0
$\begingroup$

I've read and understood the proof for the complex plane for the analytic continuation theorem. The proof relies on $E$ being both closed and open, with $\quad E = \bigcap _ { n \geq 0 } E _ { n }$ and $E _ { n } = \{ z \in \Omega | f ^ { ( n ) } ( z ) = 0 \}$, so it can only be $\varnothing$ or $\Omega$ (the open set where the holomorphic function is defined).

While I get this, my intuition (I'm a physicist) of the analytic continuation theorem was that functions couldn't be zero on open subsets because the derivatives of $f$ on $x$ and $y$ are connected. I don't know how to connect this with the proof above.

Related to this question, does analytic continuation work on $\mathbb{R}^1$ functions? what about $\mathbb{R}^2$ functions? It should work on $\mathbb{R}^2$ because it's isomorphic to $\mathbb{C}$ but I might be wrong.

This questions might be super basic to most mathematicians but as a physicist I struggle quite a bit. Maybe because the're so basic I didn't found a lot on the internet.

Thanks!

$\endgroup$
  • $\begingroup$ Analytic continuation is a concept in functions of complex variables. Functions of real variables have no such concept. $\endgroup$ – herb steinberg Feb 9 at 19:19
  • $\begingroup$ @herbsteinberg Why do you say that ? If $f$ is analytic for $x \in (a,b) \subset \mathbb{R}$ you can ask for some function $g$ analytic for $x \in (c,d) \supset (a,b)$ such that $g|_{(a,b)} = f$. Those two functions will be naturally analytic on some complex open sets $U,V \subset \mathbb{C}$ containing those real intervals. $\endgroup$ – reuns Feb 9 at 19:22
  • $\begingroup$ The main idea of analytic continuation is the picture on the right there. Complex analytic means around every point of the domain the function is given by a power series in $z$. So it is obvious that the analytic continuation of $0$ is unique, thus the analytic continuation of $f$ on some given connected domain if it exists is unique. The other main idea is that holomorphic functions (complex differentiable on a domain) are analytic. $\endgroup$ – reuns Feb 9 at 19:31
  • $\begingroup$ I agree with your statement, but the definition of analytic continuation is designed to get around singularities. For example $ln(z)$ has a singularity at $z=0$. If you confine the domain to real $z=x$, then the function is defined only for $x\gt 0$. Using analytic continuation, it will be defined for all complex $z$, except $z=0$. $\endgroup$ – herb steinberg Feb 9 at 19:33
  • $\begingroup$ @herbsteinberg Let $f : (1,\infty) \to \mathbb{R}, f(x)=\ln x$ and $g \to \mathbb{R}: (0,\infty), g(x)=\ln x$. Then $g$ is the unique analytic continuation of $f$ to $(0,\infty)$ and there are no analytic continuation of $f$ to any open set containing $[0,\infty)$. Where is the problem ? Also $\frac1x$ is the unique meromorphic contiuation of $f'$ to $\mathbb{R}$. If the meromorphic continuation of $h'$ to $\mathbb{R}$ has no simple poles it implies a unique meromorphic continuation of $h$ to $\mathbb{R}$. $\endgroup$ – reuns Feb 9 at 19:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.