# 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.

Definition 7.1:

A topological space ${\displaystyle X}$  is called connected if and only if the only nonempty set which is simultaneously open and closed is ${\displaystyle X}$ .

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

Theorem 7.2:

Let ${\displaystyle X=\mathbb {R} ^{n}}$  with the standard Euclidean topology. Then any nonempty open subset ${\displaystyle O\subseteq X}$  is connected (w.r.t. to the subspace topology induced by the standard Euclidean topology) if and only if for all ${\displaystyle x,y\in O}$  there exists a continuous function ${\displaystyle \gamma :[0,1]\to O}$  such that ${\displaystyle \gamma (0)=x}$  and ${\displaystyle \gamma (1)=y}$  (such sets are also called path-connected, and the continuous map ${\displaystyle \gamma }$  is called a path).

Proof:

Assume first that ${\displaystyle O}$  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 ${\displaystyle X}$  is connected if and only if there do not exist open ${\displaystyle U,V\subset X}$  such that ${\displaystyle U\neq \emptyset }$ , ${\displaystyle V\neq \emptyset }$ , ${\displaystyle U\cap V=\emptyset }$  and ${\displaystyle U\cup V=X}$  (that is, we may not split up ${\displaystyle X}$  into two nonempty open disjoint subsets).

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

Further, we have the following:

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

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

Claim: The set ${\displaystyle [0,1]}$  is connected.

Proof:

Let now ${\displaystyle O}$  be connected, and let ${\displaystyle x}$  be an arbitrary point within ${\displaystyle O}$  (${\displaystyle O}$  was assumed to be nonempty). We define

${\displaystyle S:=\{y\in O|{\text{there exists a path from }}x{\text{ to }}y\}}$ .

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

Now we can state and prove the identity theorem.

Theorem 7.3:

An accumulation point of a sequence ${\displaystyle (z_{n})_{n\in \mathbb {N} }}$  of complex numbers is a point ${\displaystyle c\in \mathbb {C} }$  such that for all ${\displaystyle \epsilon >0}$ , ${\displaystyle B_{\epsilon }(c)}$  contains infinitely many distinct points

Let ${\displaystyle O\subseteq \mathbb {C} }$  be a connected set, and let ${\displaystyle f,g:O\to \mathbb {C} }$  be holomorphic. If there exists a sequence ${\displaystyle (z_{n})_{n\in \mathbb {N} }}$  that has an accumulation point such that ${\displaystyle \forall n\in \mathbb {N} :f(z_{n})=g(z_{n})}$ , then ${\displaystyle f\equiv g}$ .

Proof:

## Riemann's theorem

Theorem 7.6:

Let ${\displaystyle f:O\setminus \{z_{0}\}\to \mathbb {C} }$  be a holomorphic function, where ${\displaystyle z_{0}}$  is an interior point of ${\displaystyle O}$  and assume that ${\displaystyle f}$  is bounded in a neighbourhood of ${\displaystyle z_{0}}$ . Then there exists a holomorphic function ${\displaystyle {\tilde {f}}:O\to \mathbb {C} }$  which extends ${\displaystyle f}$ , that is,

${\displaystyle {\tilde {f}}\upharpoonright _{O\setminus \{z_{0}\}}=f}$ .

Thus, ${\displaystyle {\tilde {f}}}$  is an extension of ${\displaystyle f}$  to ${\displaystyle z_{0}}$ .

Proof:

Consider the function

${\displaystyle g(z):=(z-z_{0})^{2}f(z)}$ .

We claim that ${\displaystyle g}$  is holomorphic in ${\displaystyle O}$ . Indeed, for ${\displaystyle z\neq z_{0}}$  it is differentiable by the product rule, and for ${\displaystyle z_{0}}$  we have

${\displaystyle \lim _{h\to 0}{\frac {(z_{0}+h-z_{0})^{2}f(z_{0}+h)-(z_{0}-z_{0})^{2}f(z)}{h}}=\lim _{h\to 0}hf(z_{0}+h)=0}$ , i.e. ${\displaystyle g'(z_{0})=0}$ .

Hence, we may write ${\displaystyle g}$  as a Taylor series

${\displaystyle g(z)=\sum _{n=0}^{\infty }c_{n}(z-z_{0})^{n}}$

where ${\displaystyle c_{0},c_{1}=0}$  due to ${\displaystyle g(z_{0})=0}$  and ${\displaystyle g'(z_{0})=0}$ . Thus,

${\displaystyle g(z)=(z-z_{0})^{2}\sum _{n=2}^{\infty }c_{n}(z-z_{0})^{n-2}}$ ,

and hence for ${\displaystyle z\neq z_{0}}$ , ${\displaystyle z}$  in the domain of convergence of the power series, we have

${\displaystyle f(z)=\sum _{n=2}^{\infty }c_{n}(z-z_{0})^{n-2}}$ ,

where the latter expression makes sense also at ${\displaystyle z_{0}}$  and is complex differentiable there. This means that

${\displaystyle {\tilde {f}}(z)={\begin{cases}f(z)&z\neq z_{0}\\c_{2}&z=z_{0}\end{cases}}}$

is the desired continuation of ${\displaystyle f}$  to ${\displaystyle z_{0}}$ .${\displaystyle \Box }$

## Exercises

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