Let \(f_1,\ldots,f_n\) be analytic functions on the unit disk \(\Delta\) and suppose \[ \sum_{i=1}^n |f_i(z)|^2=1 \qquad(z\in\Delta). \]
Show that all \(f_i\) are constant by taking the Laplacian of both sides.
Show that all \(f_i\) are constant by applying the maximum principle, or the open mapping theorem, to suitable linear combinations of the functions.
Let \(H:\mathbb C^n\to\mathbb R\) be a strictly convex smooth function. Using method (a) or (b), prove that if \[ H(f_1(z),\ldots,f_n(z)) \] is constant on \(\Delta\), then each \(f_i\) is constant.
Solution
Write \[ F=(f_1,\ldots,f_n):\Delta\longrightarrow\mathbb C^n. \]
1. For every analytic function \(h\),\[ \Delta |h|^2=4|h'|^2. \]
Using \[ \Delta=4\partial_z\partial_{\overline z} \] and the holomorphicity of \(h\), \[ \partial_{\overline z}|h|^2 =h\,\overline{h'}, \] so \[ \partial_z\partial_{\overline z}|h|^2 =h'\overline{h'} =|h'|^2. \] Multiplying by \(4\) proves the formula.
2. Part (a): every \(f_i\) is constant.
Apply the Laplacian to \[ \sum_{i=1}^n|f_i|^2=1. \] By 1, \[ 0 =\Delta 1 =\sum_{i=1}^n\Delta|f_i|^2 =4\sum_{i=1}^n|f_i'|^2. \] Each summand is nonnegative, hence each \(|f_i'|^2\) vanishes identically. Thus every \(f_i'=0\), so every \(f_i\) is constant on the connected disk.
3. Part (b): the maximum principle gives the same conclusion.
Fix \(z_0\in\Delta\) and define \[ h(z)=\sum_{i=1}^n\overline{f_i(z_0)}\,f_i(z). \] This is holomorphic. By Cauchy–Schwarz and the hypothesis, \[ |h(z)| \le \left(\sum_i|f_i(z_0)|^2\right)^{1/2} \left(\sum_i|f_i(z)|^2\right)^{1/2} =1. \] At \(z=z_0\), \[ h(z_0)=\sum_i|f_i(z_0)|^2=1. \] Thus \(h\) attains its maximum modulus at an interior point. The maximum principle gives \(h\equiv1\).
Now \[ \begin{aligned} \|F(z)-F(z_0)\|^2 &=\|F(z)\|^2+\|F(z_0)\|^2 -2\operatorname{Re}\langle F(z),F(z_0)\rangle\\ &=1+1-2\operatorname{Re}h(z)\\ &=0. \end{aligned} \] Hence \(F(z)=F(z_0)\) for every \(z\), and all \(f_i\) are constant.
4. For part (c), fix \(z_0\in\Delta\) and put \(p=F(z_0)\).If \(dH_p=0\), then \(F\) is already constant.
For a differentiable convex function, \[ H(q)\ge H(p)+dH_p(q-p) \] for every \(q\in\mathbb C^n\cong\mathbb R^{2n}\). If \(dH_p=0\), then \(H(q)\ge H(p)\) for all \(q\), so \(p\) is a global minimizer. Strict convexity allows at most one global minimizer. Since \(H(F(z))=H(p)\) for every \(z\), we must have \(F(z)=p\) for every \(z\).
5. Assume now that \(dH_p\neq0\). On the level set\[ L=\{q:H(q)=H(p)\}, \]the real-linear functional \(dH_p\) has a strict maximum at \(p\).
Convexity gives \[ H(q)\ge H(p)+dH_p(q-p). \] If \(q\in L\), this says \[ dH_p(q-p)\le0. \] If equality held for some \(q\neq p\), then the graph of \(H\) would agree with its supporting affine function at both endpoints \(p\) and \(q\). Convexity would then force \(H\) to be affine on the segment joining them, contradicting strict convexity. Therefore \[ dH_p(q)<dH_p(p) \] for every \(q\in L\setminus\{p\}\).
6. Part (c): every \(f_i\) is constant.
Every real-linear functional on \(\mathbb C^n\) is the real part of a complex-linear functional. Hence choose a complex-linear \(\Lambda:\mathbb C^n\to\mathbb C\) such that \[ \operatorname{Re}\Lambda=dH_p. \] The function \[ g=\Lambda\circ F \] is holomorphic on \(\Delta\). Since \(F(\Delta)\subset L\), 5 gives \[ \operatorname{Re}g(z) \le \operatorname{Re}g(z_0) \] for every \(z\in\Delta\). The harmonic function \(\operatorname{Re}g\) therefore attains an interior maximum and is constant by the maximum principle for harmonic functions.
Thus equality holds in the supporting-hyperplane inequality for every \(F(z)\in L\). By the strictness proved in 5, this forces \[ F(z)=p \] for every \(z\). Hence each component \(f_i\) is constant.