Derivation of the Dirac-Delta function property: $\delta(bt)=\frac{\delta(t)}{\mid b \mid}$

Question

Considering the case such that $b \gt 0$ and $b \in \mathbb{R^+}$ and making the substitution $t'=bt$, it follows that $$\int_{t=-\infty}^{t=\infty}f(t)\delta(t)\mathrm{d}t =\color{red}{\int_{t'=-\infty}^{t'=\infty}f\left(\cfrac{t'}{b}\right)\delta(t')\cfrac{\mathrm{d}t'}{b} =\cfrac{1}{b}f(0)}$$ I have marked in $\color{red}{\mathrm{red}}$ the part for which I do not understand, namely where did $\color{red}{f(0)}$ come from?

Could someone please explain the steps in showing that the equality marked $\color{red}{\mathrm{red}}$ is true?

Many thanks

Answer 1

The common definition of the delta function is $$ \int_{-\infty}^{\infty} g(x) \delta(x) \, dx = g(0), \tag{1} $$ for any continuous $g$.

Now let's look at what $\delta(bt)$ does. We have no formula for this, so let's do that substitution, $t'=bt$, so $dt'=b \, dt$ and the limits remain unchanged because $b>0$. Then we have $$ \int_{-\infty}^{\infty} f(t) \delta(bt) \, dt = \frac{1}{b}\int_{-\infty}^{\infty} f(b^{-1}t') \delta(t') \, dt' = \frac{1}{b}f(0/b) = \frac{1}{b}f(0), $$ where the second equality has used the definition (1).

Answer 2

I suggest that you should think of $\delta$ a bit differently than you do, namely either as a distribution or as a measure. I will discuss the first point of view. The notation with integrals are handy but often lead to confusion when the "integrand" is not a real function.

So, what is a distribution? Well, you could say it is something that eats nice functions (typically belonging to some nice space, like infinitely differentiable functions with compact support, $C_0^{+\infty}(\mathbf R)$) and spits out numbers. If $\phi$ is such a function and $T$ is a distribution, then this is usually denoted by $T(\phi)$ or $\langle T,\phi\rangle$. We say that $T$ acts on the test function $\phi$. Distributions should also be linear, so $T(\phi+\psi)=T(\psi)+T(\phi)$ and $T(a\phi)=aT(\phi)$. If we want to use fancy names, we could say that a distribution $T$ is a linear functional on $C_0^{+\infty}(\mathbf R)$.

Main example of distribution

Let $f$ be a function such that the integral $$ \int_{-\infty}^{+\infty} f(x)\phi(x)\,dx $$ makes sense for all $\phi\in C_0^{+\infty}(\mathbf R)$ (such a function is usually said to be locally integrable), then there is a natural distribution $T$ associated with $f$, namely $$ T(\phi)=\int_{-\infty}^{+\infty}f(x)\phi(x)\,dx. $$ Without going into details, more or less all the functions you usually meet (with not too strong singularities) can be considered as distributions in this way, and sometimes one even does not distinguish between the function $f$ and the distribution $T$. However, it is possible to define distributions that does not correspond to any function as above, so distributions are in some sense more general then functions.

Important point

A distribution $T$ is not (unlike functions) defined at specific points $x\in\mathbf R$, so that it does not make sense to talk about $T(x)$. So, how could one define something that should correspond to typically "point-wisy" operations as differentiations $f'$, a dilation $f(bx)$ ($b\neq 0$) or a shift $f(x-b)$?

Getting closer to your question

Let us concentrate on how $T(bx)$ is defined for distributions, since this is what the question is about. Before doing so, let us see what happens in the case of when $T$ is given as in the Main example above. Thus, we want to somehow relate $T$ above with $$ \int_{-\infty}^{+\infty}f(bx)\phi(x)\,dx. $$ The change of variables $x\mapsto u=bx$ then transforms this integral into $$ \frac{1}{|b|}\int_{-\infty}^{+\infty}f(u)\phi(u/b)\,du. $$ But this integral can be written as $$ \frac{1}{|b|}T(\phi(\cdot\,/b)), $$ i.e. the factor $1/|b|$ times the action of our original distribution $T$ acting on the dilated test function. As is common in distribution theory (this is done for derivatives of distributions, shifts, Fourier transforms, ...), we want distributions to work like functions, and make a definition:

Definition of dilation of distributions

Given a distribution $T$, we define its dilation $T_b$ as the distribution $$ T_b(\phi)=\frac{1}{|b|}T(\phi(\cdot\,/b)). $$

(Just a reminder: To define a distribution, we need to say how it acts on test functions.)

Over to $\delta$ and the answer to your question

The delta distribution $\delta$ is the distribution defined by $$ \delta(\phi)=\phi(0) $$ i.e. it acts on a test function by returning the value of the test function at zero. By definition, the dilation of the delta function (which you wrote $\delta(x/b)$, but which I will write $\delta_b$ according to the definition above) $$ \delta_b(\phi)=\frac{1}{|b|}\delta(\phi(\cdot\,/b))=\frac{1}{|b|}\phi(0). $$ The second line is just the definition of $\delta$. I hope this answers your question.

Final comments

1) I hope it is clear to you that there does not exist a function $f$ such that $$ \int_{-\infty}^{+\infty}f(x)\phi(x)\,dx=\phi(0) $$ for all $\phi\in C_0^{+\infty}(\mathbf R)$.

2) The choice of space $C_0^{+\infty}(\mathbf R)$ above could have been some other space.