Use the preceding two exercises to prove that a uniform limit of harmonic functions on a domain \(U\subset\mathbb C\) is harmonic.
Solution
Let \(u_n\) be harmonic on \(U\) and suppose \(u_n\to u\) uniformly. We show that every point of \(U\) has a disk on which \(u\) is the real part of a holomorphic function.
1. Fix \(z_0\in U\) and choose \(R>0\) such that\[ \overline{D(z_0,R)}\subset U. \]For each \(n\), choose a harmonic conjugate \(v_n\) of \(u_n\) on\(D(z_0,R)\), normalized by\[ v_n(z_0)=0, \]and set\[ F_n=u_n+i v_n. \]
The disk \(D(z_0,R)\) is simply connected. Every harmonic function on a simply connected planar domain has a harmonic conjugate, unique up to an additive real constant. Choosing that constant so that \(v_n(z_0)=0\) makes \(F_n\) holomorphic on the disk.
2. Let\[ D_1=D(z_0,R/2). \]For \(n,m\) define\[ c_{nm}=u_n(z_0)-u_m(z_0) \]and\[ G_{nm}=F_n-F_m-c_{nm}. \]Then \(G_{nm}\) is holomorphic on \(D(z_0,R)\) and satisfies\(G_{nm}(z_0)=0\).
1. At \(z_0\) the imaginary parts of\(F_n\) and \(F_m\) both vanish, while their real-part difference is exactly\(c_{nm}\). Hence\[ G_{nm}(z_0)=0. \]
3. By E-HM213A17-36,\[ \int_{D_1}|\operatorname{Im}G_{nm}|^2\,|dz|^2 = \int_{D_1}|\operatorname{Re}G_{nm}|^2\,|dz|^2. \]Moreover, both sides tend to \(0\) as \(n,m\to\infty\).
After translating \(z_0\) to \(0\) and rescaling \(D_1\) to the unit disk, E-HM213A17-36 applies to \(G_{nm}\) because \(G_{nm}(z_0)=0\).
Set \[ \varepsilon_{nm} =\sup_{z\in D_1}|u_n(z)-u_m(z)|. \] Uniform convergence of \(u_n\) implies \(\varepsilon_{nm}\to0\). Since \(|c_{nm}|\le\varepsilon_{nm}\), \[ |\operatorname{Re}G_{nm}(z)| =|(u_n-u_m)(z)-c_{nm}| \le2\varepsilon_{nm} \] on \(D_1\). Therefore \[ \int_{D_1}|\operatorname{Re}G_{nm}|^2\,|dz|^2 \le4\,\operatorname{Area}(D_1)\,\varepsilon_{nm}^2 \longrightarrow0. \] The equality of the two integrals gives the same conclusion for the imaginary part.
4. Consequently\[ \int_{D_1}|G_{nm}(z)|\,|dz|^2\longrightarrow0. \]
By 3, \[ \int_{D_1}|G_{nm}|^2\,|dz|^2 = \int_{D_1} \left(|\operatorname{Re}G_{nm}|^2 +|\operatorname{Im}G_{nm}|^2\right)|dz|^2 \longrightarrow0. \] Cauchy–Schwarz then gives \[ \int_{D_1}|G_{nm}|\,|dz|^2 \le \operatorname{Area}(D_1)^{1/2} \left(\int_{D_1}|G_{nm}|^2\,|dz|^2\right)^{1/2}, \] which tends to zero.
5. On the smaller disk\[ D_2=D(z_0,R/4), \]one has\[ \sup_{D_2}|G_{nm}|\longrightarrow0. \]
Apply the pointwise bound of E-HM213A17-37 to the holomorphic function \(G_{nm}\) on \(D_1\). It gives \[ |G_{nm}(z)| \le \frac{1}{\pi\,d(z,\partial D_1)^2} \int_{D_1}|G_{nm}(w)|\,|dw|^2. \] For \(z\in D_2\), \[ d(z,\partial D_1)\ge R/4. \] The integral tends to zero by 4, so the displayed bound tends to zero uniformly on \(D_2\).
6. The sequence \(F_n\) is uniformly Cauchy on \(D_2\).
Since \[ F_n-F_m=G_{nm}+c_{nm}, \] 5 gives uniform convergence of the first term to zero, while \[ |c_{nm}| \le\sup_U|u_n-u_m|\longrightarrow0 \] because \(u_n\) converges uniformly. Hence \[ \sup_{D_2}|F_n-F_m|\longrightarrow0. \]
7. The limit \(u\) is harmonic on \(D_2\), and therefore on all of \(U\).
By 6, \(F_n\) converges uniformly on \(D_2\) to a function \(F\). A uniform limit of holomorphic functions is holomorphic, so \(F\) is holomorphic on \(D_2\). Since \[ \operatorname{Re}F_n=u_n\longrightarrow u, \] we have \[ \operatorname{Re}F=u \] on \(D_2\). The real part of a holomorphic function is harmonic, hence \(u\) is harmonic near \(z_0\). Since \(z_0\in U\) was arbitrary, \(u\) is harmonic throughout \(U\).