Let \((u_n(x,y))_{n\geq1}\) be a sequence of functions that are defined and harmonic for \((x,y)\) in an open neighborhood of the upper half plane \(\RR\times\RR_{\geq0}\). Suppose that \(\frac{\partial u_n}{\partial y}(x,0)=0\) for all \(x\in\RR\), and \(u_n(x,0)\) converges to \(0\) as \(n\to\infty\) uniformly for \(x\in\RR\). Must \(u_n(x,y)\to0\) as \(n\to\infty\) for every \((x,y)\in\RR\times\RR_{>0}\)?
Solution
No. Let \(u_n=\cosh(ny)\cos(nx)/n\). Since \(u_n\) is the real part of the holomorphic function \(\cos(nz)/n\), it is harmonic on the entire plane. Then \(\frac{\partial u_n}{\partial y}(x,0)=-\sinh(0)\cos(nx)=0\), and \(u_n(x,0)=\cos(nx)/n\to0\) as \(n\to\infty\) uniformly for \(x\in\RR\). But \(u_n(0,1)=\cosh(n)/n\) does not tend to \(0\) as \(n\to\infty\).