Tempered Distribution Example

Problem Find the Fourier transform of the linear spline $f$ shown in the figure.

Solution Start by computing the second (distributional) derivative of $f$. The first derivative is $f'(x)= H(x) - H(x-1)-H(x-2) + H(x-3)$. Since $H'(x-a) =\delta(x-a)$, we have $f''(x) = \delta(x) - \delta(x-1) - \delta(x-2) +\delta(x-3)$. The next step is finding the Fourier transform of $f''$. By definition of the Fourier transform of a distribution, $\langle \widehat{f''}, \phi\rangle = \langle f'',\hat \phi \rangle> = (-1)^2\langle f,\big(\hat \phi\big)''\rangle =\langle f, -\,\widehat{x^2\phi} \rangle,$ where the last equation follows from $\mathcal F[ x^k\phi(x)]= i^k \frac{d^k }{d\xi^k}\big(\mathcal F[\phi]\big)$. Using the equation for $f''$ given above, we see that $\langle \widehat{f''}, \phi\rangle = \langle f'',\hat \phi\rangle = \hat \phi(0) - \hat \phi(1) - \hat \phi(2) +\hat \phi(3) =\int_{\mathbb R} \big(\underbrace{1-e^{ix} - e^{2ix} +e^{-3ix}}_{\widehat{f''}} \big)\phi(x)dx$ Since we also have $\widehat{ f''}(\xi)=-\xi^2 \hat f$, we arrive at $\hat f(\xi)= -\xi^{-2} (1-e^{-i\xi} - e^{-2i\xi} +e^{-3ix}\big)$

Updated 5/1/2018 (fjn)