# How do we need to read the Dirac-comb identity?

Let

• $$(\Omega,\mathcal A)$$ be a measurable space
• $$\omega\in\Omega$$
• $$\delta_\omega$$ denote the Dirac measureat $$\omega$$ on $$(\Omega,\mathcal A)$$
• $$E$$ be a $$\mathbb R$$-Banach space
• $$\mathcal M$$ denote the set of strongly $$\mathcal A$$-measurable $$f:\Omega\to E$$

If $$f\in\mathcal M$$, $$\delta_\omega(f):=\int f\:{\rm d}\delta_\omega=f(\omega)$$ is well-defined. In that sense, $$\delta_\omega$$ can be thought of as an element of the algebraic dual space $$\mathcal M^\ast$$ of $$\mathcal M$$.

I'm able to make sense of the broadly used notation $$\delta(\;\cdot\;-\omega)$$. It's just a symbol sequence refering to $$\delta_\omega$$.

However, what does the notation $$\delta(\omega'-\omega)$$ mean?

The question arose as I've tried to understand the definition of the Dirac comb. It's defined to be $$Ш_T(x):=T\sum_{k\in\mathbb Z}\delta(x-kT)\;\;\;\text{for }T>0\text{ and }x\in\mathbb R.$$ Now, I've seen the following identity: $$Ш_T(x)f(x)=T\sum_{k\in\mathbb Z}f(kT)\delta(x-kT).$$ Above $$f$$ is a function $$\mathbb R\to\mathbb R$$. $$f(x)$$ is a real number. How is $$Ш_T(x)$$ acting on a real number? Shouldn't that identity read $$Ш_T(f)=T\sum_{k\in\mathbb N}\delta(f-kT)=T\sum_{k\in\mathbb N}\delta_{kT}(f)=T\sum_{k\in\mathbb N}f(kT)?$$

(Maybe it's just a matter of rigorous notation. Many people write things like "let $$f(x)$$ be a function", which actually doesn't make sense; $$f$$ is the function and $$f(x)$$ is the value of that function at $$x$$)

It's only a matter of notation. In engineering and signal processing, notation $$f(x)$$ (such as $$\text{III}_T(x)$$) can be interpreted either as a number (corresponding to a specific $$x$$), or as a function (with the same domain of definition as that of the independent variable $$x$$). This means that notation $$f(x) g(x)$$ will commonly refer to a function over the whole range of $$x$$, i.e., $$(fg)(x)$$. Both interpretations are valid, and it's only a matter of convenience which one to consider.

The identity formula for $$\text{III}_T(x) f(x)$$ can indeed be interpreted both ways:

a) It says that the product of two functions, $$\text{III}_T(x)$$ and $$f(x)$$, both defined over $$\mathbb{R}$$, is similar to a dirac comb, however, with its dirac delta functions not of the same magnitude but "shaped" according to (the function) $$f(x)$$.

b) It says that the product of two numbers, $$\text{III}_T(x)$$ and $$f(x)$$ (with $$x$$ being a fixed number, say, $$x=3/4$$) is (as expected) a number, with the rule being $$\text{III}_T(x) f(x) = \begin{cases} f(x), & \text{if x=kT, for any k\in \mathbb{Z},} \\ 0, & \text{otherwise.} \end{cases}$$

Again, both interpretations are valid and essentially say the same thing.

Now, what does notation $$\delta(\omega'-\omega)$$ mean? Again, two interpretations:

• it is a single number, which is equal to $$1$$ if $$\omega'=\omega$$ and $$0$$ otherwise
• Treating $$\omega'$$ as the independent variable (e.g., $$\omega' \in \mathbb{R}$$) and $$\omega$$ as a fixed number, it is a function that is a shifted version (by $$\omega$$) of the function $$\delta(\omega')$$. (You could instead treat $$\omega$$ as the independent variable, the choice will be clear from context)

P.S.: Excuse my referring to Dirac deltas and combs as (normal) "functions". For engineering purposes/applications there is no issue saying this.

• So, with interpretation (a), $Ш_Tf$ is actually a measure equal to the sum of measures $Tf(kT)\delta_{kT}$ over $k\in\mathbb Z$, right? To be precise, and I just saw that, $fШ_T$ is the so-called measure with density $f$ with respect to $Ш_T$. That makes absolutely sense to me now. – 0xbadf00d Sep 29 '18 at 16:35
• If my comment to (a) is correct, this case is completely captured by measure theory. But I got some problems to understand (b): Can you explain (b) using the identification of the $\delta_{kT}$'s as distributions (continuous linear functionals)? – 0xbadf00d Sep 29 '18 at 16:35