Is it possible to have the PDF of the difference of two iid r.v.'s look like a rectangle (instead of, say, the triangle we get if the r.v.'s are taken from the uniform distribution).
i.e. is it possible for the PDF f of j-k (for two iid r.v.'s taken from some distribution) to have f(x) = 0.5 for all -1 < x < 1?
There are no restrictions on the distribution we take j and k from except that the min is -1 and the max is 1.
After some experimentation, I'm thinking this might be impossible.
Theorem: There is no distribution $\text{Dist}$ for which $A-B \sim \text{U}(-1,1)$ when $A, B \sim \text{IID Dist}$.
Proof: Consider two random variables $A, B \sim \text{IID Dist}$ with common characteristic function $\varphi$. Denoting their difference by $D=A-B$. The characteristic function of the difference is:
$$\begin{equation} \begin{aligned} \varphi_D(t) = \mathbb{E}(\exp(t D)) &= \mathbb{E}(\exp(t (A-B))) \\[6pt] &= \mathbb{E}(\exp(t A)) \mathbb{E}(\exp(-t B)) \\[6pt] &= \varphi(t) \varphi(-t) \\[6pt] &= \varphi(t) \overline{\varphi(t)} \\[6pt] &= |\varphi(t)|^2. \\[6pt] \end{aligned} \end{equation}$$
(The fourth line of this working follows from the fact that the characteristic function is Hermitian.) Now, taking $D \sim \text{U}(-1,1)$ gives a specific form for $\varphi_D$, which is:
$$\begin{equation} \begin{aligned} \varphi_D(t) = \mathbb{E}(\exp(itD)) &= \int \limits_{\mathbb{R}} \exp(itr) f_D(r) dr \\[6pt] &= \frac{1}{2} \int \limits_{-1}^1 \exp(itr) dr \\[6pt] &= \frac{1}{2} \Bigg[ \frac{\exp(itr)}{it} \Bigg]_{r=-1}^{r=1} \\[6pt] &= \frac{1}{2} \frac{\exp(it)-\exp(-it)}{it} \\[6pt] &= \frac{1}{2} \frac{(\cos(t) + i \sin(t)) - (\cos(-t) + i \sin(-t))}{it} \\[6pt] &= \frac{1}{2} \frac{(\cos(t) + i \sin(t)) - (\cos(t) - i \sin(t))}{it} \\[6pt] &= \frac{1}{2} \frac{2i \sin(t)}{it} \\[6pt] &= \frac{\sin(t)}{t} = \text{sinc}(t). \\[6pt] \end{aligned} \end{equation}$$
where the latter is the (unnormalised) sinc function. Hence, to meet the requirements for $\text{Dist}$, we require a characteristic function $\varphi$ with squared-norm given by:
$$|\varphi(t)|^2 = \varphi_D(t) = \text{sinc}(t).$$
The left-hand-side of this equation is a squared norm and is therefore non-negative, whereas the right-hand-side is a function that is negative in various places. Hence, there is no solution to this equation, and so there is no characteristic function satisfying the requirements for the distribution. (Hat-tip to Fabian for pointing this out in a related question on Mathematics.SE.) Hence, there is no distribution with the requirements of the theorem. $\blacksquare$
This is an electrical engineer's take on the matter, with a viewpoint that is more suitable for dsp.SE rather than stats.SE, but no matter.
Suppose that $X$ and $Y$ are continuous random variables with common pdf $f(x)$. Then, if $Z$ denotes $X-Y$, we have that $$f_Z(z) = \int_{-\infty}^\infty f(x)f(x+z) \ \mathrm dx.$$ The Cauchy-Schwarz inequality tells us that $f_Z(z)$ has a maximum at $z=0$. In fact, since $f_Z$ is actually the "autocorrelation" function of $f$ regarded as a "signal", it must have a unique maximum at $z=0$ and thus $Z$ cannot be uniformly distributed as is desired. Alternatively, if $f_Z$ were indeed a uniform density (remember that it is also an autocorrelation function), then the "power spectral density" of $f_Z$ (regarded as a signal) would be a sinc function, and thus not a nonnegative function as all power spectral densities must be. Ergo, the assumption that $f_Z$ is a uniform density leads to a contradiction and so the assumption must be false.
The claim that $f_Z \sim \mathcal U[-1,1]$ is obviously invalid when the common distribution of $X$ and $Y$ contains atoms since in such a case the distribution of $Z$ will also contain atoms. I suspect that the restriction that $X$ and $Y$ have a pdf can be removed and a purely measure-theoretic proof constructed for the general case when $X$ and $Y$ don't necessarily enjoy a pdf (but their difference does).
