# What functions satisfy one of these equalities: $x \cdot \Phi(x) = (\Phi \ast \Phi)(x)= \int_{- \infty}^{x} \Phi(t) dt$?

When reading about the unit step or Heaviside function and it's derivative, the ramp function I encountered the following characterisations of the Heaviside function $$\mathscr{H}$$:

1. $$x \cdot \mathscr{H}(x) = \int_{- \infty}^{x} \mathscr{H}(t) dt$$
2. $$(\mathscr{H} \ast \mathscr{H})(x) = \int_{- \infty}^{x} \mathscr{H}(t) dt$$, where $$\ast$$ is convolution.

Are there any other functions $$\Phi: \mathbb{R} \to \mathbb{R}$$ with $$\Phi \not \equiv 0$$ satisfying either of those conditions on $$\mathbb{R}$$? I wrote out the definition of convolution but didn't come any further since surely, $$\Phi$$ could to be defined piece-wisely.

Some ideas: Plugging $$x = 0$$ into (1) and (2) yields $$\int_{-\infty}^{0} \mathscr{H}(t) dt = \int_{\mathbb{R}} \mathscr{H}(t) \mathscr{H}(-t) dt = 0,$$ meaning that the function and its reflection upon the $$y$$-axis "cancel out".

Update 1: $$f$$ is not differentiable in $$x = 0$$.

Case 1: $$f$$ is differentiable and fulfils one above requirement Now we can differentiate $$x \cdot f(x) = \int_{-\infty}^{x} f(t) dt$$ to obtain $$x \cdot f'(x) + f(x) = f(x) \implies x \cdot f'(x) = 0$$ Since $$x \not\equiv 0$$ we know that $$f'(x)$$ is constant, implying that it is constant lets say $$f(x) \equiv c \in \mathbb{R}$$. (Since $$f$$ is differentiable it is continuous and therefore can't be piecewise constant with discontinuous jumps like a step function) But this is a contraction since $$\int_{-\infty}^{x} c dx = \infty$$ for all $$x,c \in \mathbb{R}$$.

Case 2: $$f$$ has an antiderivative $$F$$ defined on $$\mathbb{R}$$ Then, by the FTOC we can write $$x \cdot f(x) = F(x) - F(-\infty) \implies x \cdot f(x) + F(-\infty) = F(x).$$ Since $$f$$ fulfils the above condition we deduce $$F(-\infty) < \infty$$. We know that $$F$$ is differentiable with $$F' = f$$. Since $$F(-\infty)$$ is a constant irrelevant to the differentiability of the LHS we conclude that $$f(x)$$ is differentiable on $$\mathbb{R} \setminus \{0\}$$. Because of case 1, $$f$$ can't be differentiable in zero.

Is this correct and if yes, how can we continue from here?

I also noticed that $$x \cdot \Phi(x) = (\Phi \ast \Phi)(x)$$ will also hold for the "reverse Heaviside function" $$\Phi := \mathbb{1}_{\{x<0\}}$$.

• You may find the following interesting: $$\int_{-\infty}^x f(t)\,dt=\int_{-\infty}^\infty f(t)\mathscr{H}(x-t)\,dt=(f\ast \mathscr{H})(x).$$ Thus the first antiderivative of $f(x)$ can itself be expressed as a convolution against the Heaviside step function. (This in particular makes it obvious why $\int_{-\infty}^x \mathscr{H}(t)\,dt=(\mathscr{H}\ast \mathscr{H})(x).$) – Semiclassical Aug 20 '19 at 21:34

Suppose $$\Phi(x)$$ is a distribution, then the only solutions, $$f(x)$$, to equation 1 are multiples of the Heaviside function. This is because, as you already stated, $$x\cdot f'(x)=0$$, which has only $$\delta (x)$$, or multiples, as solutions for$$f'(x)$$. Note that $$x\cdot \delta'(x)=\delta(x)$$ not $$0$$ and similarly for higher derivatives. Of course $$\int_{-\infty}^{x} \delta(t)dt=\mathscr{H}(x)$$.
It is straightforward to show that $$f(x)=e^{icx}\mathscr{H}(x)$$ solves equation 2 for any $$c$$. Invoking the idea of $$f(x)$$ as a distribution on the space of smooth test functions which fall off rapidly at large $$x$$, we can Fourier transform equation 2. I note that distributions do not in general have convolutions with themselves, but assume that the convolution is ok. Then equation 2 becomes $$-i\tilde{f}'(\omega)=2\pi \tilde{f}^2(\omega),$$ which has solutions $$\tilde{f}(\omega) = \frac{i}{2\pi(\omega+c)},$$ where $$\tilde{f}$$ denotes the Fourier transform of $$f$$. Fourier transforming back, we get $$f(x)=e^{icx}\mathscr{H}(x)$$.