# Confusion on analytic continuation.

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!

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 $$f(z)=\frac1{z-2}$$.

• If we know that the analytic continuation exists, say $f:\Omega_{1}\longrightarrow\mathbb{C}$ has an analytic continuation $g$ on $\Omega_{2}\supset \Omega_{1}$, do we have guarantees that $f=g$ on the whole $\Omega_{2}$? This is actually what confuses me Commented Dec 7, 2020 at 16:42
• That question makes no sense (unless $\Omega_1=\Omega_2$), since the domains of $f$ and $g$ are distinct. Therefore, $f$ and $g$ cannot possibly be equal. Commented Dec 7, 2020 at 16:44
• However, it is true that if $f_1,f_2\colon\Omega_2\longrightarrow\Bbb C$ are analytic and $f_1|_{\Omega_1}=f_2|_{\Omega_1}$, then $f_1=f_2$. Commented Dec 7, 2020 at 16:45
• To your second reply, yes, by identity theorem. To the first reply, so what is the point of analytic continuation? This may be off topic, but the point of analytic continuation is that we want to extend $f_{1}$ from a smaller domain $\Omega_{1}$ analytically to a larger domain $\Omega_{2}$, but it seems that we are not getting anything from the analytic continuation. Like, yes, we have an analytic function on $\Omega_{2}\supset \Omega_{1}$ and $F|_{\Omega_{1}}=f$, but then we still cannot say anything about $f$ on $\Omega_{2}$. (Thanks for your patience :)) Commented Dec 7, 2020 at 16:48
• The whole point of those theorems that you have mentioned is to say that if an analytic continuation exists, then there is only one such analytic continuation. The problem of whether or not such an analytic continuation exists is quite harder. I suggest that you take a look here, for instance. Commented Dec 7, 2020 at 16:55

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 $$\Omega_1 \cup \Omega_2$$