symmetric function derived from a symmetric continuous pdf is positive definite?

Question

Let $K(x)$ be a continuous bounded pdf symmetric about $0$; $K(x)$ is increasing for $x<0$ and decreasing for $x>0$. If $k(x,y):=K(x-y)$, then is $k(x,y)$ a positive definite function?

Answer 1

Not necessarily. For kernels $K$ with integrable characteristic function, Bochner's theorem says that $K$ is positive definite if and only if its characteristic function is non-negative. If $K$ is the density of a sum of independent variables $U[-1, 1]$ and $N(0, 1)$, then it satisfies your conditions, but its characteristic function is

$$e^{-t^2/2}\frac{\sin(t)}{t},$$

which is integrable, and sometimes negative.

Answer 2

Edit: the current (updated) answer by Adam (with the additional assumption that the characteristic function is integrable) fills in the gap and provides a simple solution.

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Outdated: To supply @Adam's answer with more justification:

From Bachner's theorem, a function $K(·)$ is positive definite iff $K$ is the Fourier transform of a finite positive measure. By assumption, $K$ is a continuous pdf with $K(0)$ being the largest number, so $K(0)>0$ and $\tilde{K}(x):=\frac{K(x)}{K(0)}$ satisfies $\tilde{K}(0)=1$. Hence, one can show Bachner's theorem is equivalent to that $\tilde{K}(x)$ is the Fourier transform of some probability measure. By a change of variable, this is equivalent to $\tilde{K}(x)=\int e^{itx} d\mu(x)$ for some probability measure $\mu$, i.e. $\tilde{K}$ is a characteristic function.

Define $\phi(x):=\int e^{itx}K(t)dt$, which is a characteristic function derived from density $K$. We want to show if the above is true then the real part of $\phi(x)\geq 0$. Define $\tilde{\phi}(x):=\int e^{itx}\tilde{K}(t)dt$. Since $\phi(x)=K(0)\tilde{\phi}(x)$, it is equivalent to show $\tilde{\phi}(x)\geq 0$ (the key is to use the inversion formula given that $\tilde{K}$ is a characteristic function).

It is known that a characteristic function is bounded and continuous ($\phi$ is, hence so is $\tilde{\phi}$). Hence, if the real part of $\tilde{\phi}(x)<0$ at some $x\in R$, then there exists some neighborhood around $x$, say $(a,b)$, such that $(a,b)$ is a continuity set of $\mu$ (i.e. $\mu(\{a\}),\mu(\{b\}) = 0$; this holds since continuity points for probability measures on $R^n$ are dense), and satisfies the real part of $\int_a^b \tilde{\phi}(x) dx<0$. However, $$\int_a^b \tilde{\phi}(x) dx = \int\frac{e^{itb}-e^{ita}}{it}\tilde{K}(t) dt = \int\frac{e^{-ita}-e^{-itb}}{it}\tilde{K}(t) dt \; \left(\text{let t=-t and use symmetry of}\, \tilde{K}\right)$$ $$= 2\pi \left[\frac{1}{2\pi} \lim_{T\rightarrow\infty} \int_{-T}^T\frac{e^{-ita}-e^{-itb}}{it}\tilde{K}(t) dt \right]$$

Since $\tilde{K}$ is a characteristic function derived from $\mu$ and $(a,b)$ is a continuity set for $\mu$, we can apply the inversion formula and get that the above expression equals $2\pi \mu((a,b])$, which then implies $\mu((a,b])<0$, which is a contradiction. Hence, $\tilde{K}(x)=\int e^{itx} d\mu(x)$ implies the real part of $\phi(x)\geq 0$.