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Usually when we try to show a function is not a characteristic function, we would prove it is not uniformly continuous. I am wondering if there is any other way to show $\cos(t^2)$ is not a characteristic function. $%fooling edit check$

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If we can show that $E(X^2)=0$, how does that imply $\phi$ is not a char fun? thanks but $$-\varphi^{\prime\prime}(t)|_{t=0}=0.$$ is true since $\phi''=2\sin(t^2)+4t^2 \cos(t^2)$ – user16227 Sep 17 '11 at 21:36
For a characteristic function $\varphi$, isn't the set of points $t$ with $\varphi(t)=1$ supposed to be contained in a discrete subgroup of $\mathbb R$ ? But in this case, the group generated is dense. – GEdgar Sep 17 '11 at 21:59
show that it is not positive definite – user16232 Sep 17 '11 at 23:22

Hint: Put $\varphi(t)=\cos(t^2)$. If $X$ is a random variable with this "characteristic function", we'd have $$\mathbb{E}(X^2)=-\varphi^{\prime\prime}(t)|_{t=0}=0.$$

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I know this is a very old post, but I was looking for answers to this same question and I have a doubt with this argument. Couldn't it be that $X$ doesn't have a second order moment? Related – Duronman Apr 6 '15 at 21:38
@Duronman All moments exist since $cos(t^2)$ is infinitely differentiable at 0. – simonzack May 19 '15 at 3:00

Introduce the notation $c(t)=\cos(t^2)$. By Bochner's theorem, every determinant $D(t)$ should be nonnegative, where $$ D(t)=\text{det}\begin{pmatrix}c(0) & c(t) & c(2t)\\ c(-t) & c(0) & c(t)\\ c(-2t) & c(-t) & c(0)\end{pmatrix}. $$ Since $c(t)=c(-t)$ for every $t$, $D(t)=(1-c(2t))E(t)$ with $E(t)=1-2\cos(t^2)^2+\cos(4t^2)$. For $t=1$, $E(1)=1-2\cos(1)^2+\cos(4)=-0.24...$ hence $D(1)<0$.

Alternatively, $E(t)=-6t^4+O(t^8)$ when $t\to0$. Or, $E(t)=-2\sin^2(t^2)(1+2\cos(2t^2))$.

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