# How to prove that the derivative of Heaviside's unit step function is the Dirac delta?

Here is a problem from Griffith's book Introduction to E&M.

Let $$\theta(x)$$ be the step function $$\theta(x) = \begin{cases} 0, & x \le 0, \\ 1, & x \gt 0. \end{cases}$$

The question is how to prove $$\frac{d\theta}{dx} = \delta(x)$$.

I think since the function is discontinuous at $$x = 0$$, there is no definition of $$\frac{d\theta}{dx}$$ at the point $$x = 0$$ at all. Thus, how could we show the equivalence of a equation if the left part of equation is not defined at the point $$x = 0$$?

• That equality is an equality of distributions, as in en.wikipedia.org/wiki/Distribution_%28mathematics%29 If you read any introduction to distributions, you'll find the claim you want yo prove as the very first example of hopw differentiation works for distributions. Dec 11, 2010 at 12:41
• $\delta(x)$ is "undefined" at 0 too! Dec 11, 2010 at 12:56
• About 6 minutes in here provides a good explanation youtu.be/ECslmuGlu-U Oct 8, 2020 at 22:57

This is a place where physicists and mathematicians would phrase the question differently. A mathematician would say that $$d \theta/dx$$ is undefined at $$0$$, and that $$\delta$$ is not a function. However, as Mariano says, the statement is true "in the sense of distributions".

What does that mean? A distribution is a gadget which takes as input a smooth function $$g(x)$$, which is zero for $$|x|$$ sufficiently large, and returns some real number. When we think of an ordinary function as a distribution, that makes us think of $$f$$ corresponding to the gadget $$F: g \mapsto \int f(x) g(x) dx$$. Notice that changing $$f$$ at a finite number of points leaves $$F$$ unaltered. From a physical point of view, if $$f$$ is something like the value of an electric field at a point, we would never know if it had a finite discontinuity at some point, so the gadget $$F$$ captures everything that is physically measurable about $$f$$.

Now, how can we see differentiation in terms of distributions? By integration by parts, we have $$\int f'(x) g(x) dx = - \int f(x) g'(x) dx$$. So, for any distribution $$F$$, we define the derivative of $$F$$ to be the gadget $$g \mapsto -F(g')$$.

Now, let $$F$$ correspond to $$\theta$$, so $$F(g) = \int_{-\infty}^0 g(x) dx$$. The Dirac delta distribution is $$\delta(g) = g(0)$$. I leave it to you to show that $$F'(g) = \delta(g)$$, with the definitions above.

I'd be curious to see how a physicist would answer this question. I suspect that I have acted like a Frenchman.

• Thanks, I have correct the typo. Dec 11, 2010 at 14:49
• You mean "$g(x)$, which is zero for $|x|$ sufficiently large" (missing absolute value). Dec 11, 2010 at 15:11
• That's a great quote from Goethe! :) Dec 11, 2010 at 17:31
• A physicist would probably say that $\theta'=\delta$ is equivalent to $\theta(x)-\theta(a)=\int_a^x\delta(x)\,dx$ and both sides are equal to $[0\in (a,x)]$ (with a sign switch if $x<a$).
– A.S.
Nov 26, 2015 at 20:26
• This is an old answer, but I just stumbled on it. In the penultimate paragraph, when you take $F$ to be $\theta$, shouldn't $F(g)=\int_0^\infty g$, not $\int_{-\infty}^0g$ as you have it? I think you jumped ahead, thinking of $F'(g)$ too soon. Indeed, taking the distributional derivative of $F$ as I think it should be gives $F'(g)=-F(g')=\int_{-\infty}^0g'=g(0)-g(-\infty)=g(0)$. Feb 28, 2017 at 21:08

By definition, $\delta(x)$ satisfies $$\int_{-\infty}^\infty f(x)\delta(x) dx=f(0)$$ for any continuous function $f$ on $\mathbb{R}$.

By definition of derivative in distribution theory, $$\int_{-\infty}^\infty f(x)\theta'(x) dx=-\int_{-\infty}^\infty f'(x) \theta(x) dx$$ for any $C^1$ function $f(x)$ which vanishes outside of a bounded interval.

