Appendix: Scalar Product for Inner Analytic Functions

Given two inner analytic functions $w_{1}(z)$ and $w_{2}(z)$, we consider the complex contour integral over the circle $C_{0}$ of radius $\rho_{0}$, with $0<\rho_{0}<1$, given by

$$(w_{1}\vert w_{2}) = \frac{1}{2\pi\mbox{\boldmath$\imath$}} \oint_{C_{0}}dz\, \frac{1}{z}\, w_{1}^{*}(z)w_{2}(z).$$ (61)

Since the integrand in this expression is not analytic, the integral depends on the circuit, and therefore on $\rho_{0}$. Therefore, what we have here is in fact a one-parameter family of integrals. We will show that for each value of $\rho_{0}$ this integral defines a scalar product within the space of inner analytic functions, which induces in that space a positive-definite norm. If we write the integral in terms of the integration variable $\theta$, with constant $\rho_{0}$, we get for this scalar product

$$(w_{1}\vert w_{2}) = \frac{1}{2\pi} \int_{-\pi}^{\pi}d\theta\, w_{1}^{*}(\rho_{0},\theta)w_{2}(\rho_{0},\theta).$$ (62)

If we now make both $w_{1}(z)$ and $w_{2}(z)$ equal to $w(z)=u(\rho,\theta)+\mbox{\boldmath$\imath$}v(\rho,\theta)$, we get

$$(w\vert w) = \Vert w\Vert^{2}$$

$$= \frac{1}{2\pi} \int_{-\pi}^{\pi}d\theta\, \vert w(\rho_{0},\theta)\vert^{2}$$

$$= \frac{1}{2\pi} \int_{-\pi}^{\pi}d\theta\, \left[ u^{2}(\rho_{0},\theta) + v^{2}(\rho_{0},\theta) \right]$$

$$\geq 0,$$ (63)

which is a manifestly real and positive quantity, that is zero if and only if $w(\rho_{0},\theta)=0$ for all $\theta$, which in turn is equivalent to $w(\rho,\theta)=0$ for all $\theta$ and all $\rho$ within the open unit disk, because all zeros of an analytic function must be isolated, unless it is the identically zero function. Therefore, for each value of the parameter $\rho_{0}$ the real quantity $\Vert w\Vert$ is a positive-definite norm on the space of all inner analytic functions which, as was observed in [#!CAoRFI!#], forms a vector space over the field of complex numbers. That vector space is thus seen to constitute a complex Hilbert space, with this scalar product and the associated positive-definite norm.

We can also see from the equation above that the scalar product and the norm reduce naturally to the corresponding definitions for the real functions $u(1,\theta)$ and $v(1,\theta)$ on the unit circle, when we take the $\rho_{0}\to 1_{(-)}$ limit, thus establishing a close correspondence between these two identical real Hilbert spaces on the unit circle and the complex Hilbert space on the unit disk. In addition to this, for any value of $\rho_{0}$ within the open interval $(0,1)$ we also have a pair of identical real Hilbert spaces with the real functions $u(\rho_{0},\theta)$ and $v(\rho_{0},\theta)$ on the circle of radius $\rho_{0}$.

We may now show that the Taylor basis of functions around the origin, which is complete to generate the whole space of inner analytic functions, and which consists of the set of non-negative powers

$$\left\{ \rule{0em}{2.5ex}z^{k}, k\in\{0,1,2,3,\ldots,\infty\} \right\},$$ (64)

is in fact an orthogonal basis according to this definition of the scalar product. If we make $w_{1}(z)=w_{k_{1}}(z)=z^{k_{1}}$ and $w_{2}(z)=w_{k_{2}}(z)=z^{k_{2}}$, we get

$$(w_{k_{1}}\vert w_{k_{2}}) = \frac{1}{2\pi} \int_{-\pi}^{\pi}d\theta\, \left(z^{k_{1}}\right)^{*}z^{k_{2}}$$

$$= \rho_{0}^{k_{1}+k_{2}}\, \frac{1}{2\pi} \int_{-\pi}^{\pi}d\theta\... ...\imath$}k_{1}\theta}\,{\rm e}^{\mbox{\boldmath\scriptsize$\imath$}k_{2}\theta}.$$ (65)

Using now the first result shown in Equation (18) we obtain the orthogonality relation for the Taylor basis,

$$(w_{k_{1}}\vert w_{k_{2}}) = \rho_{0}^{k_{1}+k_{2}}\, \delta_{k_{1},k_{2}}.$$ (66)

