Positive-Definite Functions and Fourier Transforms - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T09:44:14Z http://mathoverflow.net/feeds/question/62949 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/62949/positive-definite-functions-and-fourier-transforms Positive-Definite Functions and Fourier Transforms Alex R. 2011-04-25T18:19:36Z 2011-04-26T01:19:00Z <p>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. </p> <p>My question is: is there a high-brow explanation for why positive definiteness and Fourier transforms go hand-in-hand? </p> <p>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. </p> <p>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 $1-\mbox{Re} \phi(2t) \leq 4(1-\mbox{Re} \phi(t))$ this implies convergence to a degenerate distribution. It just seems rather mysterious to me how this kind of local regularity becomes global. </p> <p><strong>Edit:</strong></p> <p>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. </p> <p>Let me thus add an additional question: what is it about positive definiteness that adds such regularity conditions? </p> http://mathoverflow.net/questions/62949/positive-definite-functions-and-fourier-transforms/62957#62957 Answer by Alain Valette for Positive-Definite Functions and Fourier Transforms Alain Valette 2011-04-25T19:21:21Z 2011-04-25T19:21:21Z <p>What about the fact Fourier transform converts convolution into ordinary (pointwise) product, and therefore converts positive-definiteness into ordinary positivity?</p> http://mathoverflow.net/questions/62949/positive-definite-functions-and-fourier-transforms/62991#62991 Answer by Dima Shlyakhtenko for Positive-Definite Functions and Fourier Transforms Dima Shlyakhtenko 2011-04-26T01:19:00Z 2011-04-26T01:19:00Z <p>Perhaps the phenomenon you are asking about is: why is the definition of a positive-definite function natural? </p> <p>One answer is that positive-definite functions are exactly coefficients of group representations, in the following sense. If $\pi : \mathbb{R}\to U(H)$ is a unitary representation of $\mathbb{R}$ on some Hilbert space $H$, and $h\in H$ is a vector, then the function $$t\mapsto \langle \pi (t) h, h\rangle$$ is positive-definite. Conversely, given a positive-definite function $\phi$, there exists a Hilbert space $H$, a vector $h\in H$ and a unitary representation $\pi$ of $\mathbb{R}$ on $H$, for which $\phi(t)=\langle \pi(t)h,h\rangle$. </p> <p>Indeed, the $n\times n$ matrix occurring in the definition of a positive definite function is nothing more than the Gramm matrix of inner products $\langle \pi (t_i) h, \pi (t_j) h\rangle$; and positivity of this matrix is just a reflection of the fact that the inner product of $H$, restricted to the linear span of $\pi(t_i)h: i=1,\dots,n$ is positive-definite. </p> <p>The Fourier transform goes from the functions on the group to functions on the space of irreducible unitary representations of the group, and thus switches positivity and complete positivity.</p>