Proving that two integrals are proportional to each other (Fourier Analysis) I want to show that there exists a constant $C>0$ such that for all functions $f\in S(\mathbb{R})$,
$$\int_{\mathbb{R}}\int_{\mathbb{R}}\frac{|f(x+h)+f(x-h)-2f(x)|^2}{|h|^3}dxdh=C\int_{\mathbb{R}} |f'(x)|^2dx.$$
My idea was to use Plancherel's theorem to obtain the equivalent equality (taking Fourier transform with respect to $x$)
$$\int_{\mathbb{R}}\int_{\mathbb{R}}\frac{|(e^{2\pi ih\xi}+e^{-2\pi ih\xi}-2)\hat{f}(\xi)|^2}{|h|^3}d\xi dh=C\int_{\mathbb{R}}\xi^2|\hat{f}(\xi)|^2d\xi.$$
I noticed that the left hand side of the equality I want to prove is similar to the limit definition of a second derivative but I don't see how that would help.
I don't know how to continue from here or even if I'm on the right track. Any help would be appreciated.
Note: I'm seeing this question in the context of Fourier transforms.
 A: Proceed from what you have obtained:
It remains to show that
$$
\int_{\mathbb R} \frac{|e^{2\pi i h\xi}+e^{-2\pi i h\xi}-2|^2}{|h|^3}dh = C \xi^2\qquad (*)
$$
Note that
$$
e^{2\pi i h\xi}+e^{-2\pi i h\xi}-2 = -4\sin^2(\pi h\xi)
$$
and thus we are looking at
$$
\int_{\mathbb R} \frac{16\sin^4(\pi h\xi)}{|h|^3}dh.
$$
By change of variable $x = h\xi$, we see that
$$
\int_{\mathbb R} \frac{16\sin^4(\pi h\xi)}{|h|^3}dh = 16\xi^2 \int_{\mathbb R} \frac{\sin^4(\pi x)}{|x|^3}dx.
$$
(Check this for $\xi > 0$ and $\xi < 0$ separately!)
The integral
$$
\int_{\mathbb R} \frac{\sin^4(\pi x)}{|x|^3}dx
$$
is clearly convergent, call it $K$. It follows that $C = 16K$ in $(*)$.
A: Nice answer by @user58955. To show $$\int_{\Bbb R}{16\sin^4\pi h \xi\over |h|^3}dh=C\xi^2$$for some $C>0$, define $$F(\xi)=\int_{\Bbb R}{16\sin^4\pi h \xi\over |h|^3}dh$$hence by substituting $u=h\xi$, for $\xi\ge 0$ we obtain $$F(\xi){=\int_{\Bbb R}{16\sin^4 \pi u\over |{u\over \xi}|^3}d{u\over \xi}
\\=\int_{\Bbb R}{16\xi^3\sin^4 \pi u\over |u|^3}d{u\over \xi}
\\=\int_{\Bbb R}{16\xi^2\sin^4 \pi u\over |u|^3}d{u}
\\=\xi^2\cdot \underbrace{\int_{\Bbb R}{16\sin^4 \pi u\over |u|^3}d{u}}_{=C}
\\=C\xi^2
\\=F(1)\cdot \xi^2
}$$therefore $$F(\xi)=F(1)\xi^2\quad,\quad \xi\ge 0$$ and since $F(\xi)=F(-\xi)$, we finally prove what we want. The constant $C$ is
$$C=\int_{\Bbb R}{16\sin^4 \pi u\over |u|^3}d{u}$$
