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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.

Rodrigo de Azevedo
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Guillermo
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  • shouldn't it be $$1 = \int_{\mathbb R^d} | f(x) |^2 dx = \color{red}{(-1)^d} \int_{\mathbb R^d} x \cdot \frac{d}{dx} | f(x) |^2 dx$$? – ViktorStein Mar 03 '20 at 12:39

1 Answers1

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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 $\sqrt{\frac{n}{4\pi}}$.

robjohn
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  • Why is it $$ |\nabla f |_2 = 2 \pi |\xi \hat f |_2 \quad ?$$ What definition of Fourier transform are you using? – Brainstorming Jan 18 '15 at 19:21
  • 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. – robjohn Jan 18 '15 at 19:24
  • 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! – Yu Zhao Apr 07 '18 at 20:20
  • @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. – robjohn Apr 07 '18 at 21:48
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    Oh I see I'm sorry I thought $f$ took values in $\mathbb R^n$. Thank you for the clarification! – Yu Zhao Apr 07 '18 at 22:05
  • @robjohn Hi Rob. I hope all is well. I beleive that by integrating by parts, we have $$\int_{\mathbb{R}^n}\vec r\cdot \nabla |f|^2,d^n \vec r=-n \int_{\mathbb{R}^n}|f|^2,d^n \vec r$$What happened to the minus sign? – Mark Viola Oct 10 '21 at 17:57
  • @MarkViola: sorry for the missed minus sign. Does that look better? – robjohn Oct 10 '21 at 20:25
  • @robjohn That should do it. – Mark Viola Oct 10 '21 at 21:49