Confusion on analytic continuation.

Question

The definition of analytic continuation of holomorphic function is stated as follows:

Let $f_{1}$ and $f_{2}$ be two analytic functions on two domains (open and connected) $\Omega_{1}$ and $\Omega_{2}$ such that $\Omega_{1}\cap\Omega_{2}\neq\varnothing$. If $f_{1}$ and $f_{2}$ agree on $\Omega_{1}\cap \Omega_{2}$, we say $f_{2}$ is the analytic continuation of $f_{1}$ on $\Omega_{2}$, and vice versa.

A smaller version of it is that:

If $f$ is analytic on a domain $D\subset\mathbb{C}$ and $F$ is analytic on a bigger domain $E\subset\mathbb{C}$ such that $f=F$ on $D\subset E,$ then $F$ is the analytic continuation of $f$ on $E$.

From what I read, this kind of technique allows us to define a function in a smaller domain and extend it analytically to a larger domain. But I don't understand why this definition allows us to do so.

What confuses me is that the definition only guarantees $f=F$ on the intersection $\Omega_{1}\cap\Omega_{2}$, so perhaps $f\neq F$ on $\Omega_{2}$, then how do I know $f$ is analytic on $\Omega_{2}\setminus\Omega_{1}$?

I tried to use the identity theorem as follows:

Let $f$ and $g$ be two holomorphic functions on a domain $D$ such that $f=g$ on a subset $S\subset D$ that contains a limit point, then $f=g$ on the whole $D$.

But this seems backward. By the hypothesis of analytic continuation, we only have $f=g$ on $S$, and $g$ is analytic on $D$, we do not really know if $f$ is analytic on the whole $D$ (this is the purpose of analytic continuation, right? to extend $f$ analytically to the whole $D$.)

Am I overthinking this and confusing myself?? I guess we should have, say $f_{1}=f_{2}$ on the whole $\Omega_{1}\cup\Omega_{2}$, but I don't know how to prove it.

Edit 1: (Some Clarification, Possible Answer and Reference)

I am sorry if I am asking a confusion (bad) question. My confusion is that, even though the analytic continuation exists, I don't think that means anything helpful. It only gives us an analytic function $F$ on a bigger domain $\Omega_{2}$ such that $F|_{\Omega_{1}}=f$ for $\Omega_{1}\subset\Omega_{2}$. But it does not say anything about $f$, $f$ is still in $\Omega_{1}$. So I do not understand why analytic continuation can extend the domain on which $f$ is analytic.

The book "Complex Analysis and Applications" by Hemant Kumar Pathak, has a chapter about analytic continuation.

As Jose suggested, it does not make sense to say $f=F$ on $\Omega_{2}$, because $f$ is on $\Omega_{1}$.

The book explains that if we have an analytic continuation of $f_{1}$ from $\Omega_{1}$ into $\Omega_{2}$ via $\Omega_{1}\cap\Omega_{2}$, then the aggregate value of $f_{1}$ in $\Omega_{1}$ and $f_{2}$ in $\Omega_{2}$ can be regarded as a single function $f(z)$ analytic in $D_{1}\cup D_{2}$ such that $$f(z)=\left\{ \begin{array}{ll} f_{1}(z), z\in D_{1}\\ f_{2}(z), z\in D_{2} \end{array} \right.$$

This actually clarifies things out. This is like what we did when we want to remove the singularity: if $f_{1}$ has a removable singularity at $z_{0}$, then we actually extend $f_{1}$ to $f$ by defining $$f(z)=f_{1}(z), z\neq z_{0}\ \ \text{and}\ \ f(z_{0})=\lim_{z\rightarrow z_{0}}f_{1}(z).$$

Thus, we are actually extending $f_{1}(z)$ to $f(z)$, not to $f_{2}(z)$. We sort of complete $f_{1}(z)$ into $\Omega_{2}$ by defining $f(z)$.

I hope my explanation can help other people who study complex analysis and find analytic continuation confusing.

Feel free to add anythings more!

Answer 1

Those theorems are not about extending analytic functions, in the sense that they are not about the possibility of extending such a function. What they say is that you can extend an analytic function in, at most a single way. So, they are about uniqueness of extensions, not about their existence.

To be more precise, they say that if $\Omega_1$ and $\Omega_2$ are domains, with $\Omega_1\subset\Omega_2$, and if $f\colon\Omega_1\longrightarrow\Bbb C$ is an analytic function, then there is at most an analytic function $F\colon\Omega_2\longrightarrow\Bbb C$ whose restriction to $\Omega_1$ is $f$. But it is perfectly possible that there is none! That's the case if, for instance, $\Omega_1=D(0,1)$, $\Omega_2=\Bbb C$ and $f\colon\Omega_1\longrightarrow\Bbb C$ is defined by $f(z)=\frac1{z-2}$.

Answer 2

The function $f_1(z)=1 / z$ is holomorphic function on the extended full plane except $z=0$ which is a first-order pole.

The function $f_2(z)=\sum_{n=0}^{\infty} i ^{n-1}(z- i )^n$ is holomorphic function on the closing circle $|z- i |<1$. It is easy to find that its sum function is $1 / z$.

The function $f_3(z)=\int_0^{\infty} e ^{-z t} d t$ is holomorphic function in the region $\operatorname{Re} z>0$, and in this region $f_3(z)=1 / z$.

From the above analysis, the analytic regions of the three functions overlap with each other, and they are all equal to $1 / z$ in the overlapping region, so they are analytic continuation of each other.

From this example, we can see that $f_1$ and $f_2$ are analytic continuation, but this does not mean that they can also be analytic in each other's domain.and it is not necessarily the case that $f_1$ is equal to $f_2$ on the whole $\Omega_1 \cup \Omega_2$