# Question about the dirac $\delta$-function

I have a basic question about the dirac $\delta$-function based on the beginning of Chapter 1 of these notes.

The dirac $\delta$-function can be defined heuristically as the function that is $0$ everywhere except as $x = 0$, where it is $\infty$.

But formally, this is not the definition of the functional (since it's not a function). Here is my question:

Since $\delta$(x) is not defined for each $x$, how can we talk about the product $(f(x)-f(0))\delta(x)$? The author says this is identically $0$. Why? If we are using the heuristic definition of $\delta$, then when $x \neq 0$, $\delta(x) = 0$ so the product is $0$, and if $x = 0$, then we get $(f(0) - f(0))\cdot \infty$, but who is to say this is equal to $0$? If $0 \cdot \infty$ always equaled $0$, then under this heuristic definition, we should have $\int \limits_{\Bbb R} \delta(x) \,dx = \int \limits_{\Bbb R - \{0\}} \delta(x) \,dx + \int \limits_{ \{0\} } \delta(x)\,dx = 0 + 0 \cdot \infty = 0$, but this integral is by definition equal to $1$. But even so, what if we aren't using the heuristic definition?

• To avoid confusions the distributions like $\delta$ is to be understood as a linear functionals on the test function space, i.e. only inside an integral. One can give a precise definition of $g(x)\delta(x)$ as a distribution if $g$ is continuous. It is just a symbol, not a point-wise multiplication. Like $\frac{dy}{dx}$ is not a fraction.
– A.Γ.
Commented Jul 24, 2015 at 20:10
• @A.G. What do you mean only inside an integral? I was under the assumption that $\delta$ as a functional is by definition $<\delta, \phi> := \phi(0)$ for a test function $\phi$. If that's true, it would agree with the idea that we cannot talk about $\delta(x)$ in an integral because it's not a function. (Then again, we define $\int \limits_{\Bbb R} \phi(x) \,dx = 1$, so I guess we can talk about it inside of an integral....) Commented Jul 24, 2015 at 20:12
• It is just a matter of definition. There are many definitions of an integral, the most general is exactly a linear functional. It is not very important. What is important is that a distribution is not a point-wise function, it is a result of an action on the test functions.
– A.Γ.
Commented Jul 24, 2015 at 20:20
• @A.G. Yes, exactly. The symbolic notation is only used formally because the linear Dirac Functional shares many properties of the integral. And the notation is the point that can be confusing to those applying it. Commented Jul 24, 2015 at 20:20
• @Urgje What exactly does $\displaystyle \int \,dx f(x) \mu_{0}(dx)$ mean? You are taking one integral with respect to two measures? Commented Jul 25, 2015 at 13:58

See THIS ANSWER, where I provided a Primer on the Dirac Delta.

The heuristic statement $\delta(x)(f(x)-f(0))=0$ means that for each test function $f$, the functional $D[f(x)-f(0)]=0$, where $D[\cdot]$ is the Dirac Delta functional.

We write the functional for $D$ formally as

$$D[\cdot]=\int_{-\infty}^{\infty}\delta(x)[\cdot]dx \tag 1$$

But the right-hand side of $(1)$ is not an integral. Rather, it shares many of the same properties with integrals and is therefore useful notation. But it is only notation.

So, for a test function $f(x)$, we have

$$D[f(x)]=f(0)$$

and therefore

$$D[f(x)-f(0)]=f(0)-f(0)=0\tag 2$$

Finally, we interpret $(2)$ formally and write

$$\delta(x)(f(x)-f(0))=0$$

Text books that heuristically discuss the Dirac Delta will often give the curiously nonsensical point-wise definition of $\delta(x)$

$$\delta(x)= \begin{cases} 0,&x\ne 0\\\\ \infty,&x=0 \end{cases}$$

which obviously is meaningless even with the additional condition that $\int_{-\infty}^{\infty}\delta(x)\,dx=1$.

This "hand-waving" description can be made rigorous by defining a family of functions $\delta_n(x)$ with the properties that

$$\lim_{n\to \infty}\delta_n(x)= \begin{cases} 0,&x\ne 0\\\\ \infty,&x=0 \end{cases}$$

and

$$\lim_{n\to \infty}\int_{-\infty}^{\infty}\delta_n(x)\,dx=1 \tag 3$$

One may then write, $\delta(x)\sim \lim_{n\to \infty}\delta_n(x)$ with the interpretation provided by $(3)$. Examples of such families of functions include the pulse function

$$\delta_n(x)= \begin{cases} n/2,&-\frac{1}{n}\le x\le \frac{1}{n}\\\\ 0,&\text{otherwise} \end{cases}$$

and the Gaussian function

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

In this answer here, I discussed the regularization used in potential theory for the $\mathscr{R}^3$ Dirac Delta $\delta(\vec r)$. There, the Dirac Delta is written

\begin{align} \delta(\vec r)&\sim \lim_{a\to 0}\delta_{a}(\vec r)\\\\ &=\lim_{a\to 0} \frac{3a^2}{4\pi(r^2+a^2)^{5/2}} \end{align}

where $\lim_{a\to 0}\int_{\mathscr{R}^3}f(\vec r)\,\delta_{a}(\vec r)\,dV=f(0)$.

