Bochner's theorem states that a positive definite function is the Fourier transform of a finite Borel measure. As well, an easy converse of this is that a Fourier transform must be positive definite.

My question is: is there a high-brow explanation for why positive definiteness and Fourier transforms go hand-in-hand?

As I understand it, positive definiteness imposes wonderfully strong regularity conditions on the function. We immediately deduce that the function is bounded above at its value at 0, that it is non-negative at 0 and that continuity at 0 implies continuity everywhere.

A leading example I have in mind comes from probability. One can show (Levy's Theorem) that a sum of iid rv converges weakly to some probability distribution by considering the product of characteristic functions and showing that its tail converges to 1 around an interval containing 0, so by positive definiteness and by the identity $1-\operatorname{Re} \phi(2t) \leq 4(1-\operatorname{Re} \phi(t))$ this implies convergence to a degenerate distribution. It just seems rather mysterious to me how this kind of local regularity becomes global.

**Edit:**

To be a little more specific, I understand that the Radon Nikodym derivative is positive and $e^{ix}$ is positive definite. I am more interested in consequences of positive-definiteness on the regularity of the function. For example, if one takes the 2x2 positive definite matrix associated with the function and considers its determinant, it follows that $|f(x)|\leq |f(0)|$. If I take the 3x3 positive definite matrix, I can conclude that if $f$ is continuous at 0, it is then continuous everywhere. My issue is that these types of arguments give me no intuition at all as to what positive definiteness is.

Let me thus add an additional question: what is it about positive definiteness that adds such regularity conditions?

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