Liouville’s theorem

We can use the Cauchy estimates from last time to prove a very beautiful theorem:

Theorem 1 If {f\in{\mathcal H}({\mathbb C})} is bounded (i.e., {|f|<M} for some {M\in{\mathbb R}}), then {f} is constant.

Proof:Expand {f} as a power series at 0. We know that this representation is valid everywhere in {{\mathbb C}}. Write {f(z)=a_0+a_1z+a_2z^2+\cdots}. Recall that

\displaystyle a_n=\frac{f^{(n)}(0)}{n!}.

The Cauchy estimates tell us that

\displaystyle |a_n|=\frac{f^{(n)}(0)}{n!}\le\frac{M}{r^n}

for every positive {r} less than the radius of convergence of {f}. Of course, the radius of convergence is infinite, so this estimate is valid for all {r>0}. Thus, for {n>0}, {|a_n|} can be bounded by arbitrarily small positive numbers. That is, {a_n=0} for all {n>0}. In other words, {f(z)=a_0}, a constant. \Box

It’s worth mentioning that trigonometric functions such as {\sin} and {\cos} are holomorphic on all of {{\mathbb C}} (often called entire) functions. Hopefully this doesn’t bother you, even in light of Liouville’s theorem. If it does, you’re remembering that {\sin x} is always between {1} and {-1}. You are correct for {x\in{\mathbb R}}, but Liouville’s theorem doesn’t say anything about functions on {{\mathbb R}}. Indeed, if you plug in complex numbers, you’ll see that {\sin z} can be arbitrarily large. In fact, look at {\sin(ix)} for {x\in{\mathbb R}}. You can use Wolfram Alpha to see that on the imaginary axis, {\sin z} takes on arbitrarily large values.

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