Complex Analysis/Identity theorem, Liouville-type theorems, Riemann's theorem

After now having established the main tools of complex analysis, we may deduce the first corollaries from them, which are theorems about general holomorphic functions; for, it is the case that holomorphic functions follow a wide range of rules just by their holomorphy, and this and the following chapter cover some of them.

Connectedness and the identity theorem
For the famous identity theorem, we first need to frame the intuitively plausible notion of a connected set in mathematically precise terms.

It seems somewhat implausible that this is actually the "right" definition of connectedness. The following theorem shows that in fact, if the topological space $$X$$ equals $$\mathbb R^n$$ (and thus in particular if $$X = \mathbb R^2 = \mathbb C$$) and the subset is open, then there is a very intuitive characterisation of connectedness.

Proof:

Assume first that $$O$$ is path-connected and not connected. We first prove the following claim, which is actually an important alternative characterisation of connectedness:

Claim: A topological space $$X$$ is connected if and only if there do not exist open $$U, V \subset X$$ such that $$U \neq \emptyset$$, $$V \neq \emptyset$$, $$U \cap V = \emptyset$$ and $$U \cup V = X$$ (that is, we may not split up $$X$$ into two nonempty open disjoint subsets).

Proof of claim: Suppose first that $$X$$ is connected, and assume for a contradiction that we can write it as $$X = V \cup U$$ with $$U, V$$ nonempty, open, disjoint. Then $$V = X \setminus U$$ is open, closed, nonempty and not the whole space, contradiction. Let now $$X$$ not be connected. We pick a nontrivial open and closed set $$S \subseteq X$$, and $$X = S \cup (X \setminus S)$$ contradicts the condition of the claim.

Further, we have the following:

Claim: If $$X$$ is connected and $$f: X \to Y$$ is a continuous function between topological spaces, then $$f(X)$$ is connected.

Proof of claim: Assume otherwise. Then there exists $$S \subset f(X)$$ open, closed, nontrivial. By the continuity of $$f$$, $$f^{-1}(S)$$ is open, closed, nontrivial, contradiction.

Claim: The set $$[0, 1]$$ is connected.

Proof:

Let now $$O$$ be connected, and let $$x$$ be an arbitrary point within $$O$$ ($$O$$ was assumed to be nonempty). We define
 * $$S := \{y \in O | \text{there exists a path from } x \text{ to } y\}$$.

This set is certainly open, because if there exists a path from $$x$$ to $$y$$ and $$B_\epsilon(y)$$ is any ball around $$y$$ contained in $$O$$, then in fact $$B_\epsilon(y) \subseteq S$$ since we may connect $$z \in B_\epsilon(y)$$ to $$y$$ using a straight line and $$y$$ to $$x$$ using the path we have. But it is also closed, since the complement is open for the same reason: Pick $$w \in O \setminus S$$ and $$\eta > 0$$ such that $$B_\eta(w) \subseteq O$$; if any point in $$B_\eta(w)$$ is in $$S$$, then, since $$B_\eta(w)$$ is convex, the previous argument applies and $$w \in S$$, contradiction. Since $$O$$ is connected and $$S$$ nonempty ($$x \in S$$), $$S = O$$ and thus, by connecting any two points $$a, b$$ by first connecting $$a$$ and $$x$$ and then $$x$$ and $$b$$, $$O$$ is path-connected.

Now we can state and prove the identity theorem.

Proof:

Riemann's theorem
Proof:

Consider the function
 * $$g(z) := (z - z_0)^2 f(z)$$.

We claim that $$g$$ is holomorphic in $$O$$. Indeed, for $$z \neq z_0$$ it is differentiable by the product rule, and for $$z_0$$ we have
 * $$\lim_{h \to 0} \frac{(z_0 + h - z_0)^2 f(z_0 + h) - (z_0 - z_0)^2 f(z)}{h} = \lim_{h \to 0} h f(z_0 + h) = 0$$, i.e. $$g'(z_0) = 0$$.

Hence, we may write $$g$$ as a Taylor series
 * $$g(z) = \sum_{n=0}^\infty c_n (z - z_0)^n$$

where $$c_0, c_1 = 0$$ due to $$g(z_0) = 0$$ and $$g'(z_0) = 0$$. Thus,
 * $$g(z) = (z - z_0)^2 \sum_{n=2}^\infty c_n (z - z_0)^{n-2}$$,

and hence for $$z \neq z_0$$, $$z$$ in the domain of convergence of the power series, we have
 * $$f(z) = \sum_{n=2}^\infty c_n (z - z_0)^{n-2}$$,

where the latter expression makes sense also at $$z_0$$ and is complex differentiable there. This means that
 * $$\tilde f (z) = \begin{cases}

f(z) & z \neq z_0 \\ c_2 & z = z_0 \end{cases}$$ is the desired continuation of $$f$$ to $$z_0$$.

Exercises

 * 1) Prove that if $$f$$ is an entire function such that $$\forall z \in \mathbb C: |f(z)| < \frac{1}{|\operatorname{Im}(z)|}$$, then $$f$$ is constant.
 * 2) Prove that every uncountable subset of $$\mathbb R^2$$ as an accumulation point. Conclude that if for uncountably many $$z \in \mathbb C$$ we have $$f(z) = g(z)$$, then $$f \equiv g$$.