For every $b\geq a>0,c>0$ positive real numbers we will define the following functions:
$$\Delta_c(x)=\begin{cases} c-|x| & |x|<c \\ 0 & |x|\geq c\end{cases}$$ $$\Delta_{b\geq a}(x)=\begin{cases} 2a & |x|<b-a \\ a+b-|x| &b-a\leq|x|\leq b+a \\ 0 & |x|> a+b\end{cases}$$
Prove without direct computation of the integrals that for every $b\geq a>0$:
$$8\int\limits_{-\infty}^{\infty}\Delta_a(x)\Delta_b(x)\ dx=\int\limits_{-\infty}^{\infty}\left(\Delta_{b\geq a}(x)\right)^2\ dx$$
The only hint I was given is to use Fourier Transform, but I really don't get why. Computing the Fourier Transform of the functions seems like a much harder work than computing the integrals I need to prove equal themselves.
- I did manage to prove the case $b=a>0$. It is done easily, since the equation becomes:
$$8\int\limits_{-\infty}^{\infty}\left(\Delta_a(x)\right)^2\ dx=\int\limits_{-\infty}^{\infty}\left(\Delta_{2a}(x)\right)^2\ dx$$
From the R.H.S integral you can easily get to the L.H.S one by the simple substitution $x=2u$, and then integrating with respect to $u$.
So actually I just need help with the $b>a>0$ case.
Thank you
Hint: since inner products on $\Bbb R$ are determined by their norms viz.$$\langle u|v\rangle=\frac14\left(\Vert u+v\Vert^2-\Vert u-v\Vert^2\right),$$it suffices to build on the case you've managed by verifying$$\int_{-\infty}^\infty(\Delta_{b\ge a}(x))^2dx=\frac14\int_{-\infty}^\infty\left[(\Delta_{2(b+a)}(x))^2-(\Delta_{2(b-a)}(x))^2\right]dx.$$So start by simplifying the RHS's integrand.