# Let $f(x):=(\sin(x)/x)^2$ and $g(x):=e^{i2x}f(x)$. Show that $f*g=0$

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.