Completeness Relation

Let us now prove the completeness of the Fourier basis. In this context the concept of completeness is that of a basis within a vector space. We will first give a simple and direct proof of completeness, which is however subject to a slight limitation regarding the vector space for which the basis is shown to be complete, using the analytic structure within the open unit disk, and later establish the relation of the concept of completeness with the so-called completeness relation. The proof of completeness using the completeness relation is not subject to any such limitation.

In this section we will prove the following completeness theorem.

Theorem 1   : The basis of real functions $\left\{\rule{0em}{2ex}1,\cos(k\theta),\sin(k\theta), k\in\{1,2,3,\ldots,\infty\}\right\}$, is complete to represent the space of all integrable real functions defined on the unit circle.

The proof consists of establishing that, given an arbitrary integrable real function $\psi(\theta)$ on the unit circle, which is orthogonal to all the elements of the Fourier basis, according to the scalar product defined in Equation (25), it then follows that $\psi(\theta)$ must be zero almost everywhere. Note that the orthogonality to the elements of the basis means that $\psi(\theta)$ is such that all its Fourier coefficients, as defined in Equation (4), are zero.

Proof 1.1   :

Let $\psi(\theta)$ be a real function on the unit circle which can be obtained as the $\rho\to 1_{(-)}$ limit of an inner analytic function. We assume that it is orthogonal to all the elements of the basis, so that all its Fourier coefficients are zero, that is, we assume that for this function we have $\alpha_{0}=0$, $\alpha_{k}=0$ and $\beta_{k}=0$, for all $k\in\{1,2,3,\ldots,\infty\}$. Since we thus have all the Fourier coefficients of $\psi(\theta)$, we may use the construction presented in [#!CAoRFI!#] in order to determine the corresponding inner analytic function. However, since all the Fourier coefficients are zero, it follows at once from the step of that construction given in Equation (5) that for $\psi(\theta)$ the complex coefficients $c_{k}$ are zero for all $k$. Therefore, the power series $S(z)$ constructed in the next step of the process, given in Equation (6), is identically zero and thus converges trivially to the identically zero complex function $w_{\psi}(z)\equiv 0$ on the whole complex plane.

The analyticity region of $w_{\psi}(z)$ includes the unit circle, and therefore the series converges to zero there. Since on the one hand the series converges to zero, and on the other hand we know that for $\rho=1$ it necessarily converges to the restriction of $w_{\psi}(z)$ to the unit circle, it follows that the restriction, including both real and imaginary parts, must be zero everywhere on the unit circle. Therefore it follows that $\psi(\theta)$ and the identically zero real function coincide everywhere on the unit circle, and therefore we conclude that $\psi(\theta)=0$ everywhere on that circle. This establishes that the Fourier basis is complete for the space of all integrable real functions defined on the periodic interval, which can be obtained as the $\rho\to 1_{(-)}$ limits of inner analytic functions. This completes the first version of the proof of Theorem 1, which is valid for the vector space of real functions just described.

Note that, since all possible inner analytic functions are given by convergent power series within the open unit disk, and since these power series can be understood as expansions of those inner analytic functions in the Taylor basis of functions given by the non-negative powers $\left\{\rule{0em}{2ex}z^{k},k\in\{0,1,2,3,\ldots,\infty\}\right\}$, we may say that this Taylor basis is complete for the space of all inner analytic functions. Since the proof of the completeness of the Fourier basis given above was obtained from the complex-analytic structure within the open unit disk, it becomes clear that the completeness of the Fourier basis on the unit circle is a consequence of the completeness of the Taylor basis within the open unit disk. This adds to the relationship between the Fourier basis on the unit circle and the Taylor basis on the unit disk, which was first established during the discussion involving the orthogonality of the Fourier basis, in Section 3. In addition to all this, within the spaces generated by either basis one may define scalar products that induce positive-definite norms, thus making them both Hilbert spaces, as is discussed in Appendix A.

Let us now turn to the usual completeness relation. Let us first write it down and then exhibit its usefulness. The relation can be understood as the expression, as a Fourier series, of the Dirac delta ``function'' defined with respect to a point given by the angle $\theta_{1}$ on the unit circle, which we examined in great detail in [#!CAoRFII!#], and which we denote by $\delta(\theta-\theta_{1})$. As we have shown in [#!CAoRFII!#], using the usual rules for the manipulation of the delta ``function'', one finds that the corresponding Fourier coefficients are given by

$$\alpha_{0} = \frac{1}{\pi} \int_{-\pi}^{\pi}d\theta\, \delta(\theta-\theta_{1})$$