Now $$-\int_{-\infty}^\infty f'(x) \theta(x) dx=-\int_0^\infty f'(x) dx= -f(\infty)+f(0)=f(0)=\int_{-\infty}^\infty f(x)\delta(x) dx.$$

So $\theta'(x)=\delta(x)$.

• Could you explain why $f(\infty)$ is zero? Dec 12, 2010 at 4:46
• $f(\infty)=\lim_{x\to\infty}f(x)$ is 0 because $f$ is assumed to vanish outside of a bounded interval, i.e. equal to 0 for all $x$ outside of a bounded interval. This is one of the assumption in distribution theory.
– TCL
Dec 12, 2010 at 4:52

I'd like to add an answer that will try to justify the statement with an example. Consider the function sequence f_n = \left\{\begin{align} 0 \ \ &\text{ if } |x|>1/n \\ n^2x^2/2 + nx + 1/2 &\text{ if } -1/n\leq x \leq 0\\ -n^2x^2/2 + nx + 1/2 &\text{ if } 0<x\leq 1/n \end{align}\right..
These functions are differentiable everywhere, and they approximate the unit step function better and better as $n$ gets larger. If we look at the derivatives of these functions, we'll see that they are a sequence of tent functions that are getting narrower and taller, as Griffith uses to introduce the delta function.

Actually, with an appropriate mode of convergence, when a sequence of differentiable functions converge to the unit step, it can be shown that, their derivatives converge to the delta function. That's why, one may take the derivative of the unit step function to be defined as the limit of the derivatives, which is the delta function.

• That is very illuminating! Thank you. I believe that your function fn should be 1 for x > 1/n. Feb 6, 2023 at 10:09

Of course, as it was stated above, the equality holds in the sense of distributions. In order to approach this same issue, i.e. showing that the equality follows from the theory of distributions, one can define the derivative of a function-type distribution (a distribution which in fact is a function in $L^1_{\text{loc}}(\mathbb R)$), as the limit, in the sense of distributions, of the usual incremental ratio, as follows: $$\theta(x) = \begin{cases} 0\quad \text{ if } x\le 0\\ 1\quad \text{ if }x>0. \end{cases}$$ Then the functions $$\delta_\varepsilon (x) \equiv \frac{\theta(x)-\theta(x-\varepsilon)}{\varepsilon}$$ for, say, $\varepsilon>0$ are in fact $$\delta_\varepsilon(x) = \begin{cases} 0\quad\text{ if }x<\varepsilon\\ \dfrac{1}{\varepsilon}\quad\text{ if }0< x\le\varepsilon\\ 0\quad\text{ if }x\le0 \end{cases} =\varepsilon^{-1}\chi_{(0,\varepsilon]}(x) =\varepsilon^{-1}\chi_{(0,1]}\left(\frac{x}{\varepsilon}\right),$$ where $\chi_{B}$ is the characteristic function of $B$. Now, for any $\varphi\in\mathscr D(\mathbb R)$, $$\lim_{\varepsilon\to0}\langle \delta_\varepsilon, \varphi\rangle =\lim_{\varepsilon\to0}\int_{0}^{1}\varphi(y\varepsilon)dy=\varphi(0) = \langle\delta,\varphi\rangle.$$ The same holds when $\varepsilon<0$, as one can check explicitly. Thus $\theta'=\lim_{\varepsilon\to0}\delta_\varepsilon=\delta$, in the sense of distributions $\mathscr D'$.

• Although the question is old, this post is a nice contribution to the topic. Since you're here, I'd like to ask you a question considering that you are a Physicist. The original question stated that he got the exercise from Griffith's book and I don't think most physicist view it from distributional perspective. Do you have an idea about how physicists generally "prove" it? Jan 13, 2016 at 15:52
• @BigbearZzz Actually this type of result is often justified via integration by parts, which is kind of right, even though one avoids a systematic treatment of distribution theory. The widespread knowledge is that whenever there is a unit jump in the function, the derivative gets a delta contribution. Jan 13, 2016 at 16:15
• I love how you used the word "justified" there instead of "proved". Some physicist I know use the words interchangeably :< Jan 13, 2016 at 16:19