Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $f(x)$ is a $d$-dimensional real function and $\int_{R^{d}}|f(x)|^2dx=1$. Show that

$$ (\int_{R^{d}}|x|^2|f(x)|^2dx)(\int_{R^{d}}|\xi|^2|\hat f(\xi)|^2d\xi)\geq\frac{d^2}{16\pi^2}$$

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

share|improve this question
add comment

1 Answer

up vote 3 down vote accepted

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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