TITLE: How to prove that the derivative of Heaviside's unit step function is the Dirac delta?
QUESTION [30 upvotes]: Here is a problem from Griffith's book Introduction to E&M.

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

REPLY [4 votes]: 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 $\mathscr D'$.