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

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?

• (1) As written, I don't think the reasoning is valid. (2) Yes, if you consider the fact that the Dirac delta function is not a function and $\delta(x) f(x)$ is not Riemann integrable. On the other hand, to actually conclude that the mean converges to "$dA(0)$" you don't need to reason "via" the delta function. You can do it rigorously and directly (at least when $dA(\omega)$ is replaced by $f d\omega$ where $d\omega$ is the Lebesgue measure and $f$ is some continuous function.) – Willie Wong May 5 '15 at 13:10
• @Willie Wong As to your comment (1): Can you explain you basis for why you think the reasoning is invalid? It is valid for a Riemann integral. You can tell by the question I suspect the logic, but I want to understand what is wrong. Comment (2): Since $dA(\omega)$ is part of a Fourier-Stieltjes integral it cannot "just" be replaced by $fd\omega$. – P T May 5 '15 at 19:24

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.

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 $$\lim_{T \to\infty} \int\mathrm{d}A(\omega) g_T(\omega) = A(\{0\})$$

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 $x\in E_0 \setminus E_{i(\epsilon)}$