distributionsmathematical-statisticsprobabilityself-studystatistical significance
Suppose that X is a real-valued random variable. Suppose that there exists a constant M > 0 such that the support of X lies entirely in the interval $[-M,M]$. Let $\phi$ denote the characteristic function of X. Show that $\phi$ is infinitely differentiable.
If infinitely differentiable is equivalent to absolutely continuous, then
$$\int_{-M}^{M}| \phi(t)|\;dt < \infty$$
Density functions are found with the inverse Fourier transform. The density function of the distribution, if such a density exists, will be given by
$$f(t) = \frac{1}{2\pi}\int_{\mathbb{R}}e^{-itx}\phi(x) dx = \frac{1}{2\pi}\int_{\mathbb{R}}e^{-itx}\left((1-x^2/2)e^{-x^2/4}\right) dx .$$
This integral can be split into two, each of which has an integrand of the form
$$\exp(-Q_t(x))x^{2k}$$
where $Q_t$ is a quadratic form with negative leading term and $k$ is a non-negative integer. This makes each integrand a Schwartz (rapidly decreasing) function, assuring its integrability for any $t$. The integrability proves it is continuous; the rapid decrease proves it is absolutely continuous. The integrals are readily performed by completing the square in the exponential, reducing them to multiples of even moments of the Gaussian distribution. The result is
$$f(t) = \frac{2}{\sqrt{\pi}} t^2 e^{-t^2}.$$
The continuity of $f$ confirms the earlier conclusion of absolute continuity of the distribution.
The square of this (symmetric) variable has a Gamma$(3/2, 1)$ distribution.
Alternatively, one might recognize that
$$\phi(t) = -2\left(-\frac{1}{2} + \frac{t^2}{4}\right)e^{-t^2/4} = (-i)^2\frac{d^2}{dt^2}2e^{-t^2/4}$$
is proportional to the second derivative of the Gaussian $e^{-t^2/4}$, implying (since the operator $-id/dt$ on characteristic functions is equivalent to multiplication of distribution functions by the variable) that the density $f(x)$ exists and is proportional to $x^2$ times the density whose c.f. is $2e^{-t^2/4}$. That is immediately recognizable as a Gaussian (Normal) distribution with density proportional to $e^{-x^2}$. At this point all one has to do is work out the normalizing constant of $2/\sqrt{\pi}$ via integration or by computing the variance of a Normal distribution with standard deviation $\sqrt{1/2}$.
Best Answer
I'm not sure what you mean exactly by support here, but if X lies in $[-M,M]$ a.s. then the following works:
First, have a look at the statement of Theorem 6.6 (pg 107) of: http://stat.uconn.edu/~boba/stat6894/probabilityI.pdf
As X is bounded a.s., $E|X|^k$ is finite for every $k>0$. Hence, by the theorem, $\phi^{(k)}(t)$ exists for every $k>0$.