Fourier transform of derivative that exists almost everywhere We consider the function  
$$f (x) = 
     \begin{cases}
       1+x &  -1\leq x\leq 0\\
       1-x &  0\leq x\leq 1 \\
       0   & \text{otherwise}
     \end{cases}$$
and the formula 
$$F(f'')(\xi)=-\xi^2F(f)(\xi)$$
I have used the fact that $f$ is twice-differentiable almost everywhere (it has a problem at $x=0,-1$ and $1$) and that this wouldn't affect the integral in the Fourier transform to get that the Fourier transform of $f$ is $0$ on all points that are not $0$, which is false. 
What would be the error and how would be the correct way to use that formula?
 A: The formula $F(f'')(\xi)=-\xi^2F(f)(\xi)$ is valid when the derivative $f''$ is the distributional derivative of $f$. When $f\in C^1$, the classical derivative $f'$ represents the distributional derivative.  When $f$ is absolutely continuous, $f'$ exists a.e. and still represents the distributional derivative. Beyond that is the danger area; "almost everywhere" is not enough, as a single point may be the support of a distribution.
Here $f$ is Lipschitz continuous, hence absolutely continuous. So 
$$f' (x) = 
     \begin{cases}
       1 \quad &  -1 < x< 0\\
       -1 &  0< x < 1 \\
       0   & |x|>1
     \end{cases}$$
correctly represents the distributional derivative. 
However, $f'$ is not continuous. Its derivative is $0$ a.e., but since $f'$ has jumps, there's a singular part of the derivative. Specifically,
$$f '' (x) = \delta (x+1) - 2 \delta (x)  + \delta (x-1)$$ 
as  Rodrigo de Azevedo pointed out. To see why, notice that
$$f ' (x) = H (x+1) - 2H(x)  + H(x-1)$$ 
where $H$ is the Heaviside function, and recall that $H'=\delta$, the Dirac delta.
