Positive harmonic function with harmonic reciprocal must be constant

Question

Let $f(z)$ be a positive harmonic function on the unit disk such that $\frac{1}{f(z)}$ is also harmonic. Show $f(z)$ must be constant.

Answer 1

Hint: $0=\Delta(f\cdot 1/f)=f\Delta(1/f)+2\langle \nabla f,\nabla(1/f) \rangle+\Delta(f)(1/f).$

Answer 2

Note that $$\frac{\partial^2}{\partial x^2}(\frac{1}{f})=\frac{\partial}{\partial x}(-\frac{f_x}{f^2})=-\frac{f_{xx}}{f^2}+2\frac{f_x^2}{f^3}.$$ This implies that $$\tag{1}\Delta (1/f)=(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2})(\frac{1}{f})=-\frac{f_{xx}+f_{yy}}{f^2}+2\frac{f_x^2+f_y^2}{f^3} =-\frac{\Delta f}{f^2}+2\frac{f_x^2+f_y^2}{f^3}.$$ Since $f$ and $1/f$ are both harmonic, $\Delta (1/f)=\Delta f=0$, which implies by $(1)$ that $$2\frac{f_x^2+f_y^2}{f^3}=0,$$ or equivalently, $$f_x^2+f_y^2=0$$ which implies that $f$ is constant.

Answer 3

A different (I think more elegant) approach: $\displaystyle \pi r^2=\int_{B(0,r)}1=\int_{B(0,r)}\frac{\sqrt{f}}{\sqrt{f}}\le\bigg(\int_{B(0,r)}f\bigg)^{1/2}\bigg(\int_{B(0,r)}\frac{1}{f}\bigg)^{1/2}=\pi r^2\frac{\sqrt{f}}{\sqrt{f}}=\pi r^2\,,$ hence we must have equality in the Cauchy-Schwarz inequality so that $\displaystyle \sqrt{f}=\frac{c}{\sqrt{f}}\,,$ so $f=c\,.$