5
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ "to" or "two"? - visitors are discourage by such mistakes! $\endgroup$
    – Moti
    Apr 11 '20 at 19:19
  • $\begingroup$ @Moti I fixed that already $\endgroup$
    – Ray Bern
    Apr 13 '20 at 10:45
6
+100
$\begingroup$

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 $(*)$.

$\endgroup$
1
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.