Continuity of derivative of Fourier transform Let $f, \hat{x}f \in L^{1}(\mathbb{R})$.
Show that $\mathcal{F}(f) \in \mathcal{C}^{1}(\mathbb{R})$ and $(\mathcal{F}(f))^{'} = \mathcal{F}(-i\hat{x}f)$, where
$$\mathcal{F}(f)(p) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty}e^{-ipx}f(x)dx$$
I am struggling with the first part. How do I show that $\mathcal{F}(f)(p)$ has indeed continuous derivative? Can I simply use Leibniz integral rule to differentiate under the integral sign?
If so, is the argument posted below correct?
$\forall p\in \mathbb{R}$
$$|\frac{d}{dp}(\mathcal{F}(f)(p))| = |\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-ipx}(-ix)f(x)dx| \leq \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}|xf(x)|dx = \frac{1}{\sqrt{2\pi}}\lVert \hat{x}f \rVert_{1}$$
Hence
$$\lVert \frac{d}{dp}(\mathcal{F}(f)(p)) \rVert_{\infty} \leq \frac{1}{\sqrt{2\pi}}\lVert \hat{x}f \rVert_{1}$$

You are in the good direction, but here you are only showing that $\mathcal{F}(f)$ has bounded derivative, not that this derivative is continuous. You can now use the fact that $\mathcal{F}(f)' = \mathcal{F}(-ixf)$ and the fact that the Fourier transform of $L^1$ functions is continuous to get that $\mathcal{F}(f)$ is continuous, and so $\mathcal{F}(f)∈ C^1$.
