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" $\Phi := \mathbb{1}_{\{x<0\}}$.

  • 2
    $\begingroup$ 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 $\int_{-\infty}^x \mathscr{H}(t)\,dt=(\mathscr{H}\ast \mathscr{H})(x).$) $\endgroup$ Aug 20, 2019 at 21:34

1 Answer 1


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 $f(x)=e^{icx}\mathscr{H}(x)$.


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