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.
The Fourier transform of the derivative of a function $f\in L_1(\mathbb{R})$
fourier analysisfourier transform
Best Answer
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.