Derivative of this indicator function

Question

What is the partial derivative w.r.t $x$ of this function: $${{1}_{\tau>ax+by}}$$ where $a$, $b$ $\in (0,1)$. I am thinking it would be like $\frac{\partial {{1}_{\tau>ax+by}}}{\partial x}=\delta(\frac{\tau-by}{a}-x),$ where $\delta$ is a Dirac-delta function. Please clarify whether I am correct or wrong ?

Answer 1

Let $f = 1_{\tau > ax + by}$. Assuming $a > 0$, $$f = 1_{x < \frac{\tau - by}{a} }$$

For $\phi \in D(\mathbb R)$,

$$\langle \frac{\partial f}{\partial x}, \phi \rangle = - \langle f, \phi' \rangle = - \int_{-\infty}^{\infty} f(x) \phi'(x)dx =- \int_{-\infty}^{\frac{\tau - by}{a}} \phi'(x)dx = -\phi \left( \frac{\tau - by}{a} \right)$$

If $a < 0$, then we get $ \phi \left( \frac{\tau - by}{a} \right)$.

Hence $f'$ is (depending on the sign of $a$) $\pm$ the Dirac function of mass $c:= \frac{\tau - by}{a}$