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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.

1 Answers 1

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

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    Oh I see I'm sorry I thought $f$ took values in $\mathbb R^n$. Thank you for the clarification!2018-04-07