Find all harmonic functions \(u:\DD\to \CC\) such that \(u(1/2) = 2\) and \(\abs{u(z)}\geq 2\) for all \(\abs{z} \leq 1\).
For complex-valued \(u\) the conditions cut out no family simpler than the conditions themselves, which is why the solution treats real-valued \(u\). For every real harmonic \(h\) on \(\DD\) with \(h(1/2)=0\), \(u=2+ih\) satisfies both, since \(\abs{u}^2=4+h^2\). The real part need not be constant either: for \(0<\eps\le1\) and \(K^2\ge4\eps\), the harmonic function \(u=2+\eps\qty{(x-\tfrac12)^2-y^2}+iKy\) has \(u(1/2)=2\) and \(\Re u\ge2-\eps y^2\ge1\) on \(\overline\DD\), so \(\abs{u}^2\ge(2-\eps y^2)^2+K^2y^2\ge4+(K^2-4\eps)y^2\ge4\).
Solution
Assume \(u\) is real-valued; the remark treats complex-valued \(u\). Since \(\DD\) is connected, \(u\) is continuous and \(\abs u\ge2\), either \(u\ge2\) on \(\DD\) or \(u\le-2\) on \(\DD\). As \(u(1/2)=2\), \(u\geq 2\) on \(\DD\), so \(u\) attains its minimum at the interior point \(1/2\). By the minimum principle for harmonic functions, \(u\equiv2\). The only such function is the constant \(\boxed{u\equiv2}\).