On the characteristics function of smooth compactly supported distributions

Question

My question is concerned with the Fourier transform of a density function of a continuous random variable (or characteristics function). In a book of Kim Lai Chung with the title "A course in Probability theory", page 192, the author wrote and proved that "for each $0<\alpha\leqslant 2$, the function ${\phi _\alpha }\left( t \right) = {e^{ - {{\left| t \right|}^\alpha }}}$ is a characterictics function". My question: "Given the function ${\psi _\beta }\left( t \right) = {e^{ - k{{\left| t \right|}^\beta }}}$, $t\in\mathbb{R}$, with $0<\beta <1$, $k>0$. Whether or not $\psi_\beta(t)$ is the characteristics function of a smooth compactly supported distribution?" (Actually I can prove that the answer is NO in case $1\le \beta\le 2$, and so here I just focus the question on the case $0<\beta<1$).

Answer 1

A compactly supported random variable has finite expectation. If $\Psi_{\beta}$ were the characteristic function of a compactly supported random variable $X$, then $\Psi_{\beta}$ would have to be differentiable at $t=0$ and $$E(X)=\frac{1}{i}\frac{d}{dt}\Psi_{\beta}(t) |_{t=0}\,. $$ But in case $0<\beta<1$, $\Psi_{\beta}$ is not differentiable at $t=0$.

Answer 2

Hints:

1. Since $t \mapsto e^{-k |t|^{\beta}}$ is integrable, it follows from the Riemann-Lebesgue lemma that the distribution admits a continuous density (with respect to Lebesgue measure), i.e. there exists $p_{k,\beta} \geq 0$ such that $$\int e^{\imath \, t x} p_{k,\beta}(x)\, dx = e^{-k |t|^{\beta}}. \tag{1}$$
2. Show that $$p_{k,\beta}(x) = k^{-d/\beta} p_{1,\beta}(k^{-1/\beta} x).$$
3. Conclude from the continuity of $p_{k,\beta}$ and the fact that $p_{k,\beta}(0)>0$ that $p_{k,\beta}(x)>0$ for any $x \in \mathbb{R}^d$.

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Solution (2nd step): By $(1)$ and change of variables, we have $$\begin{align*} \int e^{\imath \, t x} k^{-d/\beta} p_{1,\beta}(k^{-1/\beta} x) \, dx &\stackrel{y:= k^{-1/\beta}x}{=} \int e^{\imath \, (t k^{1/\beta}) y} p_{1,\beta}(y) \, dy \\ &\stackrel{(1)}{=} \exp \left(-1 \cdot |k^{1/\beta}t|^{\beta} \right) = \exp(-k |t|^{\beta}). \end{align*}$$ Since the Fourier transform determines measures uniquely, we conclude that $$p_{k,\beta}(x) = k^{-d/\beta} p_{1,\beta}(k^{-1/\beta}x)$$ for (Lebesgue-)almost all $x \in \mathbb{R}^d$. From the continuity of $p_{k,\beta}$ it follows that the claim holds for all $x \in \mathbb{R}^d$.

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Remark: The given characteristic function is the characteristic function of the so-called $\beta$-stable symmetric distribution.