Does existence of a PDF imply the integrability of the characteristic function?

Question

If the characteristic function of a random variable is integrable, then the random variable has a probability density function. Can anything be said about the other direction? That is, if a random variable has a probability density function (even a smooth one), is the characteristic function integrable or can anything at all be said about its decay?

Answer 1

In general, the characteristic function will not be integrable. Because if it was, it would follow (via Fourier inversion) that the PDF is actually a continuous function. So any discontinuous PDF (i.e. not a.e. to a continuous function) will do, for example $1_{[0,1]}$.

Conversely, a cheap criterion is the following: If the PDF $f$ has $k$ derivatives which are all integrable, then iterated partial integration shows

$$ g(\xi)=E(e^{i X \xi})=\int e^{i x \xi} f(x)\, dx = (-1)^k \cdot (i\xi)^k \cdot \int f^{(k)}(x) e^{i x \xi}\,dx. $$

Since characteristic functions are always bounded, this implies $$ |g(\xi)| \lesssim (1+|\xi|)^{-k}, $$ which for $k\geq 2$ yields integrability of the characteristic function.

Very similar arguments also hold in higher dimension.

What I used here is the general heuristic/fact that smoothness of a function corresponds to decay of its Fourier transform.