The difference of two i.i.d random variables by convolution Let $X$ and $Y$ be i.i.d. and uniformly distributed random variables on $[1/2,1]$. Furthermore, we define $Z:=X-Y$ and want to calculate the pdf by convolution as follows:
$f_Z(z)=\int^{\infty}_{-\infty}f_X(x)f_Y(x-z)\mathrm{d}x$
First, note that $Z$ can take values from $-1/2\le z \le 1/2$ and $f_Y$ is zero outside $1/2\le x -z \le 1$. Thus we can write the convolution 
$f_Z(z)=\int^{\infty}_{-\infty} 1/4 \chi_{[1/2,1]}\chi_{[1/2+z,z+1]}\mathrm{d}x$,
where $\chi$ is the indicator function (also denoted $\mathbb{1}$). 
I now divide the interval of $z$ into $-1/2\le z\le 0$ and $0\le z \le 1/2$ and obtain
$f_Z(z)=\begin{cases} \int^1_{1/2+z}\frac{\mathrm{d}x}{4}=\frac{0,5-z}{4},& \ \ 0\le z \le 1/2 \\
\int^{z+1}_{1/2}\frac{\mathrm{d}x}{4}=\frac{0,5+z}{4},& \ \ -1/2\le z \le 0\end{cases}$
This density is wrong and I could not figure out the problem for hours.
 A: Let $X,Y$ be uniformly distributed on $[a,b]$. Then the pdf of $Z = X-Y$ is
$$p(z) = \int_\mathbb{R} dx\int_\mathbb{R} dy \ p(x) p(y) \ \delta(z-(x-y)) =$$
$$ = \frac{1}{(b-a)^2} \int_\mathbb{R} dx \int_\mathbb{R} dy \ \mathbb{1}^x_{[a,b]} \mathbb{1}^y_{[a,b]} \delta(z-(x-y))=$$
$$  = \frac{1}{(b-a)^2} \int_\mathbb{R} dx \int_\mathbb{R} dy \ \mathbb{1}^x_{[a,b]} \mathbb{1}^y_{[a,b]} \delta(y-(x-z))=$$
$$  = \frac{1}{(b-a)^2} \int_\mathbb{R} dx \ \mathbb{1}^x_{[a,b]} \mathbb{1}^{x-z}_{[a,b]} =$$
The product of the two indicators is equivalent to the conditions:
$$a \leq x \leq b \quad \wedge \quad z+a\leq x \leq z+b.$$
For different values of $z$, these conditions can be reduced to a single inequality, for instance if $0 \leq z \leq b-a$, then the inequality $a+z \leq x $ is "stronger" than $a \leq x$, and the inequality $x \leq b+z$ is "weaker" than $x \leq b$. So for those values of $z$, you have the product of the indicators simplifying to:
$$\mathbb{1}^x_{[a,b]} \mathbb{1}^{x-z}_{[a,b]} = \mathbb{1}^x_{[a+z,b]}, \quad 0 \leq z \leq b-a.$$
Similarly, for another set of values of $z$ (negative $z$, but not too negative), you'll have a different total constraint, and if $z +a \geq b$ or $z+b \leq a$, the two indicators will be totally disjoint and give zero.
To help visualize this, you can draw a picture like this:
$$...---[--*---------]--*---...$$
where the square brackets denote the interval $[a,b]$, and the stars are $a+z$ and $b+z$ respectively. 
In your solution you had the wrong numerical factor, $1/4$ instead of a $4$.
