Uniqueness of analytic continuation to assign sum values

by Eddy   Last Updated January 14, 2018 13:20 PM

A recent Mathologer video about the sum $1+2+\ldots = -1/12$ reawakened some unease I have about using analytical continuation to assign values to divergent sums. Specifically, the idea that the analytic continuation of different functions gives the same result, or perhaps it doesn't and its something else that I've not understood. I'll lay out below what I think must be true, to see if anyone knows if it is true or why it doesn't have to be.


Let $\Omega_1,\Omega_2\subset \mathbb{C}$ be open and connected, and $u_k:\Omega_1 \rightarrow \mathbb{C},v_k:\Omega_2 \rightarrow \mathbb{C}, k\in\{1,2,\ldots\}$ be analytic. Define $$ f(z) = \sum_{k=1}^\infty u_k(z) \\ g(z) = \sum_{k=1}^\infty v_k(z) $$ and suppose that $u_k,v_k$ are such that both $f$ and $g$ are convergent and analytic on some (possibly different) open subset of $\mathbb{C}$. Denote the unique analytic continuation of the functions by $\tilde{f},\tilde{g}$. In addition suppose that there exists $z_1\in \Omega_1,z_2\in \Omega_2$ such that, for a given sequence $a_1,a_2,\ldots$ $$ u_k(z_1)=a_k \\ v_k(z_2)=a_k $$ Then we can prove that $\tilde{f}(z_1) = \tilde{g}(z_2)$.


It seems to me that, if the above is true, the analytic continuation of a function has meaning with regard to infinite sums, if not then it doesn't.

Is the above true?

If yes: prove it (or, intuitively justify it, link to a proof, reference a book, etc)

If no: why are the results obtained by analytically continuing the Riemann zeta function the ones we go with?

Answers 1

Yes, the statement is true; if two analytic functions are defined on a connected domain, and if they agree on a open subset, then they agree on their whole domain. And in fact it is enough that they agree on a single convergent sequence and its limit. This is called the identity theorem of complex analysis. In other words, analytic continuations are unique.

This is true whether your not they are given by series representations, although of course all analytic functions admit Taylor series representation, by definition.

The core of the proof is that $f^{-1}(0)$ is obviously closed, but it is also open since it contains a disk of convergence of $f$. But a set that is both open and closed is a connected component, and since the domain is connected, it is all of the domain.

January 14, 2018 13:15 PM

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