Pointwise convergence of Fourier transforms (Q689681)

Let \(f(x)\) be a complex-valued function, \(f\in L^ 1(R)\), and \(F(t)\) denotes the Fourier transform of \(f(x)\). The following symmetric partial integral and auxiliary expression are considered: \[ \begin{aligned} \delta_ \nu(f,x) & = \int^ \nu_{-\nu} F(t)e^{-ixt}dt,\quad x\in R,\;\nu\in R^ +;\\ I_ \nu(f,\lambda,x) & = {\lambda\over \lambda-1} \int_{\nu\leq| t|\leq \lambda\nu}\left(1-{| t|\over \lambda\nu}\right) F(t)e^{-ixt} dt,\quad\nu>0,\;\lambda>1,\;x\in R.\end{aligned} \] Main results: Theorem 1. If \(f\in L^ 1\), \(x\) is a Lebesgue point for \(f\), then (1) \(\lim_{\nu\to\infty} S_ \nu(f,x)= f(x)\) if and only if (2) \(\lim_{\lambda\downarrow 1}\limsup_{\nu\to\infty} I_ \nu(f,\lambda,x)= 0\). Theorem 2. Let \(r\in N\). If \(f(x)\) and \(x^ r f(x)\) both belong to \(L^ 1\), \(x\neq 0\) is a Lebesgue point for \(f\), and (3) \(\lim_{\lambda\downarrow 1}\limsup_{\nu\to\infty} \int_{\nu\leq | t|\leq \lambda\nu} | F^{(r)}(t)| dt\) \(=0\), then (1) holds. Two sufficient conditions for the fulfillment of (3) are given.