# 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 $X$  is called connected if and only if the only nonempty set which is simultaneously open and closed is $X$ .

It seems somewhat implausible that this is actually the "right" definition of connectedness. The following theorem shows that in fact, if the topological space $X$  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.

Theorem 7.2:

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

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.$\Box$

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.$\Box$

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.

Theorem 7.3:

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

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

Proof:

## Riemann's theorem

Theorem 7.6:

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

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

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

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

$\lim _{h\to 0}{\frac {(z_{0}+h-z_{0})^{2}f(z_{0}+h)-(z_{0}-z_{0})^{2}f(z)}{h}}=\lim _{h\to 0}hf(z_{0}+h)=0$ , 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 $z_{0}$ .$\Box$

## 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 $f\equiv g$ .