Descriptive complexity of analytic continuation

Question

Consider the set of complex power series $$ f(z)=\sum_{n=0}^\infty a_nz^n $$ that have radius of convergence $1$ and can be analytically continued to the neighborhood of some point on the unit circle. What is the descriptive complexity of this set when viewed as a subset of the Polish space of complex sequences? One can ask similar questions for Dirichlet series $$ g(s)=\sum_{n=1}^\infty\frac{a_n}{n^s} $$ A crude estimate of "having radius of convergence $1$" using the radius formula is $\Pi^0_3$. I initially hoped the analytic continuation part might be true analytic, which would add to the examples of naturally occuring non-Borel sets, but then convinced myself that it is actually Borel (but it could still be a good example of statements of high quantifier complexity). To see that it is Borel, first note that for any point $c$ in the open unit disk we can compute $\sum_{n=0}^\infty b_n(z-c)^n$, the power series expansion of $f(z)=\sum_{n=0}^\infty a_nz^n$ centered at $c$, in a Borel way, because we can compute $f^{(n)}(z)$ easily and then plug in $z=c$. Then $f(z)$ can be analytically continued to some point on the unit circle iff there exists $q>0$ and a sequence $(c_i:i<\omega)$ of points in the open disk with rational coordinates that converges to some point on the boundary (this can be expressed in a Borel way by saying the sequence is Cauchy) and such that the radius of convergence of $f(z)$ centered at $c_i$ is at least $q$. This feels somewhat like $\Sigma^0_4$ but at this point my ability of quantifier counting is overwhelmed...plus there could well be a much simpler way to express this using theorems from complex variables.

Answer 1

For an analytic function given by its power series, existence of an analytic continuation to some open set intersecting the unit circle is $Σ^0_2$.

To see this, for an analytic function, its power series at $z_1$ is uniformly computable from its power series at $z_0$ ($\sum_{n=0}^∞ a_n (z-z_0)^n$), $r>|z_1-z_0|$, and $M≥\sup(|a_n| r^n)$. And an analytic continuation to a given point (with the function analytic in an open neighborhood) can be decomposed as a finite number of hops recentering the series, so its existence is $Σ^0_2$ (each intermediate $z_i$ can be rational, so we can encode them along with corresponding $r$ and $M$ into an integer, and the condition on each $M$ is $Π^0_1$).

As a side note, by Cauchy's inequality, $$ \sup(|a_n| r^n) ≤ \sup_{|z-z_0|<r} |f(z)|. $$ Also, given $M$ and $r$, the power series at $z_0$ is uniformly computably interconvertible with the function encoded (in a standard manner) as a continuous function on $\{z:|z-z_0|<r\}$.

We also have:

• Having radius of convergence $>r$ is $Σ^0_2$-complete; it is equivalent to $$ \exists r_1>r\, \exists M \,∀n \, |a_n|r_1^n≤M. $$
• Having radius of convergence exactly $r$ (including exactly 1) is $Π^0_3$-complete. To see this, given a $Π^0_3$ formula $∀n ∃m ∀k \, φ(n,m,k)$, for each $n$ we add some terms to the power series that are small enough for distances $<1-1/n$, but large at distance (say) $1-1/(2n)$, one term for each $m'$ such that $∃m<m' \, ∀k \, φ(n,m',k)$ fails.
• The conjunction of having an above continuation and having radius of convergence exactly 1 is thus $Π^0_3$, and it is also $Π^0_3$-complete. To see this, build the power series centered at -1/2, radius of convergence >1/2, and all coefficients nonnegative reals, and such that it blows up at some positive real $z<1$ iff the $Π^0_3$ formula fails (and otherwise it blows up at 1), and then recenter the series at 0.