Let \(D_n\) be the union of the open unit disks centered at \(0\) and at \(2-1/n\). Let \[ f_n:(\Delta,0)\longrightarrow(D_n,0) \] be the Riemann map normalized by \(f_n'(0)>0\). Prove that \[ f_n(z)\longrightarrow z \] uniformly on compact subsets of \(\Delta\). In particular, show that the image of the limiting map is smaller than the limit of the images of the \(f_n\).
Solution
The maps \(f_n\) converge locally uniformly to the identity on \(\Delta\).
1. The family \((f_n)\) is normal, and its derivatives at the origin satisfy\[ f_n'(0)\ge1. \]
Every \(D_n\) is contained in the disk \(B(0,3)\), so \[ |f_n(z)|<3 \qquad(z\in\Delta). \] Hence \((f_n)\) is a normal family.
Also the first unit disk is contained in every \(D_n\): \[ \Delta\subset D_n. \] Therefore the inverse Riemann map restricts to a holomorphic self-map \[ h_n:=f_n^{-1}|_\Delta:\Delta\to\Delta \] fixing \(0\). By the Schwarz lemma, \[ |h_n'(0)|\le1. \] Since \(h_n'(0)=1/f_n'(0)\) and \(f_n'(0)>0\), this gives \[ f_n'(0)\ge1. \]
2. Every locally uniformly convergent subsequence of \((f_n)\) converges to the identity.
Take any subsequence. By 1 it has a further subsequence, still denoted \((f_{n_k})\), converging locally uniformly to a holomorphic map \[ g:\Delta\to\mathbb C. \] Then \[ g(0)=0, \qquad g'(0)=\lim_{k\to\infty}f_{n_k}'(0)\ge1. \] In particular, \(g\) is nonconstant.
Set \[ K=\overline\Delta\cup\overline{B(2,1)}. \] For every \(w\in D_n\), either \(w\in\Delta\) or \(w\in B(2-1/n,1)\). In the latter case its distance from \(\overline{B(2,1)}\) is at most \(1/n\). Thus \[ \sup_{w\in D_n}\operatorname{dist}(w,K)\le\frac1n. \] For each fixed \(z\in\Delta\), the points \(f_{n_k}(z)\) therefore converge to a point of \(K\), so \[ g(\Delta)\subset K. \]
Because \(g\) is nonconstant, \(g(\Delta)\) is open and connected. But \[ \operatorname{int}K =\Delta\,\sqcup\,B(2,1), \] the disjoint union of the two open tangent disks. Since \(0\in g(\Delta)\), connectedness forces \[ g(\Delta)\subset\Delta. \] Thus \(g\) is a holomorphic self-map of \(\Delta\) fixing \(0\). The Schwarz lemma gives \[ |g'(0)|\le1. \] Combined with \(g'(0)\ge1\) and the positive normalization, this yields \[ g'(0)=1. \] The equality case of the Schwarz lemma now gives \[ g(z)=z. \]
3. The whole sequence converges to the identity locally uniformly.
If not, there would be a compact set \(L\Subset\Delta\), an \(\varepsilon>0\), and a subsequence such that \[ \sup_{z\in L}|f_n(z)-z|\ge\varepsilon \] along that subsequence. Normality would give a further locally uniformly convergent subsequence, which by 2 must converge to the identity, a contradiction.
4. The image of the limiting map is strictly smaller than the limiting images.
The limiting map is the identity, so its image is only \[ \Delta. \] On the other hand, \[ 2\in B(2-1/n,1)\subset D_n=f_n(\Delta) \] for every \(n\ge2\), because \(|2-(2-1/n)|=1/n<1\). Thus the fixed point \(2\) belongs to all sufficiently large images \(f_n(\Delta)\) but not to the image of the limiting map. In particular, \[ \operatorname{im}(\lim f_n)=\Delta \] is strictly smaller than the set-theoretic lower limit of the images, and hence than the geometric limit of the domains.