- Let \(f _ { n } : D \to ( - \infty , 1 ) , n = 1 , 2 , . . . ,\) be an increasing sequence of harmonic functions on the unit disk D such that \(f _ { n } ( 0 ) \to 1\) as \(n \to \infty\) . (I.e., \(f _ { n } ( z ) \leq f _ { n + 1 } ( z ) < 1 , \forall n \geq 1 . )\) Prove that \(f _ { n } ( z ) \to 1\) as \(n \to \infty\) , for all \(z \in D\)
Solution
Put \(u_n=1-f_n\). Each \(u_n\) is positive and harmonic on \(D=\{|z|<1\}\), and \(u_n(0)\to0\).
1. Values of a positive harmonic function on an interior disk are bounded by its central value.
Let \(u>0\) be harmonic on \(D\), and fix \(0\leq r<R<1\). The Poisson representation on the radius-\(R\) disk gives, for \(|z|\leq r\), \[ u(z)=\frac1{2\pi}\int_0^{2\pi} \frac{R^2-|z|^2}{|Re^{it}-z|^2}u(Re^{it})\,dt \] [@SS03]. It applies since \(u\) is harmonic on a neighborhood of this closed disk. The kernel satisfies \[ 0<\frac{R^2-|z|^2}{|Re^{it}-z|^2} \leq\frac{R+|z|}{R-|z|}\leq\frac{R+r}{R-r}. \] Using positivity of the boundary values and the mean-value identity at zero therefore gives \[ 0<u(z)\leq\frac{R+r}{R-r}\frac1{2\pi} \int_0^{2\pi}u(Re^{it})\,dt =\frac{R+r}{R-r}u(0). \]
2. Apply the common bound to the sequence.
For fixed \(r<1\), choose any \(R\) with \(r<R<1\). Step 1 applied separately to \(u_n\) yields \[ \sup_{|z|\leq r}|1-f_n(z)| =\sup_{|z|\leq r}u_n(z) \leq\frac{R+r}{R-r}u_n(0)\longrightarrow0. \] Every point of \(D\) lies in such a disk. Hence \(f_n(z)\to1\) for every \(z\in D\), as required; the estimate also proves uniform convergence on compact subsets. Monotonicity is compatible with the proof but is not needed once positivity of \(1-f_n\) and its central limit are known.