I have the following function:
$f(t) = \begin{cases} 0, |t| > 2 \\ 1, |t| < 1 \\ 2 - |t|, 1 \leqslant |t| \leqslant 2. \end{cases}$
And I need to check whether it is a characteristic function.
I have checked basic properties ($f(0) = 1$, $|f(t)| \leqslant 1$, and it probably is uniformly continuous? I'm not that well-versed with this one). I don't know how to proceed from there. Either there's some obvious random variable with this characteristic function (if so, then how do I find such random variable, is there some sort of an algorithm? I can't just guess) or I forgot about some important property or maybe made a mistake.
Suppose for the sake of contradiction that $f$ is the characteristic function of some r.v $X$.
$f$ is $\mathcal C^\infty$ on a neighborhood of $0$, and $f'(0)=f''(0)$. So $E(X)=E(X^2)=0$, hence $X=0$ a.s.
But the characteristic function of the constant $0$ is not $f$, so $f$ is not a characteristic function.