How does a Dirac delta function operate on a Fourier-Stieltjes integral?

Question

Consider a stationary complex random function $\zeta(t)$ represented as a Fourier-Stieltjes integral $$\zeta(t) = \int_{-\infty}^{+\infty} e^{i\omega t}dA(\omega)$$ where $dA(\omega)$ is the random amplitude. Find the mean value of $\zeta(t)$ using a temporal average. A temporal average is taken over a finite interval $T$, and then the limit is found as $T$ goes to infinity $$\overline {\zeta(t)} = \lim_{T\to \infty} \frac 1T \int_{-T/2}^{+T/2}\zeta(t) dt$$ Substitute the Fourier-Stieltjes integral into the equation and perform the $t$ integration to get $$\overline {\zeta(t)} = \lim_{T\to \infty} \frac 1T \int_{-\infty}^{+\infty} dA(\omega) \int_{-T/2}^{+T/2} e^{i\omega t}dt = \lim_{T\to \infty} \frac 1T \int_{-\infty}^{+\infty} dA(\omega) \frac {2sin(\omega T/2)} {\omega} $$ Take the limit using the following expressions
$$ \lim_{T\to \infty} \frac {2\sin(\omega T/2)}{\omega} \rightarrow 2\pi\delta(\omega) \quad \mathrm{and} \quad \lim_{T\to \infty} \frac 1T = \lim_{T\to \infty} \frac {\Delta\omega} {2\pi}\rightarrow \frac {d\omega}{2\pi} $$ which seems to give for the temporal average $$\overline {\zeta(t)} = \int_{-\infty}^{+\infty} dA(\omega) \ \delta(\omega) \ d\omega = (???) dA(0) $$ Two questions arise: 1) Is the limiting process used above valid in the context of a Fourier-Stieltjes integral? 2) If the limit is valid, does the Dirac delta function operate on a Fourier-Stieltjes integral the same as it operates on a Reimann integral?

Answer 1

Let me expand a bit more on why what you wrote is wrong.

1. Since you are asking whether your reasoning is valid, you cannot just pretend that you can "Riemann integrate" the expression $$ \int f(x) \delta(x) \mathrm{d}x $$ The "limit" you wrote, where $$ \tag{1} \lim_{T\to \infty} \sin(\pi \omega T) / \pi \omega "\to" \delta(\omega) $$ holds only in the weak sense. And that the correct, and rigorous statement, which does not involve anything that is not Riemann integrable, is that for a continuous function of compact support $f$ on $\mathbb{R}$ you have that $$ \lim_{T\to \infty} \int f(\omega) \frac{\sin (\pi \omega T)}{\pi\omega} \mathrm{d}\omega = f(0) \tag{2}$$

   In particular, for arbitrary Riemann integrable (but not necessarily continuous) functions, the above limit need not converge, and when it does converge, it does not have to converge to $f(0)$. (Consider the family of functions $f_a(x)$ such that $f_a(x) = +1$ if $x \in (0,1)$ and $-1$ if $x\in(-1,0)$, and $f_a(0) = a$ and $f_a(x) = 0$ otherwise. For every fixed $T$ the integral $\int f_a(\omega) \frac{\sin(\pi \omega T)}{\pi\omega} \mathrm{d}\omega$ exists and is independent of $a$. On the other hand if you integrate $f_a$ against the singular Dirac measure concentrated at $0$, you will get an integral $= f_a(0) = a$.

   This goes to illustrate that you cannot say that the "Dirac delta function operator on the Riemann integral" in the sense you want, since for arbitrary Riemann integrable function $f$ you actually cannot make use of the weak limit (1) to conclude that the limits of the integral exist and that the integral converges to the integral of $f$ with respect to the Dirac measure.

2. Interestingly, you can note here that (2) is without the $\frac1T$ normalization that you introduced in your question. In particular, if your measure $\mathrm{d}A(\omega)$ is absolutely continuous with respect to the Lebesgue measure, and that the Radon-Nikodym derivative (the probability distribution function if you will) $\lambda$ is continuous with compact support, you actually have that the quantity $\bar{\zeta}$ as you defined it must vanish identically.

