I figured out the answer while writing down the question. Here it is:
$(f^\ast(-t)\ast g(t))(\tau) = \int_{-\infty}^\infty{f^\ast(-t)g(\tau-t)dt} = -\int_{\infty}^{-\infty}{f^\ast(t)g(\tau+t)dt} = \int_{-\infty}^{\infty}{f^\ast(t)g(\tau+t)dt} = (f(t)\star g(t))(\tau)$
In the second step, the substitution $t\to -t$ took place.
we have:
$f(t) = \begin{cases} 2 & 0< t< 3 \\ 0 &\text{elsewhere} \end{cases}$
$g(t) = \begin{cases} 2 & 0< t< 1 \\ 0 & \text{elsewhere} \end{cases}$
so let's take a look over the intervals $\Bbb R,(0,3),(0,1)$:
Using those interval we can see $4$ cases: $(-\infty,0],(0,1),[1,3),[3,\infty)$, I want to point out that there are 4 cases but we will use $[1,3),[3,\infty)$ as one:
if $\tau\in(-\infty,0]$ then $g(\tau)=0$
if $\tau\in[1,3)\cup[3,\infty)=[1,\infty)$ then $g(\tau)=0$
so we can reduce the integral: $\int_{-\infty}^{+\infty} f(t - \tau) g(\tau) \ d\tau\to \int_{0}^{1} f(t - \tau) g(\tau) \ d\tau$
now in the interval $(0,1),\ g(\tau)$ is a constant, so we can take it out:$\int_{0}^{1} f(t - \tau) g(\tau) \ d\tau\to2\int_{0}^{1} f(t - \tau)d\tau$
now we have $g*f(t)=2\int_{0}^{1} f(t - \tau)d\tau$, set $\omega=t-\tau\implies d\omega=-d\tau,\omega(0)=t,\omega(1)=t-1$ so we get: $2\int_{0}^{1} f(t - \tau)d\tau\to-2\int_{t}^{t-1} f(\omega)d\omega\to2\int_{t-1}^{t} f(\omega)d\omega$
Can you solve this kind of intervals?
I learnt that playing with functions can help a lot for intuition.
for example: try: $t=-5,t=0.5,t=1,t=2,t=3.5,t=5$ and see what happens. after that try to find intervals for which the function gives the same value, see how it connects to the original $4$ cases i presented, what is the same and what is different.
Best Answer
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.