Liouville's Theorem (Complex Analysis)

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Theorem

Let $f: \C \to \C$ be a bounded entire function.


Then $f$ is constant.


Proof

By assumption, there is $M \ge 0$ such that $\cmod {\map f z} \le M$ for all $z \in \C$.

For any $R \in \R: R > 0$, consider the function:

$\map {f_R} z := \map f {R z}$

Using the Cauchy Integral Formula, we see that:

$\displaystyle \cmod {\map {f_R'} z} = \frac 1 {2 \pi} \cmod {\int_{\map {C_1} z} \frac {\map f w} {\paren {w - z}^2} \rd w} \le \frac 1 {2 \pi} \int_{\map {C_1} z} M \rd w = M$

where $\map {C_1} z$ denotes the circle of radius $1$ around $z$.

Hence:

$\displaystyle \cmod {\map {f'} z} = \cmod {\map {f_R'} z} / R \le M / R$

Since $R$ was arbitrary, it follows that $\cmod {\map {f'} z} = 0$ for all $z \in \C$.

Thus $f$ is constant.

$\blacksquare$


Remark


In fact, the proof shows that, for a nonconstant entire function $f$, the maximum modulus $\displaystyle \map M {r, f} := \max_{\cmod z = r} \cmod {\map f z}$ grows at least linearly in $r$.


Source of Name

This entry was named for Joseph Liouville.


Sources