# Indicator function of triangle function - Bochner's theorem

Let $\theta(t) = (1-|t|)^+$ for $t \in \mathbb{R}$. In order to show that \begin{align} \int_{-\infty}^\infty \theta(t) e^{-i \xi t}\ dt = \bigg(\frac{sin\big(\frac{\xi}{2}\big)}{\frac{\xi}{2}}\bigg)^2, \qquad (*) \end{align} I want to compute the Fourier transform of $\phi(t) = \mathbb{1}_{\{-1/2 \leq t \leq 1/2\}}$ and relate this to $\theta$.

It could be shown that the Fourier transform $\hat{f}$ of $\phi$ equals \begin{align} \hat{f}(\xi) = \frac{\sin(\pi \xi)}{\pi \xi}. \qquad (**) \end{align} Making the substitution $u = \frac{1}{2}t$ in $(*)$ gives \begin{align} \int_{-\infty}^\infty \theta(t) e^{-i \xi t}\ dt = 2 \int_{-1/2}^{1/2} (1-|2u|)^+ e^{-2i \xi u}\ du \qquad (***) \end{align} and could be worked out. However, I do not see how $(**)$ could be related to $(***)$. Any help is appreciated!

You can write $\theta$ as $1_I\ast1_I$ for an interval $I$. Say $I=[-a,a]$. Then you have to compute $$\int_{-a}^a1_{[-a,a]}(x-t)dt.$$
• $$\mathbb{1}_{[-1/2,1/2]} * \mathbb{1}_{[-1/2,1/2]} (x) = \int_{\max(-1/2, x-1/2)}^{\min(1/2, x+1/2)} dt$$ $$= \begin{cases} 0 \qquad \text{if x \geq \frac{1}{2}} \\ 1-x \qquad \text{if 0 \leq x < \frac{1}{2}} \\ 1+x \qquad \text{if \frac{-1}{2} < x <0} \\ 0 \qquad \text{if x \leq \frac{-1}{2}}\end{cases}$$ could be plugged in $(***)$. To make ends meet, how do we have to use the Fourier transform of $\phi$? – iJup Apr 7 '17 at 18:54
• Okido, I will do! @Julián, I solved my question. Thanks! I used that $\int_\mathbb{R} (\phi * \phi)(t) e^{-i \xi t}\ dt = (\int_\mathbb{R} \phi(t)e^{-i \xi t}\ dt)^2$. – iJup Apr 8 '17 at 13:01