And finally in this answer here, I analyze the family of functions $\delta_{\epsilon}(x)=\frac{1}{\sqrt{\pi\,\epsilon}}e^{-\tan^2(x)/\epsilon}$ that describes the "train" of Dirac Deltas

$$\sum_{\ell =-\infty}^{\infty}\delta(x-\ell \pi)\sim \lim_{\epsilon \to 0}\frac{1}{\sqrt{\pi\,\epsilon}}e^{-\tan^2(x)/\epsilon}$$

• Well, what does $\int \limits_{\Bbb R} \delta(x) \,dx$ mean formally? How are we integrating something that is not a function? We define this integral to be $1$, but what does the integral mean? Commented Jul 24, 2015 at 20:14
• It is not an integral. It is a functional that takes functions and maps them into numbers. Commented Jul 24, 2015 at 20:17
• I'm very confused. $<\delta, \phi(x)>$ is defined to be $\phi(0)$, and from your answer it sounds like writing $\int \limits_{\Bbb R} \delta(x) \phi(x)\,dx$ is just alternate notation for $< \delta, \phi(x)>$ and doesn't really mean the integral. Then why do we define $\int \limits_{\Bbb R} \phi(x) \,dx = 1$? We shouldn't have to. It should follow directly from the definition of $<\delta, f(x)>$, since $<\delta, f(x)> = f(0)$, and if $f(x)$ is the constant function $1$, then $f(0) = 1$, so $<\delta, 1> = 1$. Commented Jul 24, 2015 at 20:20
• So you're saying we can find actual functions $g_{n}$ such that they each integrate to $1$, and if we consider their integral against any test function $\phi(x)$, then the limit of these integrals can be thought of as the integral against the limiting function, which isn't really a function? I'm just trying to see if I understand what you are saying. You wrote $\delta(x) ~ \lim_{n \to \infty} g_{n}(x)$. Is that because actually $\lim_{n \to \infty} <g_{n}, \phi> = \phi(0)$? Commented Jul 24, 2015 at 20:45
• You're welcome! It has been my pleasure. And thank you for that nice compliment. I'm at home recovering from surgery 3 and a half weeks ago. You just made my weekend! Much appreciative Commented Jul 24, 2015 at 21:37

To say that $(f(x) - f(0)) \delta(x)$ is identically $0$ means that if we integrate $(f(x) - f(0)) \delta(x)$ against any test function, we get the same thing as if we integrate $0$ against that test function. Let $\varphi$ be a test function. Then \begin{align} & \int_{-\infty}^\infty \varphi(x)\Big( (f(x)-f(0)) \delta(x) \Big) \, dx \\[10pt] = {} & \int_{-\infty}^\infty \Big(\varphi(x)(f(x)-f(0))\Big) \delta (x)\, dx \\[10pt] = {} & \left. \varphi(x)(f(x)-f(0)) \vphantom{\frac 1 1} \,\right|_{x=0} = \varphi(0)(f(0)-f(0)) = \cdots \end{align}

• Well, at the end of the day, as Dr. MV said, the integral notation is just that: notation. $\delta(x)$ is just notation for the linear functional taking a test function $\phi(x)$ to $\phi(0)$. Then for any test function $\phi(x)$, $\phi(x) - \phi(0)$, under this functional, this function is sent to $\phi(0) - \phi(0) = 0$. Thus, $\delta(x)(\phi(x) - \phi(0))$, which is just notation for the image of $\phi(x) - \phi(0)$ under the linear functional $\delta$, is just $0$. Commented Jul 24, 2015 at 21:20
• How did you get that $\displaystyle \int \limits_{-\infty}^{\infty} \Big ( \phi(x)(f(x) - f(0)) \Big ) \delta(x) \,dx = \phi(x)(f(x) - f(0))$ (evaluated at $x = 0$)? Commented Jul 25, 2015 at 1:04
• @user46944 : You would need to show that $x\mapsto\varphi(x)(f(x)-f(0))$ is itself a test function. That would depend on knowing something about what sort of function $f$ is. Then use the fact that for every test function $\psi$ we have $\displaystyle\int_{-\infty}^\infty \psi(x)\delta(x)\,dx = \psi(0)$. That is simply the definition of the Dirac delta function $\delta$. ${}\qquad{}$ Commented Jul 25, 2015 at 2:24
• Well $f(x)$ was already a test function in the notes, so $\phi(x)(f(x) - f(0))$ is a test function if $\phi$ is. Commented Jul 25, 2015 at 3:11

As the other answers testify, there are many useful viewpoints on "what's really happening" with Dirac's $\delta$. Also, there is the chronic confusion of whether "formal" means that something is really true/correct for trivial reasons, or, rather, that it is a suggestive heuristic that perhaps cannot be made legitimate but is useful.

