Show that $$\phi * \phi(x)=\begin{cases} >0 , \operatorname{if} x\in(0,4\pi)\\ = 0, \operatorname{otherwise} \end{cases}$$ where $\phi: \mathbb R \to \mathbb R$ and $\phi(t)=\begin{cases} 1-\cos{(t)} , \operatorname{if} t\in[0,2\pi]\\ 0, \operatorname{otherwise} \end{cases}$
My idea:
$$\phi * \phi(x)=\int_{\mathbb R} \phi(x-y)\phi(y)dy$$
And for $$\phi(x-y)=[1-\cos{(x-y)}]\chi_{[0,2\pi]}(x-y)$$ and the event $$\chi_{[0,2\pi]}(x-y)=\chi_{[-x,-x+2\pi]}(-y)=\chi_{[x,x-2\pi]}(y) \quad\quad(*)$$.
Now using (*):
$$\int_{\mathbb R} \phi(x-y)\phi(y)dy=\int_{\mathbb R} [1-\cos{(x-y)}]\chi_{[x,x-2\pi]}(y)[1-\cos{(y)}]\chi_{[0,2\pi]}(y)dy$$
But now how do I go about calculating $$\chi_{[x,x-2\pi]}(y)\chi_{[0,2\pi]}(y)$$ If $x \notin (0,4\pi)$, then $[x,x-2\pi]\cap[0,2\pi]=\varnothing$ but I somehow believe that my bound $[x,x-2\pi]$ is wrong. Any ideas?
Since $\;\phi=0\;$ outside $\;[0,2\pi]\;$ , we get:
$$\phi * \phi(x)=\int_{\mathbb R} \phi(x-y)\phi(y)dy=\int_0^{4\pi}\left(1-\cos(x-y)\right)\left(1-\cos y\right)\,dy=$$
$$=\int_0^{4\pi}\left(1-\cos y-\cos(x-y)+\cos y\,\cos(x-y)\right)\,dy=4\pi-\overbrace{\left.\sin y\right|_0^{4\pi}+\left.\sin(x-y)\right|_0^{4\pi}}^{=0}+$$
$$+\int_0^{4\pi}\cos y\,\cos(x-y)\,dy=4\pi+\frac12\int_0^{4\pi}\left(\cos(2y-x)+\cos x\right)\,dy=$$
$$=4\pi+\left.\frac12\left(\frac12\sin(2y-x)+4\pi\cos x\right)\right|_0^{4\pi}=2\pi\left(2+\cos x\right)>0,\,\,\forall\,x\in\Bbb R$$