$Exercise \oplus Problem 1$

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

Prove that :

\frac{d \hat{f} (ξ)}{d ξ} = \hat{- 2 π i x f (x)}

$P r o o f .$

Since

\frac{d \hat{f} (ξ)}{d ξ} = lim_{h \to 0} \int_{R} f (x) e^{- 2 π i x ξ} \cdot \frac{1}{h} \cdot (e^{- 2 π i x h} - 1) d x

as

h \to 0

| \frac{d \hat{f} (ξ)}{d ξ} - (\hat{- 2 π i x f (x)}) | = | \int_{R} f (x) e^{- 2 π i x ξ} \cdot [\frac{1}{h} \cdot (e^{- 2 π i x h} - 1) + 2 π i x] d x |

\forall ε > 0

, Since

f (x), x f (x) \in S (R)

,there exists

N

such that

\int_{| x | \geq N} | f (x) | d x < ε

and

\int_{| x | \geq N} | x f (x) | d x < ε

And there exists

h_{m} > 0

such that when

| h | < h_{m}

,

| \frac{1}{h} \cdot (e^{- 2 π i x h} - 1) + 2 π i x | < \frac{ε}{N}

Hence

\begin{aligned} | \frac{d \hat{f} (ξ)}{d ξ} - (\hat{- 2 π i x f (x)}) | & = | \int_{R} f (x) e^{- 2 π i x ξ} \cdot [\frac{1}{h} \cdot (e^{- 2 π i x h} - 1) + 2 π i x] d x | \\ = | \int_{| x | \geq N} f (x) e^{- 2 π i x ξ} \cdot [\frac{1}{h} \cdot (e^{- 2 π i x h} - 1)] d x \\ + \int_{| x | \geq N} f (x) e^{- 2 π i x ξ} \cdot 2 π i x d x \\ + \int_{- N}^{N} f (x) e^{- 2 π i x ξ} \cdot [\frac{1}{h} \cdot (e^{- 2 π i x h} - 1) + 2 π i x] d x | \\ \leq \int_{| x | \geq N} | f (x) e^{- 2 π i x ξ} \cdot [\frac{1}{h} \cdot (e^{- 2 π i x h} - 1)] | d x \\ + \int_{| x | \geq N} | f (x) e^{- 2 π i x ξ} \cdot 2 π i x | d x \\ + \int_{- N}^{N} | f (x) e^{- 2 π i x ξ} \cdot [\frac{1}{h} \cdot (e^{- 2 π i x h} - 1) + 2 π i x] | d x \\ < 2 π (1 + o (h)) ε + 2 π ε + 2 ‖ f ‖_{\infty} ε \end{aligned}

Hence finished the proof.

点这返回我的主页

点这返回题目目录

点这返回我的主页

点这返回题目目录