# Theorem: derivative of the Fourier transform

1. Jun 19, 2010

### eliotsbowe

Hello, I'm dealing with the proof of the theorem below:

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

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

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

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

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

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

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

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

Last edited: Jun 19, 2010
2. Jun 19, 2010

### mathman

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

|eix - 1|=|eix/2 - e-ix/2|=2|sin(x/2)|≤|x|

3. Jun 19, 2010

### eliotsbowe

Thanks a lot!!!