On the eigenfunction expansions associated with the Schrödinger operator having spherically symmetrical potential
Let \(\Omega\) be a bounded domain in \({\mathbb{R}}^ 3\), \(x_ 0\in \Omega\) and \(0<q(r)\) a function on ]0,\(\infty [\) such that \(\int^{\infty}_{0}q^ 2(r)r^ 2dr<\infty.\) Consider the Friedrichs extension \(\hat L\) of \(L=-\Delta +q(| \cdot -x_ 0|),\) \(D(L)=C_ 0^{\infty}(\Omega)\), having discrete spectrum. Let \(0<\lambda_ 1\leq \lambda_ 2\leq..\). be the sequence of eigenvalues of \(\hat L\) and \((u_ n)_{n\in {\mathbb{N}}}\) the corresponding complete orthonormal system of eigenfunctions, i.e. \(\hat Lu_ n=\lambda u_ n\), \(<u_ n,u>=\delta_ n.\) The expansion \((*)\quad f=\sum^{\infty}_{n=1}<u_ n,f>u_ n\) is said to be Lebesgue summable to a at the point \(x\in \Omega\) if \[ \lim_{| x| \downarrow 0}\sum^{\infty}_{n=1}([\sin (\sqrt{\lambda_ n}\cdot | x|)]/(\sqrt{\lambda_ n}\cdot | x|))<u_ n,f>u_ n(x)=a \] holds true. The result of the present paper states that the expansion (*) of a function \(f\in W^ 1_ 2(\Omega)\) is Lebesgue summable at the point \(x_ 0\) if and only if \(\lim_{r\downarrow 0}\int_{\theta}f(x_ 0+r\theta)d\theta\) exists. Here \(f(x_ 0+r\theta)d\theta\) denotes the integral of f over the sphere of radius r with respect to the normalized Lebesgue measure, i.e. \(\int_{\theta}d\theta =1\).
complete orthonormal system of eigenfunctions
scientific article; zbMATH DE number 3886460