Two fun facts

Well, it seems that my 15 minutes are up, so let’s get back to some complex analysis. Here are two fun facts about holomorphic functions:

Theorem 1 Suppose {f\in {\mathcal H}(U)}, and {U} is connected. Then if {f} constant in a neighborhood of some point {z_0\in U}. Then {f} is constant on all of {U}.

Proof: Since the set {\{f(z_0)\}} is closed, so is it’s preimage. Let {V} denote this preimage. It suffices to show that {V} has no boundary. Then {V} is open, closed, and non-empty. Since {U} is connected, it follows that {U=V}.

Suppose {V} has a boundary, and let {v} be some point on the boundary. Expand a power series about {v}:

\displaystyle f(z)=a_0+a_1(z-v)+a_2(z-v)^2+\cdots.

Since {f} is holomorphic at {v}, this power series must have a non-zero radius of convergence so it is valid on some part of the interior of {V}. As we know, the only power series with constant image is the constant series {f(z)=a_0}. But now this series is valid in a small ball about {v}. This contradicts the assumption that {v} was on the boundary of {V}, so {V} must be open, proving the theorem.

\Box

Theorem 2 Suppose {f\in{\mathcal H}(U)}, and {U} is connected. Suppose we have a sequence of points {z_1,z_2,\dots} converging to {z_\infty\in U} such that {f(z_1)=f(z_2)=\cdots}. Then {f} is constant on {U}

Proof: By continuity of {f}, {f(z_1)=f(z_2)=\cdots=f(z_\infty)}. Expanding {f} as a power series about {z_\infty}, write

\displaystyle f(z)=a_0+a_1(z-z_\infty)+a_2(z-z_\infty)^2+\cdots

Since we have a sequence of points converging to {z_\infty} whose {f}-values all agree with {z_\infty}, we see that the derivative at {z_\infty} must be

\displaystyle f'(z_\infty)=\lim_{n\rightarrow\infty}\frac{f(z_n)-f(z_\infty)}{z_n-z_\infty}=\lim_{n\rightarrow\infty}\frac{0}{z_n-z_\infty}=0

Thus, {f'(z)=0} in a neighborhood of {z_\infty}, meaning that the power series: {f'(z)=a_1+2a_2(z-z_\infty)+3a_3(z-z_\infty)^2+\cdots} must in fact be constant. Thus, {a_n=0} for {n>0}, meaning that {f(z)=a_0} in a neighborhood of {z_\infty}.

Now we can apply the previous theorem to conclude that {f} is in fact constant of all of {U}.

\Box

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One Response to Two fun facts

  1. Pingback: Open mapping theorem « Andy Soffer

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