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Let $f\in L_1(\mathbb{R})$ (That is to say $f$ is absolutely integrable over $\mathbb{R}$) with derivative $f'\in L_1(\mathbb{R})$. The Fourier transform of $f$ is given by: \begin{align} \hat{f}(t) = \int_{-\infty}^{\infty} f(x) e^{itx} \text{d}x. \end{align} I want to prove that the Fourier transform of $f'$ is given by: \begin{align} \widehat{f'}(t) = -it \hat{f}(t). \end{align} Here is my own way to give a proof: By definition, we have \begin{align} \widehat{f'}(t) = \int_{-\infty}^{\infty} f'(x) e^{itx} \text{ d}x \end{align}. By parts, we get \begin{align} \widehat{f'}(t) = \left[f(x)e^{itx} \right]_{-\infty}^{\infty} - it \int_{-\infty}^{\infty} f(x) e^{itx} \text{ d}x. \end{align} I wonder why the first term in the preceding equality vanishes? This is my problem. I appreciate any help.

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    $\begingroup$ I believe you need $f'$ to be integrable. $\endgroup$
    – copper.hat
    Mar 29, 2019 at 16:59
  • $\begingroup$ $f'$ is integrable. I forgot writing this condition. $\endgroup$ Mar 29, 2019 at 18:59

2 Answers 2

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We must suppose that $f$ is absolutely continuous for the derivative to really have any meaning. In that case, the fundamental theorem of calculus gives $f(x) = f(0) + \int_0^x f'(t)\,dt$. Since $f'$ is in $L^1$, then as $x \to \infty$, dominated convergence implies that $\int_0^x f'(t)\,dt \to \int_0^\infty f'(t)\,dt$; in particular the limit exists. Thus $\lim_{x \to \infty} f(x)$ exists. Now if this limit equals anything other than zero, $f$ would not be integrable. So we have $\lim_{x \to \infty} f(x) = 0$, and a similar argument gives $\lim_{x \to -\infty} f(x) =0$ as well. Since $e^{itx}$ is bounded, this shows that the term in question does indeed vanish.

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  • $\begingroup$ Ok, I actually proved that the limit $\lim_{x\to +\infty} f(x)$ exists by the integrability of $f'$. But how can one use the integrability of $f$ to show that the limit must be zero?. This is not clear for me. Anyway, i really appreciate your contribution. $\endgroup$ Mar 29, 2019 at 23:53
  • $\begingroup$ @HusseinEid: Try drawing a picture of a function with $\lim_{x \to \infty} f(x) = c \ne 0$. It should be immediately clear that $\int_{-\infty}^\infty |f| = \infty$, and it should take only a little bit more work to prove it. $\endgroup$ Mar 29, 2019 at 23:56
  • $\begingroup$ Ok, i will try that. $\endgroup$ Mar 30, 2019 at 12:14
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See:

Fourier Transform of Derivative

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