Problem 7

Exercise 11.7 (Product of real and imaginary part Liouville)   Let $f=u+iv$ be an entire function such that $\vert u\vert\vert v\vert$ is bounded. Prove that $f$ must be a constant function.

Proof. Expand

$$-if^2 = -i(u^2 + 2iuv - v^2) = 2uv + i(v^2-u^2)$$ (11.35)

If $\sup \vert u\vert\vert v\vert = M$, then the entire function $e^{-if^2}$ has a bound:

$$\vert e^{-if^2}\vert=\vert e^{2uv + i(v^2-u^2)}\vert = e^{2uv} \leq e^{2\vert uv\vert}=e^{2\vert u\vert\vert v\vert}\leq e^{2M}.$$ (11.36)

Therefore $e^{-if^2}$ is constant, so that its absolute value $e^{2uv}$ is also constant, which indicates $uv\equiv \pm M$. If we set $u=\pm M/v$, then the Cauchy-Riemann equations for $f$ may be applied:

$$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$$ (11.37)

$$\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$$ (11.38)

In particular, the equations yield

$$\frac{\partial u}{\partial x} = \mp \frac{M}{v^2} \frac{\partial v}{\partial x} = \frac{\partial v}{\partial y}$$ (11.39)

$$\frac{\partial u}{\partial y} = \mp \frac{M}{v^2} \frac{\partial v}{\partial y} = -\frac{\partial v}{\partial x}$$ (11.40)

Substituting the right half of the first equation into the right half of the second equation shows

$$\frac{M^2}{v^2}\frac{\partial v}{\partial x}= -\frac{\partial v}{\partial x} \iff \left(\frac{M^2}{v^2} + 1\right) \frac{\partial v}{\partial x}=0$$ (11.41)

Since $M^2/v^2+1>0$, this implies the partial derivative equals zero. The symmetry of the equations dictates that each partial derivative $\{u_x, u_y, v_x, v_y\}$ vanishes, so that $u$ and $v$ are both constant, proving that $f$ is constant. $\qedsymbol$