Proof of Uniqueness Almost Everywhere

Given a sequence of Fourier coefficients $a_{k}$ of an integrable DP real function $f(\theta)$ with zero average, let us show that they uniquely characterize that real function almost everywhere, that is, up to a zero-measure function. The point of the proof is to show that this is true independently of the convergence of the Fourier series of the function, hence including the cases in which that series diverges almost everywhere. Since two real functions with opposite parities clearly must always have different sequences of Fourier coefficients, it is enough to show that the result holds within the two sets of real functions with the same parity. Although the definition of the integral in itself will not be used directly, the argument will involve the concept of a zero-measure function, and therefore we should always think in terms of the Lebesgue integral and of the usual Lebesgue measure, for conceptual reasons as well as for the sake of generality. Hence, when we talk here of an integrable function, we mean integrable in the sense of Lebesgue.

Imagine then that we are given two different integrable DP real functions $f_{1}(\theta)$ and $f_{2}(\theta)$, both with zero average and both with the same parity, and consider their difference

$$g(\theta) = f_{1}(\theta)-f_{2}(\theta),$$

which of course has the same parity as $f_{1}(\theta)$ and $f_{2}(\theta)$, and which is also a zero-average function. Let us assume that both $f_{1}(\theta)$ and $f_{2}(\theta)$ have the same sequence of Fourier coefficients $a_{k}$, for $k=1,2,3,\ldots,\infty$. It follows that $g(\theta)$ has all its Fourier coefficients $b_{k}$ equal to zero, since from the relation above we clearly have

$$b_{k} = a_{k}-a_{k},$$

for all $k\geq 1$. This means that the function $g(\theta)$ has zero scalar product with all the elements of the Fourier basis, and thus that it is orthogonal to all of them. However, as shown in [1], that basis satisfies a completeness relation,

$$\delta(\theta-\theta_{1}) = \frac{1}{2\pi} + \frac{1}{\... ...os(k\theta_{1}) + \sin(k\theta) \sin(k\theta_{1}) \right],$$ (5)

for all $\theta$ and all $\theta_{1}$, and is therefore complete. From this it follows that $g(\theta)$, being orthogonal to all the elements of the basis, must be a zero-measure function, that is, a function which is zero-measure equivalent to the identically zero function. In order to show this, we simply multiply the equation above by $g(\theta)$ and integrate on $\theta$ over the periodic interval,

$$\begin{eqnarray*} \int_{-\pi}^{\pi}d\theta\, g(\theta) \delta(\theta-\theta_{... ...(\theta)\, \sin(k\theta) \right] \sin(k\theta_{1}) \right\}. \end{eqnarray*}$$

The integral on the left-hand side can be calculated through the use of the properties of the delta ``function'', which were given and demonstrated in [1]. The three expressions within square brackets on the right-hand side are the various Fourier coefficients of the function $g(\theta)$. The first one is zero because $g(\theta)$, just like $f_{1}(\theta)$ and $f_{2}(\theta)$, is a zero-average function. The other two sequences of coefficients are identically zero because the scalar products of $g(\theta)$ and the elements of the basis are all zero. Besides, one of the sequences contains the coefficients of the part of the basis which has the wrong parity, and thus these coefficients must all be zero by a parity argument. Therefore the whole right-hand side is zero and hence we have

$$g(\theta_{1}) = 0,$$

which is valid for almost all values of $\theta_{1}$. Therefore $g(\theta)$ is zero almost everywhere. In other words, $f_{1}(\theta)$ and $f_{2}(\theta)$ differ only by a zero-measure function, and are thus equal to each other almost everywhere.

We may conclude from this that, if $f_{1}(\theta)$ and $f_{2}(\theta)$ are to be different by more than a zero-measure function, then their sequences of Fourier coefficients must be different, and therefore the coefficients uniquely characterize the originating function, up to a zero-measure function. Observe that nothing in this argument involves the convergence of the Fourier series of the real functions. In particular, the proof of completeness of the Fourier basis presented in [1] and expressed by the relation in Equation (5) was obtained directly from the complex analytic structure within the open unit disk, and it also does not involve in any way the convergence of the Fourier series on the unit circle.