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

Solution 1:

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.

Solution 2:

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