I have a question about the Heaviside step function $\theta(\xi)$, defined by $$\theta(\xi):=\begin{cases} 1, & \xi\geq0\\ 0, & \xi<0 \end{cases}$$ I need to evaluate the square of the Heaviside step function, i.e.
$$[\theta(\xi)]^2$$
So, my question is: does the relation $$[\theta(\xi)]^2=\theta(\xi)$$ holds?
$\endgroup$ 13 Answers
$\begingroup$If you take $\theta(\xi)$ as you said, (with $\theta(0)=1$): $$\theta(\xi):=\begin{cases} 1, & \xi\geq0\\ 0, & \xi<0 \end{cases}$$ It will be a regular function (not a generalized, like $\delta(x)$), and there is not any reason to think about it unusually. so the relation holds:$$[\theta(\xi)]^2=\theta(\xi)$$
For the case of $\theta(0)={1\over2}$ as other answers say, $\theta^2(0)\neq \theta(0)$.
But if you define it as : $$\theta(\xi):=\begin{cases} 1, & \xi>0\\ \text{undefined},& \xi=0\\ 0, & \xi<0 \end{cases}$$ I will show that again$[\theta(\xi)]^2=\theta(\xi)$. As a generalized function, $\theta(\xi)$ will be defined through the following relation (where $g(t)\equiv\theta(\xi)$) : $$\int_{-\infty}^{+\infty}\phi(t)g^{(n)}(t) dt=(-1)^n \int_{-\infty}^{+\infty}\phi^{(n)}g(t)dt \, \,\,\,\,\,\,**$$ where $\phi$ vanishes at $\pm \infty$.
If $g^2(t)=g(t)$, you must have: $$\int_{-\infty}^{+\infty}\phi(t) {\frac{d^n g^2(t)}{dt^n}}dt=(-1)^n\int_{-\infty}^{+\infty}\phi^{(n)}(t)g(t)dt$$ We denote ${\frac{d^n g^2(t)}{dt^n}}$ with $f(t)$ for simplicity: $$\int_{-\infty}^{+\infty}\phi(t) f(t)dt=(-1)^n\int_{-\infty}^{+\infty}\phi^{(n)}(t)g(t)dt\, \,\,\,\,\,\,***$$ The right hand side is simply equal to $(-1)^n\int_{0}^{+\infty}\phi^{(n)}(t)dt=(-1)^{n-1}\phi^{(n-1)} (0)$.
Now, we use the $**$ equation with $g(t)=\delta (t)$: $$\int_{-\infty}^{+\infty}\phi(t)\delta^{(n)}(t) dt=(-1)^{n}\phi^{(n)} (0)$$
we conclude that in the left hand side of $***$, $f(t)$ must be the $(n-1)\text{th}$ derivative of $\delta (t)$, which implies ${\frac{d^n \theta^2(\xi)}{d\xi ^n}}={\frac{d^n \theta (\xi)}{d\xi ^n}}$ or $\theta^2(\xi)=\theta(\xi)$.
$\endgroup$ 1 $\begingroup$If you take this definition , it won´t generally be true since then $H^{2}\left(0\right)=\frac{1}{4}\neq\frac{1}{2}$. In another forum the same question was asked.
$\endgroup$ 1 $\begingroup$Yes. For every $\xi \in \mathbb{R}$ you can check that
$$\theta(\xi)^2 = \theta(\xi)$$
so you conclude that the functions $\theta^2$ and $\theta$ agree.
$\endgroup$
