For a continuous bounded pdf $K(x)$ that is symmetric about $0$ (also suppose $K(0)>0$), let $\phi(t):=\int e^{itx}K(x)dx$ be its characteristic function. Does it hold that $K(x)$ is positive definite (i.e. $\sum_{i=1}^n \sum_{j=1}^n z_i \bar{z_j}K(x_i-x_j)\geq 0$ for all complex $z$'s, real $x$'s, and $n$) iff the real part of $\phi(t)$ is nonnegative? (or at least the "only if" part holds)
Attempt: From Bochner's theorem, we have (suffice via changes of variables): $K$ is positive definite iff there exists a probability measure $\mu$ such that $\frac{K(x)}{K(0)}=\int e^{-itx} d\mu(t)$. Despite that $\mu$ and $\phi$ appear like Fourier inverses of each other, $\mu$ here is a measure while $\phi$ is a real-valued function. How do we proceed to draw a connection between $\phi$ and $\mu$?
The uniqueness of Fourier transform says if $\mu_1, \mu_2$ are finite positive measures and $F(\mu_1)=F(\mu_2)$, then $\mu_1=\mu_2$. However, do we know $\nu(A):=\int_A \phi(t) dt$ is a finite measure?
Edit: As pointed out by Adam, the above reasoning using Fourier inversion needs not to hold in general. A sufficient condition for the iff statement to hold would need an additional assumption that $\phi(t)$ is integrable.
