Null convolution The convolution of two functions $u(t),v(t)$ is defined as
$$u(t)*v(t) = \int_{-\infty}^{+\infty} u(\tau)\,v(t-\tau)\,d\tau.$$
Could you provide a simple example of two functions $u(t),v(t)$ for which their convolution is null for all $t$, without $u(t)$ or $v(t)$ being null for all $t$?
 A: Let $u$ and $v$ be Schwartz functions whose Fourier transforms $\hat u$ and $\hat v$ are bump functions with disjoint support. (To make such $u$ and $v$, start by defining their Fourier transforms, and then set $u$ and $v$ equal to the inverse Fourier transform of the functions you defined.) Then for each $\omega$,
\begin{align*}
\widehat{u\ast v}(\omega) = \hat u(\omega)\cdot\hat v(\omega) = 0.
\end{align*}
Since the Fourier transform is a linear isomorphism of the Schwartz space to itself, $u\ast v = 0$. However, each of $u$ and $v$ are not zero because their respective Fourier transforms are nonzero.
A: Complementing Alex Ortis answer. I agree that using the property that the Fourier Transform $\mathcal{F}$ is such that $\mathcal{F}\{u * v\} = \mathcal{F}\{u\} \mathcal{F}\{v\}$ is the most straightforward way to come up with an example.
For a concrete example of $u$ and $v$ for which $\mathcal{F}\{u\} \mathcal{F}\{v\} = 0$ you can use the sinc function. The Fourier transform of a sinc is a rectangle. For instance, $u(t) = \text{sinc}(\omega_1 t)$ and $v(t) = \text{sinc}(\omega_1 t) \sin(\omega_2 t)$ for appropriate choice of $\omega_1$ and $\omega_2$ should give you the example. Since their Fourier transform can be different then zero in intervals that do not overlap.
To show that, use (besides the Fourier properties mentioned above) the translation property and write $\sin(\omega_2 t) = \frac{\exp(j\omega_2 t) - \exp(-j\omega_2 t)}{2 j}$.
