$Exercise \oplus Problem 3$

你好，这里是我的个人网站数学分析的每周一题栏目(数学分析每周一题，其中数学分析指的是数学中的分析学, 主要包括微积分，实分析，复分析) ——————Alina Lagrange

Suppose that

φ \in S (R)

and

\int_{R} | φ |^{2} d x = 1

, show that

\int_{R} | x φ (x) |^{2} d x \int_{R} ξ^{2} | \hat{φ} (ξ) |^{2} d ξ \geq \frac{1}{16 π^{2}}

$P r o o f .$

As

φ \in S (R) \Rightarrow \int_{R} | φ (x) |^{2} d x = - \int_{R} x \frac{d}{d x} | φ (x) |^{2} d x

= - \int_{R} x \overset{―}{φ (x)} φ^{'} (x) + x φ (x) \overset{―}{φ^{'} (x)} d x

By Cauthy-Schwartz inequality and Plancherel formula and the Fourier transform

\begin{aligned} 1 & = \int_{R} | φ (x) |^{2} d x \\ \leq 2 \int_{R} | x ‖ φ (x) ‖ φ^{'} (x) | d x \\ \leq 2 ‖ x φ (x) ‖_{L^{2} (R)} {‖ φ^{'} (x) ‖}_{L^{2} (R)} \\ = 2 ‖ x φ (x) ‖_{L^{2} (R)} {‖ \hat{φ^{'} (x)} ‖}_{L^{2} (R)} \\ = 2 ‖ x φ (x) ‖_{L^{2} (R)} ‖ 2 π i ξ \hat{φ} (ξ) ‖_{L^{2} (R)} \\ Note : | φ |^{2} = \bar{φ} φ \end{aligned}

Hence we finished the proof.

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