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?
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.
