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.

"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. " -source

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.

Also consider $g(x) \in L^1 (\mathbb{R})$ where $g(x) = 1$ if $-1\leq x \leq 1$ and $0$ otherwise.

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 $x\in \mathbb{R}$.

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 $\bar{g}(0)=0$

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.

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!

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!

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    $\begingroup$ This is a nice question for another site; see the FAQ. The Fourier transform of $g$ is a sinc function which is not $0$ at $0$. You shouldn't ignore the jumps in $f$ when you take the derivative. $\bar{g}(\xi) \ne \xi^2 \bar{f}(\xi)$. wolframalpha.com/input/… $\endgroup$ Commented Jul 14, 2012 at 19:32

1 Answer 1


To expand on Douglas Zare comment, the problem is that the identity $\hat{f''}(\xi)=-\xi^2\hat{f}(\xi)$ holds only for regular functions (say Schwartz functions). By duality you can extend this to distributions. So the identity holds for your function $f$, but you have to consider its distributional second derivative $f''$, which is $g$ plus some Dirac deltas and derivates of Dirac deltas supported on $\{-1,+1\}$. These corrections account for the non zero value at the origin of $\hat{g}$.


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