6
$\begingroup$

Suppose $f(x)$ is a $d$-dimensional real function and $$\int_{R^{d}}|f(x)|^2 \,\mathrm d x=1$$ Show that $$ \left( \int_{R^{d}}|x|^2|f(x)|^2 \,\mathrm d x \right) \left( \int_{R^{d}}|\xi|^2|\hat f(\xi)|^2 \,\mathrm d\xi \right) \geq \frac{d^2}{16\pi^2}$$

I derived that $$1=\int_{R^{d}}x \left(\frac{d}{dx}\right)|f(x)|^2 \,\mathrm dx$$ but I lost my way. I need your help.

$\endgroup$
1
  • $\begingroup$ shouldn't it be $$1 = \int_{\mathbb R^d} | f(x) |^2 dx = \color{red}{(-1)^d} \int_{\mathbb R^d} x \cdot \frac{d}{dx} | f(x) |^2 dx$$? $\endgroup$ – Ramanujan Mar 3 '20 at 12:39
9
$\begingroup$

Consider the equation $$ \sum_{i=1}^n\frac12x_i\frac{\mathrm{d}}{\mathrm{d}x_i}|f|^2=\mathrm{Re}\left(\nabla f\cdot\overline{xf}\right)\tag{1} $$ Integrating $(1)$ over $\mathbb{R}^n$ and then integrating by parts on the left side: $$ \begin{align} \frac n2\|f\|_2^2 &=\mathrm{Re}\left(\int_{\mathbb{R}^n}\nabla f\cdot\overline{xf}\,\mathrm{d}x\right)\\ &\le\left|\int_{\mathbb{R}^n}\nabla f\cdot\overline{xf}\,\mathrm{d}x\right|\\[6pt] &\le\|\nabla f\|_2\|xf\|_2\\[9pt] &=2\pi\|\xi\hat{f}\|_2\|xf\|_2\tag{2} \end{align} $$ Thus, $$ \|\xi\hat{f}\|_2\|xf\|_2\ge\frac{n}{4\pi}\|\hat{f}\|_2\|f\|_2\tag{3} $$ The last inequality says that the $L^2$ support radius for $f$ and $\hat{f}$ cannot have a product less than $\frac{n}{4\pi}$. This inequality is sharp as can be seen using the function $f(x) = e^{-\pi x\cdot x}$, whose Fourier Transform is itself, and whose $L^2$ support radius is $\sqrt{\frac{n}{4\pi}}$.

$\endgroup$
5
  • $\begingroup$ Why is it $$ \|\nabla f \|_2 = 2 \pi \|\xi \hat f \|_2 \quad ?$$ What definition of Fourier transform are you using? $\endgroup$ – Brainstorming Jan 18 '15 at 19:21
  • $\begingroup$ I am using $$\hat{f}(\xi)=\int_{\mathbb{R}^n}f(x)\,e^{-2\pi ix\cdot\xi}\,\mathrm{d}x$$ It is the one for which $e^{-\pi x\cdot x}$ is its own Fourier Transform. $\endgroup$ – robjohn Jan 18 '15 at 19:24
  • $\begingroup$ Hi, I don't think this answer makes sense. Is $\nabla f$ the gradient of $f$? If so, the first formula doesn't seem to type check. The only reasonable interpretation of $xf$ is $x\cdot f$ which gives a scalar so the left hand side is a vector, whereas the right hand side is a scalar. Also, concerning @Brainstorming's point, we would have to use divergence instead of gradient, so that $$\sum_{i=1}^n \xi_i\hat f_i = \sum_{i=1}^n \widehat{\partial_i f_i}$$ Do you know how to fix this? Thanks! $\endgroup$ – Yu Zhao Apr 7 '18 at 20:20
  • $\begingroup$ @YuZhao: $f$ is a scalar function, so $x\cdot f$ doesn't make sense. $\nabla f$ would be a tensor if $f$ were not a scalar function. $\endgroup$ – robjohn Apr 7 '18 at 21:48
  • $\begingroup$ Oh I see I'm sorry I thought $f$ took values in $\mathbb R^n$. Thank you for the clarification! $\endgroup$ – Yu Zhao Apr 7 '18 at 22:05

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.