3. Whereas the first limit that you wrote can be interpreted to be more-or-less true in the weak sense, the second limit as far as I can tell is just pure nonsense. It appears that you are trying to use something like Heisenberg's uncertainty principle, but not that that is an inequality, not an equality.

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But let us look at how we can actually compute your $\bar{\zeta}$. Since you are working with a Fourier-Stieltjes integral, I assume your measure $A$ is finite Borel. So you have reached the point of trying to compute

$$ \lim_{T\to\infty} \int \mathrm{d}A(\omega) \frac{\sin( \omega T/2)}{\omega T / 2} $$

We note here that writing $$ g(\omega) = \sin(\omega / 2) / (\omega/2) $$ and $$ g_T(\omega) = g(\omega T) $$ we have that $$ |g_T(\omega)| \leq 1 $$ and that for every $\omega_0 > 0$, the functions $g_T |_{\mathbb{R}\setminus(-\omega_0,\omega_0)}$ converge uniformly to $0$. while $$ g_T(0) = 1.$$

Inspired by the above properties, we do the following: Since $A$ is a finite measure, we can decompose $\mathbb{R}$ in the following way.

• First isolate the set $\{0\}$; if $A(\mathbb{R}\setminus \{0\}) = 0$ then we are done. Else denote $E_0 = \mathbb{R}\setminus \{0\}$, and
• by countable additivity you have that there exists a strictly decreasing sequence $a_i$, with $a_0 = +\infty$, such that $$ A(E_0 \cap (-a_i,a_i) ) \leq \frac12 A(E_0 \cap (-a_{i-1}, a_{i-1})) $$ and define $$ E_i = E_0 \cap (-a_i, a_i), F_i = E_i \setminus E_{i+1}.$$ Note that we have $\mathbb{R}$ is the disjoint union of all $F_i$ together with $\{0\}$.

Observe that $$ |g_T |_{F_i} | \leq \min(1, 2(a_{i+1} T)^{-1} )$$ So by construction $$ \begin{align*} \left|\int \mathrm{d}A(\omega) g_T(\omega) - A(\{0\}) \right| &\leq \sum_{i = 0}^\infty \int_{F_i} \mathrm{d}A(\omega) |g_T(\omega) | \\ & \leq \sum A(F_i) \min(1, 2(a_{i+1} T)^{-1}) \\ & \leq A(\mathbb{R}) \sum 2^{-i} \min(1, 2(a_{i+1} T)^{-1}) \end{align*} $$ Now, let $T$ be fixed, and suppose $i(T)$ is the largest (which we allow to be $+\infty$) value for which $a_{i(T) + 1} \sqrt{T} \geq 2$. Then we have the sum $$ \sum 2^{-i} \min(1,2(a_{i+1} T)^{-1}) = \sum_{i = 0}^{i(T)} + \sum_{i = i(T) + 1}^\infty $$ The first sum we the summands are bounded by $2^{-i} \frac{1}{\sqrt{T}}$ so that the sum is bounded by $\frac{2}{\sqrt{T}}$. The second sum we can bound the summands by $2^{-i}$ so that the sum is bounded by $2^{-i(T)}$.

Now note that $i(T)$ grows unboundedly as $T$ increases. So we have that $$ \lim_{T\to \infty} \left| A(\{0\}) - \int \mathrm{d}A(\omega) g_T(\omega) \right| = 0 $$ or $$ \lim_{T \to\infty} \int\mathrm{d}A(\omega) g_T(\omega) = A(\{0\}) $$

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Note that the explicit computations involve $g_T$ in the proof above can be replaced by the following properties:

• $g_T(x) \leq 1$
• $g_T(0) = 1$
• For every $\epsilon > 0$ there exists $i(\epsilon)$ and $T(\epsilon)$ such that
  • $A(E_{i(\epsilon)}) < \epsilon /2$
  • $|g_T(x)| < \epsilon / 2$ for all $T > T(\epsilon)$ and $x\in E_0 \setminus E_{i(\epsilon)}$