convolution square root of uniform distribution I need to find a probability distribution function $f(x)$ such that the convolution $f * f$ is the uniform distribution (between $x=0$ and $x=1$). I would like to generate pairs of numbers with independent identical distributions, so that their sum is uniformly distributed between $0$ and $1$.
This can't be something new, and I can search on google for convolution square root but I can't seem to find the right information on probability distributions.
Can someone out there point me at the right information?

Assume that $X$ is a random variable with density $f$ and that $f\ast f=\mathbf 1_{[0,1]}$. Note that the function $t\mapsto\mathbb E(\mathrm e^{\mathrm itX})$ is smooth since $X$ is bounded (and in fact, $X$ is in $[0,\frac12]$ almost surely). Then, for every real number $t$,
$$
\mathbb E(\mathrm e^{\mathrm itX})^2=\frac{\mathrm e^{\mathrm it}-1}{\mathrm it}.
$$
Differentiating this with respect to $t$ yields a formula for $\mathbb E(X\mathrm e^{\mathrm itX})\mathbb E(\mathrm e^{\mathrm itX})$. Squaring this product and replacing $\mathbb E(\mathrm e^{\mathrm itX})^2$ by its value yields
$$
\mathbb E(X\mathrm e^{\mathrm itX})^2=\frac{\mathrm i(1-\mathrm e^{\mathrm it}+\mathrm it\mathrm e^{\mathrm it})}{4t^3(\mathrm e^{\mathrm it}-1)}.
$$
The RHS diverges when $t=2\pi$, hence such a random variable $X$ cannot exist.
