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Suppose $f$ is a holomorphic function in a simply connected open set $U$, and we know it's Taylor expansion at a point $p\in U$. We can then find a holomorphic map $g$ of $U$ to the unit disc which sends $p$ to 0, and obtain Taylor series of $f(g^{-1}(z))$ near $z=0$. Mapping back to $U$ this then yields an expansion of $f$ in a series of functions, which is convergent in all of $U$. It is essential that if we knew only first $N$ terms if the Taylor expansion, we can still obtain first $N$ terms of the final expansion.

Can we find a similar expansion for $f$ if $U$ is not simply-connected? I.e. write $f(z)=\sum_n a_n h_n(z)$, convergent in all of $U$, with $h_n$ single valued in $U$, and $a_n$ being a linear combination of first $n$ terms of Taylor expansion of $f$ at $p$? Of course, the functions $h_n$ should not depend on choice of $f$.

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  • $\begingroup$ After 2 days of thinking, I can do only "trivial" domains (say, an annulus but not a punctured disk). Do you want to know it in full generality, or you have some specific domain or class of domains in mind? $\endgroup$
    – fedja
    Commented May 12, 2019 at 22:31
  • $\begingroup$ @fedja I am actually very interested in the case of an annulus, so if you can say anything about this, I would be very grateful! $\endgroup$ Commented May 12, 2019 at 23:01
  • $\begingroup$ @fedja I guess more generally I am most interested in the case of Riemann sphere with finitely many cuts. $\endgroup$ Commented May 12, 2019 at 23:05
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    $\begingroup$ Enjoy! However, don't accept yet: I want to see the question solved for general domains. $\endgroup$
    – fedja
    Commented May 13, 2019 at 1:13
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    $\begingroup$ @fedja Tacit in the suggestion is that you would take $k^{(n)}_w(z)$ and orthonormalize (greedily) using Gram-Schmidt to get your $h_n$. Then it makes perfect sense to write down $f = \sum a_n h_n$ by interpolating the derivatives for any holomorphic function on $U$. (nb $h_n^{(k)}=0$ if $k <n$) Getting global convergence requires some kind of estimate probably, but my guess would be that there is a soft argument for that, as this argument does it for bounded functions. $\endgroup$ Commented May 13, 2019 at 3:25

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OK, here goes the "annulus".

Let's say that a domain $\Omega$ has a good approximation property at a point $w\in\Omega$ if for every $\rho>1$ there is a compact set $K=K(\rho)\subset \Omega$ and a constant $C=C(\rho)>0$ such that for every function $f$ analytic in $\Omega$ and every $m\ge 0$ there is a function $g$ analytic in $\Omega$ such that $g(z)-f(z)=O(|z-w|^{m+1})$ as $z\to w$ and $\sup_\Omega|g|\le C(\rho)\rho^m\sup_K|f|$.

Claim 1: The unit disk has a good approximation property at $0$.

Proof: Just let $K$ be the circle of radius $r\in(\rho^{-1},1)$ and take for $g$ the Taylor polynomial of $f$ of degree $m$.

Claim 2: Good approximation property is a conformal invariant on the Riemann sphere (with the understanding that if $w=\infty$, then the condition $g(z)-f(z)=O(|z-w|^{m+1})$ is replaced with $g(z)-f(z)=O(|z|^{-m-1})$, of course).

Proof: Obvious (map everything).

Claim 3: The Riemann sphere $\widehat{\mathbb C}$ with finitely many (reasonable) cuts $\Gamma_j\subset \mathbb C$ has good approximation property at $\infty$ (I assume $\Gamma_j$ are reasonable so that they are compact, pairwise disjoint, and $\widehat{\mathbb C}\setminus\Gamma_j$ is conformally equivalent to the unit disk for every $j$)

Proof: This is just a version of the simplest lemma on the separation of singularities. Fix $\rho>1$. First, each $\widehat{\mathbb C}\setminus\Gamma_j$ has good approximation property at $\infty$, so we can surround $\Gamma_j$ by some very close to it smooth contour $C_j$ that has the property of the compact $K$ in the definition of the good approximation property for $\widehat{\mathbb C}\setminus\Gamma_j$. Next, the function $f$ can be written as the sum $f(\infty)+\sum_j f_j$ where $f_j$ is the (clockwise) Cauchy integral of $f$ on any contour surrounding $\Gamma_j$ and having the evaluation point and other $\Gamma_k$ outside it. We can now readily estimate $f_j$ on $C_j$ by $\max_{C_j}|f|+|f(\infty)|+\sum_{k\ne j}\frac{\ell(C_k)}{2\pi\operatorname{dist}(C_j,C_k)}\max_{C_k}|f|$ by the triangle inequality, which yields $\max_{C_j}|f_j|\le C\max_K|f|$ where $K=\cup_j C_j$. Now, given $m$, just find an appropriate $g_j$ for each $f_j$ separately and put $g=\sum_j g_j$.

