Suppose \(f : \mathbb { C } \to \mathbb { C }\) is entire and bounded. Use Cauchy’s formula to prove that \(f ^ { \prime }\) is identically zero and hence that \(f\) is constant. This is Liouville’s Theorem.
Solution
Choose \(M\geq0\) such that \[ \abs{f(z)}\leq M \] for every \(z\in\CC\).
1. Fix \(z_0\in\CC\). For every \(R>0\),\[ f'(z_0) = \frac{1}{2\pi i} \int_{\abs{\zeta-z_0}=R} \frac{f(\zeta)}{(\zeta-z_0)^2} \,d\zeta. \]
Because \(f\) is entire, it is analytic on and inside every circle centered at \(z_0\). Cauchy’s formula for the first derivative gives the identity.
2. For every \(R>0\),\[ \abs{f'(z_0)}\leq\frac{M}{R}. \]
On \(\abs{\zeta-z_0}=R\), one has \(\abs{f(\zeta)}\leq M\). The circle has length \(2\pi R\), so step 1 gives \[ \begin{aligned} \abs{f'(z_0)} &\leq \frac{1}{2\pi}(2\pi R)\frac{M}{R^2}\\ &= \frac{M}{R}. \end{aligned} \]
3. One has \(f'(z_0)=0\).
Step 2 holds for every \(R>0\). Letting \(R\to\infty\) gives \(\abs{f'(z_0)}=0\).
4. The derivative \(f'\) is identically zero on \(\CC\).
The point \(z_0\in\CC\) was arbitrary, so step 3 applies at every point.
5. The function \(f\) is constant.
Let \(z_1,z_2\in\CC\) and define \[ h(t)=f\bigl(z_1+t(z_2-z_1)\bigr), \qquad 0\leq t\leq1. \] By the chain rule and step 4, \[ h'(t) = f'\bigl(z_1+t(z_2-z_1)\bigr)(z_2-z_1) = 0. \] Thus \(h\) is constant on \([0,1]\), and therefore \[ f(z_2)=h(1)=h(0)=f(z_1). \] Since \(z_1\) and \(z_2\) were arbitrary, \(f\) is constant.
6. Q.E.D.
Steps 4 and 5 give the two requested conclusions.