Scaling property of the Dirac- Delta function does not preserve normalization

Question

The well known scaling property for the Dirac $\delta$-function (distribution) states (see the Wikpedia entry)

$$\int_{-\infty}^\infty dx\:\delta(a x) =\frac{1}{|a|}$$

Now we can generally consider $\delta(x)$ as the limiting case of a probability density function

$$p(x,\epsilon)=\frac{1}{\epsilon \sqrt{\pi}} e^{-x^2/\epsilon^2}$$

with

$$\ \int_{-\infty}^\infty dx\:p(x,\epsilon) = 1$$

and

$$\delta(x)=\lim_{\epsilon\to 0}\:p(x,\epsilon)$$

Assume now we spread the events over a different $x$-range by doing the linear scale transformation $x\rightarrow ax$, so $p(x,\epsilon)$ transforms to

$$p(ax,\epsilon)=\frac{1}{\epsilon \sqrt{\pi}} e^{-a^2x^2/\epsilon^2}$$

However, this function is not normalized anymore, as

$$\ \int_{-\infty}^\infty dx\:p(a x,\epsilon) = \frac{1}{|a|}$$

Imagine $p(x)$ describes the way you spend a given amount of money over time, then here, by increasing the length of time ($|a|<1$), one would have miraculously gained money! Obviously, this is not possible. In order to qualify as a probability density function $p(a x)$ must be properly normalized (this is expressed in the 'Location-Scale Family' theorem of probability theory, see also this link) i.e.

$$p(ax,\epsilon)=\frac{|a|}{\epsilon \sqrt{\pi}} e^{-a^2x^2/\epsilon^2}$$

With this normalized function, the increased period of time the money is spent is compensated by the smaller amounts you have to spend, so you correctly end up with the same total spent, i.e.

$$\ \int_{-\infty}^\infty dx\:p(a x,\epsilon) = 1$$

However, if one generates the $\delta$-function from this, one finds

$$\int_{-\infty}^\infty dx\: \delta(a x)=\lim_{\epsilon\to 0}\: \int_{-\infty}^\infty dx\: p(a x,\epsilon) =1 $$

rather than $\int_{-\infty}^\infty dx\: \delta(a x)= \frac{1}{|a|}$ as stated throughout in the literature.

Can anybody see anything wrong with my argument?

EDIT: My argument above used the (by definition) normal property of the generating Gaussian distribution function if one generally interprets the Dirac $\delta$-function as the limit of a probability density function. In this sense it was an additional constraint that enables the use of a continuous mathematical function (in this case the Gaussian function) to approximate the actually discrete probability distribution. However, this can actually also be derived without resorting to probability theory by considering $\delta$ as a distribution that acts as a 'unit impulse' function when applied to a test function $f(x)$ i.e.

$$\int_{-\infty}^\infty dx\: \delta(x)f(x) = f(0) $$

Also with this interpretation, $\delta(x)$ can be interpreted as the limit of $\delta$-sequences, for instance

$$\delta_n(x)=\frac{n}{\sqrt{\pi}} e^{-n^2x^2}$$

with the notation as defined in the link. If we set $n=\frac{a}{\epsilon}$ this becomes instead

$$\delta_{a,\epsilon}(x)=\frac{a}{\epsilon\sqrt{\pi}} e^{-a^2x^2/\epsilon^2}$$

which means

$$\ \int_{-\infty}^\infty dx\:\delta_{a,\epsilon}(x) = 1 \neq \frac{1}{|a|}$$

Answer 1

Just about everything you wrote is correct. However, you assumed that to produce $\delta(ax)$, you needed to use probability densities. You don't. $\delta(x)$ is the limit of probability densities, but $\delta(ax)$ is not. So I think you are just having an interpretation issue. $\delta(ax)$ is not normalized, and isn't supposed to be: as you said, its integral should be $\frac{1}{|a|}$.

Answer 2

You define $\delta(ax)$ to be a sequence of normalized functions. But that is not the definition of $\delta(ax)$.

The definition of $\delta$ is that (with some abuse of notation) $$\int_{-\infty}^{\infty}\delta(x)\,f(x)\,dx = f(0)$$ for all functions $f$ continuous at $x=0.$ (The real definition actually limit $f$ to a smaller set of functions, but this works for now.)

The definition of scaling the argument, i.e. $\delta(ax)$ (with $a>0$) is that a variable change should work: $$ \int_{-\infty}^{\infty}\delta(ax)\,f(x)\,dx = \{ \text{ set } y = ax \} = \int_{-\infty}^{\infty}\delta(y)\,f(y/a)\,dy/a = f(0/a)/a = f(0)/a \\ = \int_{-\infty}^{\infty}\frac{1}{a}\delta(x)\,f(x)\,dx. $$ Since this should be true for all $f$ (continuous at $x=0$) we have $$\delta(ax) = \frac{1}{a}\delta(x).$$ For $a<0$ we get $$\delta(ax) = -\frac{1}{a}\delta(x)$$ so in general, for $a\neq0$ we have $$\delta(ax) = \frac{1}{|a|}\delta(x).$$

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Your formula $\int_{-\infty}^\infty dx\,p(a x,\epsilon) = \frac{1}{|a|}$ is correct and gives the correct formula for $\delta(ax)$: $$ \int_{-\infty}^\infty dx \, \delta(a x) \, f(x) = \lim_{\epsilon\to 0} \int_{-\infty}^\infty dx \, p(a x, \epsilon) \, f(x) = \{ y := ax \} = \lim_{\epsilon\to 0} \int_{-\infty}^\infty \frac{dy}{|a|} \, p(y, \epsilon) \, f(y/a) = \int_{-\infty}^\infty \frac{dy}{|a|} \, \delta(y) \, f(y/a) = \frac{1}{|a|} f(0/a) = \frac{1}{|a|} f(0) = \int_{-\infty}^\infty \frac{dy}{|a|} \, \delta(y) \, f(y). $$