Defining the convolution between two Heaviside functions.

Question

Consider $H \in L^1_{loc}(\mathbb{R})$ the Heaviside function, \begin{equation} H(x)=\begin{cases} 1, \!\!\quad x \ge 0\\ 0 \quad x <0 \end{cases}. \end{equation} My problem is

Define $H * H$, although the support of $H$ isn't compact.

I thought to use the following theorem:

Let $U:C^{\infty}_{c}(\mathbb{R}^{n}) \rightarrow C^{\infty}(\mathbb{R}^{n})$ a linear operator such that

• $T_h \, U=U\, T_h$, $ \forall h \in \mathbb{R}^{n}$, where $$T_h(\varphi)(x)=\varphi(x-h) \quad \forall \varphi \in C^{\infty}_{c}(\mathbb{R}^{n}).$$
• If $\varphi_j \rightarrow 0$ in $C^{\infty}_{c}(\mathbb{R}^{n})$ then $$\partial^{\alpha}U(\varphi_j)\rightarrow 0$$ uniformly on compacts, $\forall \alpha \in \mathbb{Z}_{+}$. Therefore there is only one $u \in \mathcal{D}'(\mathbb{R}^{n})$ such that $$U(\varphi)=u*\varphi.$$

My goal is finding the right $U:C^{\infty}_{c}(\mathbb{R})\rightarrow C^{\infty}(\mathbb{R})$ and use the theorem above.

I would like an answer.

Thank you.

Answer 1

$H*H(x)=\int_0^{x}H(x-y)H(y)dy=\int_0^{x}1 dy=x$ for all $x >0$ and $H*H(x)=0$ for $x <0$. Note that $H(x-y)H(y)=0$ except when $0\leq y \leq x$.

Answer 2

As you have apparently anticipated, there is a problem with this convolution. In fact, there is no "definition" of it that is compatible with other reasonable, mild conditions.

Are you sure that you are somehow compelled to describe it?