User robert - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T06:39:55Z http://mathoverflow.net/feeds/user/25118 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/102244/in-fourier-transforms-positive-definite-functions-bochners-theorem-and-deriva In Fourier Transforms: Positive Definite Functions, Bochner's Theorem, and Derivatives Robert 2012-07-14T17:56:42Z 2012-08-22T12:11:00Z <p>I've been reading about Bochner's Theorem lately, but when I apply it to the derivative of a function, I seem to get a contradiction with the theorem.</p> <blockquote> <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. " -<a href="http://mathoverflow.net/questions/62949/positive-definite-functions-and-fourier-transforms" rel="nofollow">source</a></p> </blockquote> <p>So consider $f(x) \in L^1 (\mathbb{R})$ where $f(x) = \tfrac{1}{2} x^2$ if $-1 \leq x \leq 1$ and $0$ otherwise.</p> <p>Also consider $g(x) \in L^1 (\mathbb{R})$ where $g(x) = 1$ if $-1\leq x \leq 1$ and $0$ otherwise.</p> <p>Now Bochner's theorem states that the Fourier transform of each of these functions should be positive definite. One well known property of Positive Definite functions $h(\xi)$ is that: $h(0) \geq |h(x)|$ for all $x\in \mathbb{R}$.</p> <p>Now consider $\bar{g}$ and $\bar{f}$ the Fourier transforms of $g$ and $f$ respectively. Since $g(x)=f''(x)$ (a.e.) then $\bar{g}(\xi) = \xi^2 \bar{f}(\xi)$ implying $\bar{g}(0)=0$</p> <p>By Bochner's Theorem we know $\bar{g}$ is positive definite, but then this implies that $\bar{g}$ is zero for all $\xi$. But one can see by the Fourier inversion theorem this would imply $g$ is zero a.e. </p> <p>Obviously this is a contradiction. I've been banging my head on this for days, can you let me know where the error in the reasoning is? I am wondering if there is a mistake in the statement '$\bar{g}(\xi) = \xi^2 \bar{f}(\xi)$' or in my understanding of positive definite functions or in my use of Bochner's theorem. Thanks in advance!</p> <p>Note, I know my examples of f and g are discontinuous. If this is the problem, it isn't actually a problem in my situation, so if it helps, consider a smooth mollification of g and let f be the anti-derivative of its antiderivative. Thanks!</p>