# Continuous derivative of the Fourier transform of a function

Define $L_{\mathrm{bc}}^{1}\left(\mathbb{R}\right)$ to be the set of functions $f:\mathbb{R}\rightarrow\mathbb{C}$ which are continuous, bounded, and satisfy
${\displaystyle \left\Vert f\right\Vert _{1}=\int_{-\infty}^{\infty}\left|f\right|<\infty}$
Then given $f\in L_{\mathrm{bc}}^{1}\left(\mathbb{R}\right)$, define its Fourier transform $\hat{f}$ to be
${\displaystyle \hat{f}\left(y\right)=\int_{-\infty}^{\infty}f\left(x\right)e^{-2\pi ixy}dx}$
In my readings of the Fourier transform I encountered details about its derivative. In particular, if $g\left(x\right)=-2\pi ixf\left(x\right) \in L_{\mathrm{bc}}^{1}\left(\mathbb{R}\right)$, then $\hat{f}$ is continuously differentiable and $\hat{f}\,^{\prime}=\hat{g}$. The proof I read of it left much to be desired, or at least much was left to the reader. I ended up putting considerable effort into a complete proof, using various resources from internet searches, and I thought I would give back with this proof itself. Link to proof: Continuous derivative of the Fourier transform of a function