This is an exercise of the book Analysis III of Amann and Escher:
Let $f(x):=(\sin(x)/x)^2$ and $g(x):=e^{i2x}f(x)$. Show that $f*g=0$.
Before to begin my definition of the Fourier transform:
$$ \hat h(\xi):=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty h(x)e^{-i\xi x}dx,\quad\text{for }h\in L_1(\Bbb R)\tag1 $$
Now some facts to use here, for $f$ and $g$ of the exercise we have $$ \hat f(\xi)=\sqrt{\frac\pi 8}(2-|\xi|)\chi_{[-2,2]}(\xi),\quad\hat g(\xi)=\hat f(\xi-2)\tag2 $$
Also I know that $$ \widehat{h*\ell}=\sqrt{2\pi}\hat h\hat\ell,\quad\text{for }h,\ell\in L_1(\Bbb R)\tag3 $$ And also I know that the Fourier transform, restricted to $L_1$, is linear and injective (with dense image in $C_0$). Then $f*g=0\iff \widehat{f*g}=0$ a.e., and from all the above we have that $$ \begin{align}\hat f\hat g(\xi)&=\frac8\pi(2-|\xi|)(2-|\xi-2|)\chi_{[-2,2]}(\xi)\chi_{[-2,2]}(\xi-2)\\ &=\frac8\pi(2-|\xi|)(2-|\xi-2|)\chi_{[-2,2]\cap[0,4]}(\xi)\\ &=\frac8\pi(2-|\xi|)(2-|\xi-2|)\chi_{[0,2]}(\xi)\\ &=\frac8\pi(2-\xi)\xi\chi_{[0,2]}(\xi) \end{align} $$
However $\hat f\hat g\neq 0$ a.e. and Im 100% sure that $(1)$, $(2)$ and $(3)$ are absolutely correct (I checked them many times, using also Wolfram Mathematica to see if I had some silly mistake).
So, what is wrong? It is possible that the exercise is wrong?
The exercise is wrong. According to Wolfram Alpha, the numerical value of $(f*g)(0)\approx\frac12$. Since $f*g$ continuous, it can thus not be $0$ almost everywhere.