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Suppose $$\int_{I_1} x^2|f(x)|^2dx\ge\frac12\int_\Bbb Rx^2|f(x)|^2dx$$ and $$\int_{I_2} x^2|\hat f(x)|^2dx\ge\frac12\int_\Bbb Rx^2|\hat f(x)|^2dx$$

for interval $I_1, I_2$ centered at origin and $f\in\mathcal S(\Bbb R)$.

How to prove that $|I_1||I_2|\ge\frac1{2\pi}$?

I use Cauchy-Schwarz inequality and Plancherel formula but it doesn't work...

Please give some hint.

Phira
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Arturo
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2 Answers2

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First of all, note that the claim does not hold for $f=0$. In the sequel, we assume that $f$ is not identical zero.

Set $I_1=[-a,a]$, $I_2=[-b,b]$ for $a,b>0$. Since $f$ is a Schwartz function, hence in particular bounded, we can assume $$\int f^2(x) \, dx = 1.$$

By the uncertainty principle and the assumption on the integrals,

$$\begin{align*} \frac{1}{16 \pi^2} &\leq \left( \int_{\mathbb{R}} x^2 \cdot f(x)^2 \, dx \right) \cdot \left( \int_{\mathbb{R}} x^2 \cdot \hat{f}(x)^2 \, dx \right) \\ &\leq 4 \left( \int_{-a}^a x^2 \cdot f(x)^2 \, dx \right) \cdot \left( \int_{-b}^b x^2 \cdot \hat{f}(x)^2 \, dx \right) \end{align*}$$

Now find a upper bound (depending on $a$, $b$) for the remaining integrals using that $\|f\|_{L^2}=1$ and Plancherel's identity.

Remark Note that there are several definitions for the Fourier transform. The constant $\frac{1}{16\pi^2}$ refers to the Fourier transform

$$\hat{f}(\xi) = \int f(x) \cdot e^{-2\pi \imath \, x \cdot \xi} \, dx$$

Depending on the definition in your book, you might have to modify this constant.

saz
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  • Oh.... I think you're right but it's exercise from Fourier analysis by Stein and Shakarchi (Ch.5 22). I need to ask about it to professor. – Arturo Dec 08 '13 at 13:13
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    @JinyongGo I guess the claim is correct if the integrals are strictly positive (i.e. $f$ is not identical zero). – saz Dec 08 '13 at 13:39
  • Then, can you give me some hint? It's from chapter of Fourier transform, so I think it's related to those thing I mentioned but no improvement at all... – Arturo Dec 08 '13 at 13:42
  • @JinyongGo I have added a hint. – saz Dec 08 '13 at 15:13
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In this answer, it is shown that in $\mathbb{R}^n$ and using the same Fourier Transform that saz mentions, $$ \|\xi\hat{f}\|_2\|xf\|_2\ge\frac{n}{4\pi}\|\hat{f}\|_2\|f\|_2\tag{1} $$ Therefore, $$ \begin{align} \frac12\|xf(x)\|_2\|\xi\hat{f}(\xi)\|_2 &\le\|xf(x)\|_{L^2(I_1)}\|\xi\hat{f}(\xi)\|_{L^2(I_2)}\tag{2}\\ &\le\frac{|I_1||I_2|}{4}\|f(x)\|_{L^2(I_1)}\|\hat{f}(\xi)\|_{L^2(I_2)}\tag{3}\\ &\le\pi|I_1||I_2|\|xf(x)\|_2\|\xi\hat{f}(\xi)\|_2\tag{4} \end{align} $$ $(2)$: Given
$(3)$: $|x|\le\frac{|I_1|}2$ and $|\xi|\le\frac{|I_2|}2$
$(4)$: $(1)$ in $\mathbb{R}^1$

Thus, we get that $|I_1||I_2|\ge\frac1{2\pi}$

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robjohn
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