## § Derivative of step is dirac delta

I learnt of the "distributional derivative" today from my friend, Mahathi. Recording this here.

#### § The theory of distributions

As far as I understand, in the theory of distributions, A distribution is simply a linear functional $F: (\mathbb R \rightarrow \mathbb R) \rightarrow \mathbb R$. The two distributions we will focus on today are:
• The step distribution, $step(f) \equiv \int_0^\infty f(x) dx$
• The dirac delta distribution, $\delta(f) = f(0)$.
• Notationally, we write $D(f)$ where $D$ is a distribution, $f$ is a function as $\int D(x) f(x) dx$.
• We can regard any function as a distribution, by sending $f$ to $F(g) \equiv \int f(x) g(x) dx$.
• But it also lets us cook up "functions" like the dirac delta which cannot actually exist as a function. So we move to the wider world of distributions

#### § Derivative of a distribution

Recall that notationally, we wrote $D(f)$ as $\int D(x)f(x) dx$. We now want a good definition of the derivative of the distribution, $D'$. How? well, use chain rule!
\begin{aligned} &\int_0^\infty U dV = [UV]|_0^\infty - \int_0^\infty V dU \\ &\int_0^\infty f(x) D'(x) dx \\ &= [f(x) D(x)]|_0^\infty - \int_0^\infty D f'(x) \end{aligned}
Here, we assume that $f(\infty) = 0$, ahd that $f$ is differentiable at $0$. This is because we only allow ourselves to feed into these distribtions certain classes of functions (test functions), which are "nice". The test functions $f$ (a) decay at infinity, and (b) are smooth. The derivation is:
\begin{aligned} &\int_0^\infty U dV = \int_0^\infty f(x) \delta(x) = f(0) \\ &[UV]|_0^\infty - \int_0^\infty V dU = [f(x) step(x)]|_0^\infty - \int_0^\infty step(x) f'(x) \\ &= [f(\infty)step(\infty) - f(0)step(0)] - step(f') \\ &= [0 - 0] - (\int_0^\infty f'(x) dx) \\ &= 0 - (f(\infty) - f(0)) \\ &= 0 - (0 - f(0)) \\ &= f(0) \end{aligned}
• Thus, the derivative of the step distribution is the dirac delta distribution.