The most specific response I have to the original question is simply about multiplying (compactly supported) distributions $u$ by smooth functions $f$ to obtain another distribution: $f\cdot u$ is the distribution defined by $(f\cdot u)(\varphi)=u(f\cdot \varphi)$ where $f\cdot \varphi$ is the pointwise multiplication, producing another test function.

Thus, in the case at hand, for any $f$ vanishing at $0$ (e.g., produced by replacing $f$ by the function $x\to f(x)-f(0)$), we have $(f\cdot \delta)(\varphi)=\delta(f\cdot \varphi)=(f\cdot\varphi)(0) = f(0)\cdot u(0)=0$. That is, such $f\cdot \delta$ is the $0$ distribution, not the number $0$.

As some answers noted, $\int_{\mathbb R}\delta(x)\cdot 1\;dx$ can be construed as $\langle \delta,1\rangle$. Rather than disclaiming the integral expression by saying it's "just formal", why not say that it is to be interpreted as the extension to the pairing between distributions and test functions of the pairing of test functions and test functions by integrating-against. It is the extension-by-continuity of that pairing, using the weak-dual topology on distributions, after all.

There is some precedent for taking the viewpoint that "formally meaningless" integrals are in fact precise, meaningful extensions-by-continuity of pairings that are literal integrals on dense subspaces. Fourier transform and inversion on $L^2(\mathbb R)$ are examples of this: the integral itself only makes sense on $L^1$, but after proving Plancherel, we extend-by-isometry/continuity, and still write the integral, even though it is not literally that integral.

There is also the possibility of thinking of "multiplication by" $\delta$ as an operator mapping test functions to distributions, as suggested in some other answers. But here one might consider writing $\delta\otimes \delta$ when it's meant as an operator. Indeed, for two distributions $\alpha,\beta$, the operator $\alpha\otimes \beta$ is $(\alpha\otimes\beta)(\varphi)=\beta(\varphi)\cdot \alpha$. These are rank-one operators given by the "smallest" possible Schwartz kernels...

Dirac's $\delta$ is a distribution, not a function per se. Formally $\langle\delta,f\rangle=f(0)$. A common choice for the space where these things live is the dual of the Schwartz functions. In physics the manipulation of these things is less rigorous in the notation. Since for some function spaces (like $L^2$) all linear functional into $\mathbb{R}$ are also functions ($\langle f,g\rangle=\int fg$) the notion of a generalized function is extended and the notation kept.

A more rigorous treatment is also to define $\int \delta f=\lim_{\epsilon\rightarrow\infty}\int f \rho_\epsilon$ where $\rho$ is a $C^\infty$ function with compact support around zero, and $\rho_\epsilon=\rho(r/\epsilon)/\epsilon$

$\delta g$ would be the distribution defined by $\langle\delta g,f\rangle=f(0)g(0)$, so if $g(0)=0$, then $\delta g$ is always zero.

• Dirac is not a function, Are you Sure ? Commented Jul 24, 2015 at 21:30
• @Cardinal The dirac $\delta$ is not a function because no function satisfies the two properties we want it to satisfy: that it is $0$ everywhere except at $x = 0$, where it is $\infty$, and that the integral of it over $\Bbb R$ equals $1$. No real valued function satisfies this, so it can't be a function. Commented Jul 24, 2015 at 21:35
• @user46944 so you think $\frac{1}{x}$ is not a function as matter of the $\infty$ Commented Jul 24, 2015 at 21:58
• So what is the difference? ! sounds weird ! Commented Jul 24, 2015 at 21:59
• what about this: ? $$\delta(x)=\frac{d}{dx}u(x)$$ Now, if $u$ is the step function, how a linear transform could bring about a distribution ? Commented Jul 24, 2015 at 22:05

One can define $$\delta(x) = \left\{ \begin{array}{rcl} |x| > \epsilon &:& 0\\\\ |x| \le \epsilon &:& \displaystyle \frac{1}{2\epsilon} \end{array} \right.$$

Whence $$\int_{-\infty}^{+\infty} \delta(x) = \int_{-\epsilon}^{+\epsilon} \frac{ 1 }{2\epsilon} dx = 1.$$ and $$\int_{-\infty}^{+\infty} f(x) \delta(x) = \int_{-\epsilon}^{+\epsilon} \frac{ f(x) }{2\epsilon} dx = f(0).$$