- Suppose that \(u : \mathbb { C } \to \mathbb { R }\) is a harmonic function such that
\[ u ( z ) \leq 1 0 \log ( | z | + 2 ) , \]
for all \(z \in \mathbb { C }\) . Prove that u is constant.
Solution
1. The harmonic function is the real part of an entire function.
The plane is simply connected, so \(u\) has a globally defined harmonic conjugate \(v\); consequently \(H=u+iv\) is entire [@SS03]. One explicit construction, writing \(z=x+iy\), is \[ v(x,y)=-\int_0^x u_y(t,0)\,dt+\int_0^y u_x(x,s)\,ds. \] Indeed, \(v_y=u_x\), while harmonicity gives \[ v_x=-u_y(x,0)+\int_0^y u_{xx}(x,s)\,ds =-u_y(x,0)-\int_0^y u_{yy}(x,s)\,ds=-u_y(x,y). \] Thus the Cauchy–Riemann equations hold everywhere and \(\operatorname{Re}H=u\).
2. The exponential of \(H\) is a polynomial.
The entire function \(F=e^H\) satisfies \[ |F(z)|=e^{u(z)}\leq e^{10\log(|z|+2)}=(|z|+2)^{10}. \] Write its entire Taylor expansion as \(F(z)=\sum_{k\geq0}a_kz^k\). Cauchy’s coefficient estimate on \(|z|=R\) gives \[ |a_k|\leq\frac{(R+2)^{10}}{R^k} \] [@SS03]. For every integer \(k>10\), the right-hand side tends to zero as \(R\to\infty\). Hence \(a_k=0\) for all such \(k\), so \(F\) is a polynomial of degree at most ten.
3. This polynomial is constant, and so is \(u\).
An exponential has no zero. By the fundamental theorem of algebra, a nonconstant complex polynomial has a zero [@SS03]. The polynomial \(F=e^H\) from step 2 must therefore be a nonzero constant. Differentiating gives \(0=F'=H'e^H\), and nonvanishing of \(e^H\) implies \(H'=0\). Thus \(H\) is constant on the connected plane, and its real part \(u\) is constant as well.