Proof of Dirac Delta's sifting property

A common way to characterize the dirac delta function $\delta$ is by the following two properties:

$$1)\ \delta(x) = 0\ \ \text{for}\ \ x \neq 0$$

$$2)\ \int_{-\infty}^{\infty}\delta(x)\ dx = 1$$

I have seen a proof of the sifting property for the delta function from these two properties as follows:

Starting with

$$\int_{-\infty}^{\infty}\delta(x-t)f(x)\ dx$$

for some “sufficiently smooth” function $f$, since $\delta(x – t) = 0$ for $x \neq t$ we can restrict the integral to some epsilon interval around $t$

$$\int_{-\infty}^{\infty}\delta(x-t)f(x)\ dx = \int_{t-\epsilon}^{t+\epsilon}\delta(x-t)f(x)\ dx$$

On this infinitesimal interval, $f$ is “approximately constant” and so we can remove it from the integral

$$\int_{t-\epsilon}^{t+\epsilon}\delta(x-t)f(x)\ dx = f(t)\int_{t-\epsilon}^{t+\epsilon}\delta(x-t)\ dx = f(t)$$

This proof seems a little too hand wavy for me. The points I find problematic are in quotations. What is meant by “sufficiently smooth” in this case? Is continuous enough? Also, how exactly is the extraction of the function from the integral done rigorously, without just assuming that it is “approximately constant”? I have seen this proof done with non-standard analysis and I understand that the delta function is by nature a rather “hand wavy” object so that a rigorous proof using these two properties may not even exist. Still I ask if anyone can perhaps make the above proof rigorous or offer a new proof without appealing to non-standard analysis.

(I’m not too sure what tags to include for this question. If anyone could retag for me that’d be much appreciated

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Well, as you mention, no truely rigorous treatment can be given with such a description of the Delta Dirac function – no such function actually satisfies those requirements. Thus, I won’t take too much effort to make the below too precise.

Where it says “sufficiently smooth”, it doesn’t actually need anything there at all! Whatever $f$ is, as long as it is finite almost everywhere, the product with that delta function will be $0$ away from a neighbourhood of $t$, so you can restrict the integral like that.

For the extraction of $f$, being continuous in a closed neighbourood of $x=t$ is enough. If $f$ is continuous through $[ t-\epsilon, t+ \epsilon ]$ then by the Extreme value theorem it attains a maximum and minimum in that interval, call them $M$ and $m$ respectively.

Then since $m \leq f(x) \leq M$ is that range, $$m=\int^{t+\epsilon}_{t-\epsilon} m \delta(x-t) dx \leq \int^{t+\epsilon}_{t-\epsilon} f(x) \delta(x-t) dx \leq \int^{t+\epsilon}_{t-\epsilon} M \delta(x-t) dx = M.$$

Now as $\epsilon \to 0$, both $m$ and $M$ go to $f(t)$ as $f$ is continuous so by the Squeeze theorem, $$\int^{t+\epsilon}_{t-\epsilon} f(x) \delta(x-t) dx \to f(t)$$ as $\epsilon \to 0$.

Theory of Distributions by J. Ian Richards and Heekyung Youn gives rigorous proofs of things like this without need for functional analysis, topology, or measure theory.

Definitely the proof you give is extremely “hand-wavy”. In particular, the two defining properties you give cannot be taken literally if one conceives of a function as something where you put in a number $x$ and get out a number $f(x)$. To take those two properties is intuitive but very very “hand-wavy”.

The simplest possible nascent delta function is a rectangular pulse:
$$\delta_h(x) = \begin{cases} 0 & |x|\geq h \\ 1/2h & -h\leq x \leq h \end{cases}.$$

For a continuous function $f,$ the sifting property of $\delta_h(x)$
is very easily proven.
$$\int_{-h}^h \delta_h(x) f(x) \,dx = \left.\frac{F(x)}{2h}\right|_{-h}^{h} = \frac{F(h) – F(-h)}{2h}$$
where $F$ is the antiderivative of $f.$ Since $f$ is continuous, its
antiderivative is continuously differentiable and the limit of $(F(h)-F(-h))/2h$ as $h\rightarrow 0$ is $f(0).$

It is true that no function has the two properties listed in the
question. But there are infinitely many families of continuous functions
$\delta_h(x)$ that are positive on
$(-h,h)$ and 0 everywhere else with integral equal to 1 for
all $h.$ These functions satisfy property (2) for all $h$ and
property (1) in the limit $h\rightarrow 0.$

It is very easy to prove that
these nascent delta functions satisfy the sifting
property in the limit $h\rightarrow 0.$ The exact same squeeze
argument from
the proof in the answer above applies since the integral of
$\delta_h(x)$ over $[-h,h]$ is 1 for every $h.$

Both the rectangular pulse and hat function are examples of this type of
nascent delta function.
A smooth example of this type of nascent delta function is the mollifier
$$\frac{K}{h}\exp\left( \frac{-1}{1 – x^2/h^2} \right)$$
where $K$ is the constant that makes the integral 1 for $h=1$ (and also
all other values of $h.$)