Since the integer powers are analytic on the whole complex plane, there is no obstruction to taking the $\rho_{0}\to 1_{(-)}$ limit, and thus we see that in this case the Taylor basis is not only orthogonal, but also normalized,

$$(w_{k_{1}}\vert w_{k_{2}}) = \delta_{k_{1},k_{2}},$$ (67)

with $\Vert w_{k}\Vert=1$ for all $k$, where the scalar product is now defined on the unit circle. If we write the inner analytic functions in terms of their Taylor series around the origin,

$$w_{1}(z) = \sum_{k=0}^{\infty} c_{1,k}z^{k},$$

$$w_{2}(z) = \sum_{k=0}^{\infty} c_{2,k}z^{k},$$ (68)

we obtain for the scalar product, since we may always integrate convergent power series term-by-term,

$$(w_{1}\vert w_{2}) = \frac{1}{2\pi\mbox{\boldmath$\imath$}} \oint_{C_{0}}dz\, \frac{1}... ..._{2}=0}^{\infty} c_{1,k_{1}}^{*}c_{2,k_{2}} \left(z^{k_{1}}\right)^{*}z^{k_{2}}$$

$$= \sum_{k_{1}=0}^{\infty} \sum_{k_{2}=0}^{\infty} c_{1,k_{1}}^{*}c_... ...ath$\imath$}} \oint_{C_{0}}dz\, \frac{1}{z} \left(z^{k_{1}}\right)^{*}z^{k_{2}}$$

$$= \sum_{k_{1}=0}^{\infty} \sum_{k_{2}=0}^{\infty} c_{1,k_{1}}^{*}c_{2,k_{2}} (w_{k_{1}}\vert w_{k_{2}})$$

$$= \sum_{k_{1}=0}^{\infty} \sum_{k_{2}=0}^{\infty} c_{1,k_{1}}^{*}c_{2,k_{2}}\, \rho_{0}^{k_{1}+k_{2}}\, \delta_{k_{1},k_{2}}$$

$$= \sum_{k=0}^{\infty} \rho_{0}^{2k} c_{1,k}^{*}c_{2,k},$$ (69)

where we identified the scalar product $(w_{k_{1}}\vert w_{k_{2}})$ and then used the orthogonality relations of the Taylor basis. So long as $\rho_{0}<1$, and so long as $c_{1,k}$ and $c_{2,k}$ are exponentially bounded, this series converges exponentially fast. We may also write the corresponding expression for the norm, if we make $c_{1,k}=c_{2,k}=c_{k}$ and $w_{1}(z)=w_{2}(z)=w(z)$,

$$\Vert w\Vert^{2} = (w\vert w)$$

$$= \sum_{k=0}^{\infty} \rho_{0}^{2k} \vert c_{k}\vert^{2},$$ (70)

with the same conditions for the convergence of the series. In all this structure, if we take the $\rho_{0}\to 1_{(-)}$ limit, the scalar product and the norm may in general diverge, unlike what happens in the case of the elements of the Taylor basis. However, so long as $\rho_{0}<1$ all the inner analytic functions have finite norms and finite scalar products with one another. In some cases, it may be possible to determine the values of these quantities on the unit circle using the $\rho_{0}\to 1_{(-)}$ limit, even if the corresponding series expressions written directly on the unit circle diverge.

Perhaps the best way to characterize this structure is as a one-parameter family of pairs of identical real Hilbert spaces, one associated to the real parts and another associated to the imaginary parts of the inner analytic functions, where the parameter is the radius $\rho_{0}$ of each circle within the unit disk, which are connected to each other by a process of analytic continuation. For each value of $\rho_{0}$ within the open interval $(0,1)$ there is a one-to-one mapping between the inner analytic functions on the open unit disk and the real functions obtained as the real parts of these inner analytic function restricted to the circle of radius $\rho_{0}$. This one-to-one mapping preserves the scalar product and the norm, as they are defined within each space. This fact is still true even in the $\rho_{0}\to 1_{(-)}$ limit, although in that case not every real object at the unit circle, resulting from the limit, is a normal real function, and although in many cases the norms and scalar products may diverge in the limit.

Note that the integral defining the scalar product of the inner analytic functions is a one-dimensional integral over the circle of radius $\rho_{0}$, despite the fact that each complex inner analytic function consist of a pair of real functions of two variables. However, this is a natural characteristic of the scalar product in this context, since it is a well-known fact that an analytic function is completely determined on a two-dimensional region of the complex plane by its values only at a one-dimensional boundary of that region. In this way, although only a one-dimensional restriction of the inner analytic function is explicitly taken into account in the integral over the circle of radius $\rho_{0}$ that defines the scalar product, that restriction still includes implicitly the whole structure of the inner analytic function within the corresponding disk of radius $\rho_{0}$. Therefore, it is perhaps arguable that the most natural definition of the scalar product is that associated to the choice $\rho_{0}=1$, despite the convergence issues that this choice may involve.