# Convolution of probability densities with “easy” result

SETTING

I am looking for two continuous probability distributions $X,Y$ with respective densities $f_X(\cdot) \text{ and } f_Y(\cdot)$, one of them with support on $[\epsilon,a-\epsilon]$, the other with support on $[-\epsilon, \epsilon]$ (with $\epsilon,a>0$, $\epsilon<\frac{a}{2}$) with properties defined below. The convolution $(f_X*f_Y)(\cdot)$ of the probability density functions has support on [0,a].

REQUIREMENT

The required properties are not very rigorous but rather soft. All densities $f_X(\cdot), f_Y(\cdot) \text{ and } (f_X*f_Y)(\cdot)$ should NOT be piecewise defined on their respective supports, i.e. I don't want to use $$f(z)=\begin{cases}...& z\in [...]\\ ... & z\in[...] \end{cases}$$ to write the functions. I do not care whether intervals are open, closed or semi-closed. I need this example for teaching/simple models where I dont want to many cases.

EXAMPLE WHICH DOESN'T WORK

An example which doesn't work is $X\sim\text{Unif}([\epsilon,a-\epsilon])$ and $Y\sim\text{Unif}([-\epsilon,\epsilon])$ as the density of $X+Y$ is $$f_{X+Y}=(f_Y*f_Y)(y)= \begin{cases} \frac{y}{2\epsilon(a -2\epsilon)} & 0<y<2\epsilon \\ \frac{1}{a-2 \epsilon} & 2\epsilon\leq y \leq a-2\epsilon \\ \frac{a-y}{2\epsilon(a -2\epsilon)} & a-2\epsilon< y <a. \end{cases}$$

Thanks in advance for suggestions or proofs that such a thing is not possible.