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Let's assume we have a function of the following form $f(x,a):=g(H(a-x))$, where $H$ is the Heaviside step function. We now would like to look at the derivative $\frac{\partial}{\partial x}\int_0^1 f(x,a)da$. (This form should explain the "generalized functions" in the title.)

The actual question now is if it is allowed to write the following: $$\frac{\partial}{\partial x}\int_0^1 f(x,a)da = \int_0^1 \frac{\partial g(y)}{\partial y} \cdot \frac{\partial H(a-x)}{\partial x} da,$$ where $y = H(a-x)$.

In a general setting the chain rule can to my understanding not be used under such circumstances, since $H(x-x)=\infty$ and thus the differentiability of $g$ at this point is a little hard to define. Additionally $H(a-a$ neither is differentiable at this point. Also when doing this we have the weird situation of the multiplication of a step function with a Dirac delta, both having discontinuities at $x=a$.

Does someone know about the rules in such points? Does there exist a chain rule for Heaviside step functions?

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Not quite. For simplicity assume $g$ is $C^1$

$g(H) = g(0)+ (g(1)-g(0))H$ whose distributional derivative is $(g(1)-g(0))\delta$.

Taking $\phi \in C^\infty_c,\int \phi=1,\phi_n(x)=n\phi(nx)$ then $H\ast \phi_n \to H$.

$g(H\ast \phi_n)\to g(H)$ in the sense of distributions, and we'll have no fear to say that

$$(g(H\ast \phi_n))'=g'(H\ast \phi_n) (H\ast \phi_n)'=g'(H\ast \phi_n) \phi_n \to (g(H))'$$ in the sense of distributions.

So $\lim_{n\to \infty}g'(H\ast \phi_n) (H\ast \phi_n)'$ is the true distributional meaning of $g'(H) H'$. It is merely saying that the correct value of $g'(H(0))$, which gives $$(g(H))'= g'(H(0)) \delta$$ is $g'(H(0))=\int_0^1 g'(x)dx$.

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