rho,\theta)$ exists almost everywhere. Whenever the $\rho\to 1_{(-)}$ limit of the integral exists, it defines the action of the delta ``function'' on that particular real object.

It is also interesting to observe that the Dirac delta ``function'', although it is not simply a conventional integrable real function, is in effect an integrable real object, even if it corresponds to an inner analytic functions that has a simple pole at $z_{1}$, which is a non-integrable hard singularity, with degree of hardness equal to one. This apparent contradiction is explained by the orientation of the pole at $z=z_{1}$. If we consider the real part $u_{\delta}(\rho,\theta)$ of the inner analytic function $w_{\delta}(z,z_{1})$, although it is not integrable along curves arriving at the singular point from most directions, there is one direction, that of the unit circle, along which one can approach the singular point so that $u_{\delta}(\rho,\theta)$ is identically zero during the approach, which allows us to define its integral using the $\rho\to 1_{(-)}$ limit<sup>3</sup>. The same is not true, for example, for the imaginary part $v_{\delta}(\rho,\theta)$ of the same inner analytic function, which generates the Fourier-conjugate function to the delta ``function'', and that diverges to infinity as $1/\vert z-z_{1}\vert$ when one approaches the singular point along the unit circle, thus generating a non-integrable real function in the $\rho\to 1_{(-)}$ limit.

In the development presented in [#!CAoRFI!#] the real functions were represented by their Fourier coefficients, and the inner analytic functions by their Taylor coefficients. We can easily do the same here, if we observe that the inner analytic function $w_{\delta}(z,z_{1})$ in Equation (4) is the sum of a geometric series,

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

$$= \frac{1}{2\pi} + \frac{1}{\pi} \sum_{k=1}^{\infty} \left( \frac{z}{z_{1}} \right)^{k}$$

$$= \frac{1}{2\pi} + \frac{1}{\pi} \sum_{k=1}^{\infty} \left[ \cos(k\theta_{1}) - \mbox{\boldmath$\imath$} \sin(k\theta_{1}) \right] z^{k}.$$ (28)

This power series is the Taylor series of $w_{\delta}(z,z_{1})$ around the origin, and therefore it follows that the Taylor coefficients of this inner analytic function are given by

$$c_{0} = \frac{1}{2\pi},$$

$$c_{k} = \frac{\cos(k\theta_{1})}{\pi} - \mbox{\boldmath$\imath$}\, \frac{\sin(k\theta_{1})}{\pi},$$ (29)

where $k\in\{1,2,3,\ldots,\infty\}$. Since according to the construction presented in [#!CAoRFI!#] we have that $c_{0}=\alpha_{0}/2$ and that $c_{k}=\alpha_{k}-\mbox{\boldmath$\imath$}\beta_{k}$, we have for the Fourier coefficients of the delta ``function''

$$\alpha_{0} = \frac{1}{\pi},$$

$$\alpha_{k} = \frac{\cos(k\theta_{1})}{\pi},$$

$$\beta_{k} = \frac{\sin(k\theta_{1})}{\pi},$$ (30)

where $k\in\{1,2,3,\ldots,\infty\}$. Note that these are in fact the results one obtains via the integrals defining the Fourier coefficients [#!FSchurchill!#],

$$\alpha_{k} = \frac{1}{\pi} \int_{-\pi}^{\pi}d\theta\, \cos(k\theta)\delta(\theta-\theta_{1}),$$

$$\beta_{k} = \frac{1}{\pi} \int_{-\pi}^{\pi}d\theta\, \sin(k\theta)\delta(\theta-\theta_{1}),$$ (31)

by simply using the fundamental property of the delta ``function''.

Having established the representation of the Dirac delta ``function'' within the structure of the inner analytic functions, in sequence we will show that the Dirac delta ``function'' is not the only singular distribution that can be represented by an inner analytic function. As we will see, one can do the same for its first derivative, and in fact for its derivatives of any order. This is an inevitable consequence of the fact that the proper inner analytic function $w_{\delta}^{0\mbox{\Large$\cdot$}\!}(z,z_{1})$ associated to $w_{\delta}(z,z_{1})$ is a member of an integral-differential chain.