\(\mathrm {Exercise \ \oplus \ Problem } \ 1 \)  

 

\( \qquad \)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) \(\qquad \ \)——————Alina Lagrange

 

Prove that : \[ \frac{d \hat f (\xi) }{ d\xi } = \widehat {-2\pi i x f(x) }\]

 

\(\mathcal{P}roof. \)

Since \[ \frac{d \hat f (\xi) }{ d\xi }=\lim_{h\to 0 } \int_\mathbb R f(x)e^{- 2\pi i x \xi } \cdot \frac 1h \cdot( e^{-2\pi i x h}-1 ) dx \] as \( {h\to 0 } \) \[ \left | \frac{d \hat f (\xi) }{ d\xi }-(\widehat {-2\pi i x f(x) } ) \right | = \left | \int_\mathbb R f(x)e^{- 2\pi i x \xi } \cdot \left [ \frac 1h \cdot( e^{-2\pi i x h}-1 ) +2\pi i x \right ] dx \right | \] \( \forall \varepsilon >0 \), Since \(f(x),xf(x) \in \mathcal S (\mathbb R ) \),there exists \( N \) such that \( \int _{|x | \geq N } | f(x)| dx <\varepsilon \) and \( \int _{|x | \geq N } | xf(x)| dx <\varepsilon \)
And there exists \( h_m >0 \) such that when \( |h| < h_m \) , \[| \frac 1h \cdot( e^{-2\pi i x h}-1 )+ 2\pi i x| <\frac \varepsilon N \] Hence \[ \begin{aligned} \left | \frac{d \hat f (\xi) }{ d\xi }-(\widehat {-2\pi i x f(x) } ) \right | & { } = \left | \int_\mathbb R f(x)e^{- 2\pi i x \xi } \cdot \left [ \frac 1h \cdot( e^{-2\pi i x h}-1 )+2\pi i x \right ] dx \right | \\&= \left | \int_{|x| \geq N } f(x)e^{- 2\pi i x \xi } \cdot \left [ \frac 1h \cdot( e^{-2\pi i x h}-1 ) \right ] dx \right. \\& \ \quad \left . + \int_{|x| \geq N } f(x)e^{- 2\pi i x \xi } \cdot 2\pi i x dx\right. \\& \quad + \left . \int_{ -N}^N f(x)e^{- 2\pi i x \xi } \cdot \left [ \frac 1h \cdot( e^{-2\pi i x h}-1 ) +2\pi i x \right ] dx \right | { } \\ & \leq \int_{|x| \geq N } \left | f(x)e^{- 2\pi i x \xi } \cdot \left [ \frac 1h \cdot( e^{-2\pi i x h}-1 ) \right ] \right | dx \\& \ \quad + \int_{|x| \geq N } \left| f(x)e^{- 2\pi i x \xi } \cdot 2\pi i x \right| dx \\&\quad + \int_{ -N}^N \left| f(x)e^{- 2\pi i x \xi } \cdot \left [ \frac 1h \cdot( e^{-2\pi i x h}-1 ) +2\pi i x \right ] \right |dx \\& < 2\pi (1+o(h)) \varepsilon + 2\pi\varepsilon+ 2\|f \|_\infty \varepsilon \end{aligned}\] Hence finished the proof.

 

 


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