§ 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:(R→R)→RF: (\mathbb R \rightarrow \mathbb R) \rightarrow \mathbb R. The two distributions we will focus on today are:
  • The step distribution, step(f)≡∫0∞f(x)dxstep(f) \equiv \int_0^\infty f(x) dx
  • The dirac delta distribution, δ(f)=f(0)\delta(f) = f(0).
  • Notationally, we write D(f)D(f) where DD is a distribution, ff is a function as ∫D(x)f(x)dx\int D(x) f(x) dx.
  • We can regard any function as a distribution, by sending ff to F(g)≡∫f(x)g(x)dxF(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)D(f) as ∫D(x)f(x)dx\int D(x)f(x) dx. We now want a good definition of the derivative of the distribution, D′D'. How? well, use chain rule!
∫0∞UdV=[UV]∣0∞−∫0∞VdU∫0∞f(x)D′(x)dx=[f(x)D(x)]∣0∞−∫0∞Df′(x) \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(∞)=0f(\infty) = 0, ahd that ff is differentiable at 00. This is because we only allow ourselves to feed into these distribtions certain classes of functions (test functions), which are "nice". The test functions ff (a) decay at infinity, and (b) are smooth. The derivation is:
∫0∞UdV=∫0∞f(x)δ(x)=f(0)[UV]∣0∞−∫0∞VdU=[f(x)step(x)]∣0∞−∫0∞step(x)f′(x)=[f(∞)step(∞)−f(0)step(0)]−step(f′)=[0−0]−(∫0∞f′(x)dx)=0−(f(∞)−f(0))=0−(0−f(0))=f(0) \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.