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 $\mathscr{H}$:


*

*$x \cdot \mathscr{H}(x) = \int_{- \infty}^{x} \mathscr{H}(t) dt$

*$(\mathscr{H} \ast \mathscr{H})(x) = \int_{- \infty}^{x} \mathscr{H}(t) dt$, 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" $\Phi := \mathbb{1}_{\{x<0\}}$.
 A: 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 $f(x)=e^{icx}\mathscr{H}(x)$.
