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

Question

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 $x$)

Answer 1

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.

Answer 2

Since $f(x) δ(x - y) = f(y) δ(x - y)$ then $$T \sum_{k\in\mathbb Z} f(kT) δ(x - kT) = T \sum_{k\in\mathbb Z} f(x) δ(x - kT) = \left(T \sum_{k\in\mathbb Z} δ(x - kT)\right) f(x) = T Ш_T(x) f(x).$$ (Your $Ш_T(x)$ is actually $T Ш_T(x)$.)

In addition, since $T δ(T(⋯)) = δ(⋯)$ for $T > 0$, then $$T Ш_T(x) = T \sum_{k\in\mathbb Z} δ(x - kT) = \sum_{k\in\mathbb Z} δ\left({x \over T} - k\right) = Ш\left({x \over T}\right).$$ So, actually $$T \sum_{k\in\mathbb Z} f(kT) δ(x - kT) = Ш\left({x \over T}\right) f(x).$$