Density inversion for a triangular characteristic function

First use $h(u)=(1-\mathrm{\mid }u\mathrm{\mid }{)}_{+}$ as a density in the space variable u. It is nonnegative Borel and $\int h=2{\int }_0^1(1-u)du=1$, so it defines a probability. Its characteristic function q has vanishing imaginary part by the oddness of $h(u)\sin (su)$. For $s\mathrm{\ne }0$, integration by parts with $1-u$ and $\sin (su)\mathrm{/}s$ yields $q(s)=2{\int }_0^1(1-u)\cos (su)du=\frac{2}{s}{\int }_0^1\sin (su)du=\frac{2(1-\cos s)}{s^2}\mathrm{.}$ All factors and derivatives are continuous on $[0,1]$, so FTC and the bridge apply. Density integration is applied to the real and imaginary positive/negative parts. At s=0, q(0)=1, and continuity follows from the characteristic-function lemma. Thus q is nonnegative everywhere and bounded by one, with $q(s)\le 4\mathrm{/}{s}^2$ for $\mathrm{\mid }s\mathrm{\mid }\ge 1$. The primitive $-1\mathrm{/}s$ on $[1,R]$, the bridge and monotone convergence give ${\int }_1^{\mathrm{\infty }}{s}^{-2}ds=1$; reflection gives the other tail. Hence q is integrable.

Apply density inversion to that probability with characteristic function q. It supplies the continuous density $H(y)=(2\pi {)}^{-1}\int e^{-isy}q(s)ds$. This equals h everywhere: if the two continuous densities differed at y, continuity would give a small interval where their difference had one strict sign, contradicting that both densities integrate to the same interval mass. In particular $1=h(0)=H(0)=(2\pi {)}^{-1}\int q$. Therefore $f=q\mathrm{/}(2\pi )$ is nonnegative and integrates to one, so defines a probability. It has exactly the displayed formula and the continuous value $1\mathrm{/}(2\pi )$ at zero.

Density integration and the identity in step 2.1 now give ${\phi }_f(t)=\frac{1}{2\pi }\int e^{itx}q(x)dx=H(-t)=h(-t)=(1-\mathrm{\mid }t\mathrm{\mid }{)}_{+}\mathrm{.}$ This includes t=0 and both endpoints t=±1, where the value is zero; outside the closed interval it is zero as well. The density value at x=0 was fixed by continuity, not division by zero. AC is inherited from the compact bridge and density inversion. The argument applied inversion only to the known law h before establishing that the triangle is a characteristic function.