[Math] Derivative indicator function

Question

I am wondering what is the derivative of the following function with respect to $x(t)$ in sense of distributions.
$$
I\left(\int_0^t x(\tau)d\tau \leq c\right)
$$
where $I$ is the indicator function and $c$ is a constant.

Answer 1

Let me try. Your mapping is:
$1 - H(x-c)$ , where $H$ is the Heaviside function, derivative is $-\delta(x-c)$.
composed with
$\int_0^t : C^\infty(\mathbb R) \to \mathbb R$, which is bounded and linear, surjective.
No way with the chain rule, in fact, there is no chain rule in this situation.

Simpler approach, but it might not answer your question: Suppose $s\mapsto x(\tau,s)$ is a smooth variation of $x(\tau)$, i.e., $x\in C^\infty(\mathbb R^2)$.
Then $$s\mapsto 1-H\Big(\int_0^t x(\tau,s)d\tau -c\Big)$$ can be viewed as a distribution in $s$. Take a test function $f(s)$ and consider $$ \int \Big(1-H\Big(\int_0^t x(\tau,s)d\tau -c\Big)\Big)(-f'(s))ds $$ Now try formal partial integration with respect to $s$. Since the left hand integrand is 1 or 0, by the fundamental theorem it will be the difference of the values of $f$ at those $s$ where this jumps. If you fix $x(\tau,s)$ you can work this out.