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

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

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

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

$$= \frac{1}{\pi}\, \sin(k\theta_{1}),$$ (26)

for $k\in\{1,2,3,\ldots,\infty\}$, so that the completeness relation is given by the Fourier expansion, that turns out to be a bi-linear form on the elements of the Fourier basis,

$$\delta(\theta-\theta_{1}) = \frac{1}{2\pi} + \frac{1}{\... ... \cos(k\theta) + \sin(k\theta_{1}) \sin(k\theta) \right],$$ (27)

which is manifestly divergent, but which can be made to converge for all values of $\theta$, so that we may recover the delta ``function'' almost everywhere, in fact everywhere but at $\theta_{1}$, through the use of the summation rule given in Equation (12),

$$\delta(\theta-\theta_{1}) = \frac{1}{2\pi} + \frac{1}{\... ... \cos(k\theta) + \sin(k\theta_{1}) \sin(k\theta) \right].$$ (28)

This is equivalent to the definition of the delta ``function'' as the $\rho\to 1_{(-)}$ limit of the real part of the inner analytic function given by

$$w_{\delta}(z,z_{1}) = \frac{1}{2\pi} - \frac{1}{\pi}\, \frac{z}{z-z_{1}},$$ (29)

as was discussed in detail in [#!CAoRFII!#]. One can use the expansion in Equation (27), possibly regulated as in Equation (28), to prove the completeness of the basis, while operating strictly in terms of real objects on or near the unit circle. Here is how this can be done.

Proof 1.2   :

If we assume that an arbitrary integrable real function $\psi(\theta)$ on the unit circle is given, which is such that its scalar products with all the elements of the basis are zero, then we have the infinite set of equations

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

$$\int_{-\pi}^{\pi}d\theta\, \cos(k\theta)\psi(\theta) = 0,$$

$$\int_{-\pi}^{\pi}d\theta\, \sin(k\theta)\psi(\theta) = 0,$$ (30)

for all $k\in\{1,2,3,\ldots,\infty\}$. We may therefore construct an infinite linear combination of all these equations, with the coefficients carefully chosen as shown below, involving an arbitrary parameter $\theta_{1}$ in the interval $[-\pi,\pi]$ and an auxiliary strictly positive real variable $\rho<1$, where the right-hand side is still zero,

$$\begin{eqnarray*} \left[ \frac{1}{2\pi} \right] \int_{-\pi}^{\pi}d\theta\, ... ...(k\theta)\psi(\theta) \hspace{1em} & = & 0 \;\;\;\Rightarrow \end{eqnarray*}$$

$$\int_{-\pi}^{\pi}d\theta\, \left\{ \frac{1}{2\pi} + \fr... ... \sin(k\theta) \right] \right\} \psi(\theta) \;\;=\;\; 0.$$ (31)

Since the expression within curly brackets in this last integral is now seen to be the regulated expansion of $\delta(\theta-\theta_{1})$ in the Fourier basis, shown in Equation (28), we may therefore take the $\rho\to 1_{(-)}$ limit and write that

$$\int_{-\pi}^{\pi}d\theta\, \delta(\theta-\theta_{1}) \psi(\theta) = 0.$$ (32)

Finally, using the rules of manipulation of the delta ``function'', when and where $\psi(\theta)$ is continuous, which it therefore must be almost everywhere, we have

$$\psi(\theta_{1}) = 0.$$ (33)

Since $\theta_{1}$ is an arbitrary value of $\theta$, we conclude that $\psi(\theta)$ is zero everywhere. This completes the second version of the proof of Theorem 1, which is valid for the vector space of all integrable real functions defined on the unit circle, regardless of whether or not they can be obtained from an inner analytic function.

Note that, in a sense, this method of proof of the completeness of the Fourier basis is a little more limited than the direct proof using the analytic structure within the open unit disk, because we must assume during the argument that $\psi(\theta)$ is continuous almost everywhere. However, since this hypothesis does get confirmed a posteriori by the result obtained, this is not a true limitation.

On the other hand, this second proof is less limited than the first one because in this case the vector space of functions for which one shows that the basis is complete is the space of integrable real functions without removable singularities defined on the interval $[-\pi,\pi]$, with no reference to whether or not these functions can be obtained as the $\rho\to 1_{(-)}$ limits of inner analytic functions.

In fact, by establishing the completeness of the Fourier basis without any recourse to the $\rho\to 1_{(-)}$ limit for the real functions, as a corollary of this second proof we have shown that there is no integrable real function on the unit circle, other that the identically zero real function, which corresponds to the identically zero inner analytic function. As a consequence of this, there is no integrable real function defined on the unit circle that cannot be represented by an unique inner analytic function.