How to integrate $\displaystyle\int\mathbf 1_{(-\frac12,\frac12)}(z-w)\mathbf 1_{(-\frac12,\frac12)}(w)dw$ ?
The integral should give a function of $z$, but I don't know how to compute.
$\displaystyle\int\mathbf 1_{(-\frac12,\frac12)}(z-w)\mathbf 1_{(-\frac12,\frac12)}(w)dw$
$\displaystyle=\int_{-1/2}^{1/2}\mathbf 1_{(-\frac12,\frac12)}(z-w)dw=\int_{-1/2}^{1/2}\mathbf 1_{(-\frac12,0)}(z-w)dw+\int_{-1/2}^{1/2}\mathbf 1_{(0,\frac12)}(z-w)dw$
How to continue ?
Thanks in advance.
Having done what you've done up to the left side of the last line, now change variables to $u=z-w$ and do essentially the same thing again.
You can also think of this without integrals: what is the measure of $(-1/2,1/2) \cap \{ w : z - w \in (-1/2,1/2) \}$? Well, the second set is $(z-1/2,z+1/2)$, so what happens? (As your work suggests, it depends on whether $z \in (-1/2,0)$ or $(0,1/2)$.)