Characteristic function with an indicator function We know that $\varphi(t) = \cos{t}$ is a characteristic function of Rademacher random variable. Is the function $\phi(t) = \cos{t} \cdot 1_{|t| \le \pi /2}$ still a characteristic function of some random variable?
The only thing we have to check is if $\phi$ is positively-defined. I tried looking for some examples contradicting it, but with no success, so I suspect it is a characteristic function. When I write the condition of positive-definiteness of $\phi$, it looks very similar to the condition for $\varphi$, but with more zeros, but I cannot deduce anything out of it.
 A: Suppose $\phi$ is the Fourier transform of a finite measure $\mu$ on $(\mathbb{R},\mathscr{B}(\mathbb{R}))$. By Lévy inversion
$$\begin{aligned}\frac{\mu(3)+\mu(5)}{2}+\mu((3,5))&=\int_{[-\pi/2,\pi/2]}\frac{e^{i5\xi}-e^{i3\xi}}{i\xi}\cos(\xi)d\xi\approx -0.18\end{aligned}$$
But $\mu(B)\geq 0$ for all Borel $B$, so $\phi$ is not the CF of a finite measure $\mu$.
A: If your given function $\phi (t) = {\Bbb E}(e^{ i t X})$ is the characteristic function of a random variable $X$ whose probability density is $f(x)dx$, then $ \phi(t)$ is the Fourier transform of $f(x)$ and the  extremely useful Fourier inversion formula allows us to recover the  function  $f(x)$ by the general identity  $$f(x) = \frac{1}{2\pi}  \int_{t=-\infty}^{t=\infty} e^{- i t x} \phi(t) dt$$  (See footnote regarding sign  conventions.)
In your example this simplifies to $f(x) =\frac{1}{2\pi} \int_{t=-\pi/2}^{t=\pi/2} \ (\cos t ) \ e^{-i t x} dt$ which turns out to be $$f(x)=\frac{ \cos(\pi x/2)}{ \pi ( 1- x^2)}$$
Plotting the graph of $f(x)$, we see that it fails to be always non-negative; thus $f(x)$ is NOT a probability density.

P.S. Wikipedia and many other sources use a slightly different convention  for the Fourier transform (engineer's version) that shuffles the constant $2\pi$  into other places. The version used above is what is used in statistics and in theoretical physics.
See e.g equations 3 and 4 in reference below.
Physicist's Fourier formulas
P.P.S. In response to a comment that raises a subtle issue I would add one clarification.
There is no a priori assumption that $X$ has a density. Since $\phi$ has compact support there is no doubt that it is the Fourier transform of a smooth function $f(x)$.  And the uniqueness theorem for Fourier transforms of generalized functions  implies that no finite measure $X$  could exist that has the same characteristic function as this $f(x)$. Thus ultimately the pivotal question is whether or not this function $f(x)$ is non-nonnegative.
The other  characteristic  function mentioned, $\cos t$ did not decay at $\infty$ so by the contrapositive  of the Riemann-Lebesgue Lemma it could not be the characteristic function of any absolutely integrable function. Instead it is  of course the Fourier transform of a pair of Dirac delta functions (point  masses) which are of course generalized functions.
