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.

  • $\begingroup$ "to" or "two"? - visitors are discourage by such mistakes! $\endgroup$
    – Moti
    Apr 11, 2020 at 19:19
  • $\begingroup$ @Moti I fixed that already $\endgroup$
    – Ray Bern
    Apr 13, 2020 at 10:45

2 Answers 2


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 $(*)$.


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

$$C=\int_{\Bbb R}{16\sin^4 \pi u\over |u|^3}d{u}$$


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