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Theorem: derivative of the Fourier transform

  1. Jun 19, 2010 #1
    Hello, I'm dealing with the proof of the theorem below:

    [itex]x(t) \in L^1(\mathbb{R}), tx(t) \in L^1(\mathbb{R}) \Rightarrow \mathfrak{F}[x(t)] \in C^1(\mathbb{R})[/itex]and[itex]\frac{\mathrm{d} }{\mathrm{d} \omega}\mathfrak{F}[x(t)](\omega) = \mathfrak{F}[-jtx(t)][/itex]


    I'm going to write down an interesting proof that I found, which is based on Lebesgue's dominated convergence theorem:

    Let [itex]X(\omega)[/itex] be the Fourier transform of x(t). Let's consider average rate of change of X:
    [itex]\frac{X(\omega + \Delta \omega) - X ( \omega)}{\Delta \omega} = \int_{-\infty}^{+\infty}\frac{x(t) e^{-j(\omega + \Delta \omega)t}}{\Delta \omega}dt - \int_{-\infty}^{+\infty}\frac{x(t) e^{-j(\omega)t}}{\Delta \omega}dt = \int_{-\infty}^{+\infty}\frac{x(t) e^{-j(\omega + \Delta \omega)t}}{\Delta \omega} - \frac{x(t) e^{-j(\omega)t}}{\Delta \omega}dt =[/itex]

    [itex]=\int_{-\infty}^{+\infty} x(t) e^{-j \omega t} [ \frac{e^{-j \Delta \omega t} - 1} { \Delta \omega }] dt[/itex]

    Now there's a step I don't understand:

    [itex]\forall \Delta \omega[/itex] we have:
    [itex]|x(t) e^{-j\omega t} \frac{e^{-j \Delta \omega t} - 1}{\Delta \omega}|=|tx(t) \frac{e^{-j \Delta \omega t} - 1}{-jt\Delta \omega }| = |tx(t)||\frac{e^{-j \Delta \omega t} - 1}{-jt\Delta \omega }| \leq |tx(t)|[/itex]

    (The thesis follows by noticing that tx(t) is a summable majorant for the integrandus, thus we can pass the [itex]\Delta \omega \to 0[/itex] limit inside the integral).

    In particular, I don't understand the disequation:
    [itex]|\frac{e^{-j \Delta \omega t} - 1}{-jt\Delta \omega }| \leq 1[/itex] , for any [itex]\Delta \omega[/itex]
    What am I missing?

    Thanks in advance.
     
    Last edited: Jun 19, 2010
  2. jcsd
  3. Jun 19, 2010 #2

    mathman

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    I use i not j. The following is for all real x.

    |eix - 1|=|eix/2 - e-ix/2|=2|sin(x/2)|≤|x|
     
  4. Jun 19, 2010 #3
    Thanks a lot!!!
     
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