When is a Fourier transform of a function continuous I have read about a following claim in a book but cannot prove it.
If a function $\psi(x)$ fulfills $\int dx|\psi(x)|^2|x|^2<\infty$, its Fourier transform $\phi(k)$ is continuous and nearly everywhere differentiable.
Can anyone explain this claim to me?
 A: You can show this directly. Suppose $\psi$ is as stated, and define
$$
    \hat{\psi}_n(s)=\frac{1}{\sqrt{2\pi}}\int_{-n}^{n}\psi(x)e^{-isx}dx.
$$
The function $\hat{\psi}_n$ is infinitely differentiable, and $\{ \hat{\psi_n} \}_{n=1}^{\infty}$ converges uniformly to $\hat{\psi}$ as $n\rightarrow\infty$ because
\begin{align}
 |\hat{\psi}_n(s)-\hat{\psi}(s)| 
  & \le \frac{1}{\sqrt{2\pi}}\int_{|x|\ge n}|\psi(x)|dx \\
  & \le \frac{1}{\sqrt{2\pi}}\left(\int_{|x|\ge n}|x\psi(x)|^2dx\right)^{1/2}
        \left(\int_{|x|\ge n}\frac{1}{x^2}dx\right)^{1/2} \\
  & \rightarrow 0 \mbox{ as } n\rightarrow\infty.
\end{align}
And $\{ \hat{\psi}_n ' \}$ converges in $L^2$ because
$$
     \hat{\psi_{n}}'(s) = \frac{1}{\sqrt{2\pi}}\int_{|x|\le n}e^{-isx}(-ix)\psi(x)dx,
$$
and the Parseval identity gives
$$
       \|\hat{\psi_n}'- \widehat{(-ix\psi)}\|=\int_{|x|\ge n}|x\psi(x)|^2dx
      \rightarrow 0 \mbox{ as } n\rightarrow \infty.
$$
Therefore,
\begin{align}
      \int_{0}^{x}\widehat{(-ix\psi)}(s)ds &= \lim_n \int_{0}^{x}\hat{\psi_n}'(s)ds \\
   & =\lim_n \hat{\psi_n}(x)-\hat{\psi_n}(0) \\
   & =\hat{\psi}(x)-\hat{\psi}(0).
\end{align}
This proves that $\hat{\psi}$ is an absolutely continuous function on every finite interval because it is the integral of $\widehat{(-ix\psi)} \in L^2(\mathbb{R})$. It follows that $\hat{\psi}$ has a derivative a.e. on $\mathbb{R}$, with $\widehat{\psi}'=\widehat{(-ix\psi)}$ a.e., and $\hat{\psi}$ is the the integral of its derivative.
