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!
Best Answer
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 $\hat{g}$.