# The Fourier transform of the derivative of a function $f\in L_1(\mathbb{R})$

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.

• I believe you need $f'$ to be integrable. Mar 29 '19 at 16:59
• $f'$ is integrable. I forgot writing this condition. Mar 29 '19 at 18:59

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.
• 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. Mar 29 '19 at 23:53
• @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. Mar 29 '19 at 23:56