Assume now that $\Omega$ is a bounded domain with good approximation property at $0$ (we can always map the Riemann sphere with cuts and a fixed point conformally to this configuration). Let $H$ be the Hilbert space of functions analytic in $\Omega$ and square integrable with respect to the area measure (you can also introduce some reasonable weight, if you feel like it). Consider the subspaces $H_m=z^mH$ (those are just subspaces of functions in $H$ vanishing to order $m$ at the origin).

Clearly, we have $H=H_0\supset H_1\supset H_2\supset\dots$ and each $H_{m+1}$ is a closed subspace of $H_m$ of codimension $1$. Let $h_m$ be the function in $H_m$ orthogonal to $H_{m+1}$ of unit norm. Then, since $\cap_m H_m$ consists of analytic functions vanishing at $0$ with all derivatives, i.e., of just $0$, the functions $h_m$ form an orthonormal basis in $H$, so every function in $H$ can be written uniquely as its Fourier series in $h_m$, which converges in $H$ and, thereby, uniformly on compact subsets of $\Omega$ with all derivatives.

Notice now that the coefficient at $h_0$ is uniquely determined by $f(0)$ (all other terms vanish at $0$), the coefficient at $h_1$ is thus determined by $f(0)$ and $f'(0)$, and so on. This allows to write a formal Fourier decomposition into $h_m$ for any formal Taylor series at $0$. The only task is to show that the resulting series converges uniformly on compact sets in $\Omega$ if that Taylor series represents a function analytic in $\Omega$.

Note that due to the high order of $0$ of $h_m$ at the origin and the uniform bound for the $L^2$ norm with respect to the area measure, for every compact set $Q\subset\Omega$, there exists $B=B(Q)>0$ and $r=r(Q)<1$ such that $\max_Q|h_m|\le Br^m$. Now choose $\rho>1$ such that $\rho r<1$ and use the good approximation property. It yields a compact $K\subset\Omega$ and, for every $m$, a function $g$ that is bounded by $C\rho^m\max_K|f|$ and has the same first $m$ derivatives as $f$ at $0$. Then the formal Fourier coefficient of $f$ at $h_m$ is the same as of $g$, but $g\in H$ with norm bounded by the square root of the area of $\Omega$ times the uniform norm, so its Fourier coefficient is bounded by $C\rho^m$. This immediately implies the geometric convergence of the formal Fourier series of $f$ on $Q$.

Now it remains to note that $Q$ was arbitrary and that the sum of the formal Fourier series of $f$ (which we now know to converge) has the same derivatives at $0$ as $f$, so it must converge to $f$ itself.

Formally this doesn't work for general domains, but I still wonder if one can find a counterexample to this approach. Of course, one needs to replace the area measure by some weight very fast decreasing near the boundary, but it is not at all clear to me what the convergence properties of the corresponding formal Fourier series will be. The simplest case to consider is the punctured unit disk. There is no hope to get the good approximation property there (no matter what reasonable norm of $g$ you want to bound geometrically), so the last trick that yields a good bound for Fourier coefficients out of nothing won't work anymore, but can we still do something? Or can somebody come up with a completely different idea?

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  • $\begingroup$ Thanks a lot! I have a couple of questions: 1) how would one go about constructing $h_m$ for a given $\Omega$? is it feasible at least in simple examples? maybe there is some natural choice of the weight under which multiplication by $z$ has a simple adjoint, i.e. a finite-order differential operator? ... $\endgroup$ Commented May 13, 2019 at 4:27
  • $\begingroup$ ... 2) I think in the example that I gave for simply-connected $\Omega$ the set of basis functions is essentially invariant under holomorphic maps, i.e. if we map $\Omega$ to $\Omega'$ and $p$ to $p'$ then the functions $h_m$ in two cases differ at most by phase. In your construction there is an explicit dependence on the measure. I am not sure how to formalize it, but do you think there is a natural choice of the measure, given $\Omega$? $\endgroup$ Commented May 13, 2019 at 4:28
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    $\begingroup$ @PeterKravchuk J.E.Pascoe suggested an alternative description of $h_n$ in comments that you can find more constructive. The reproducing kernel is still hard to find unless you know some particular convenient orthogonal basis (like powers of $z$ for an annulus centered at the origin). If you have some very particular domain in mind, we can try to optimize a bit. Also, if you have some a priori growth restriction for $f$, it may help to choose the weight and even to get bounds for the error. I was primarily concerned with theoretical possibility. The practical implementation is more delicate. $\endgroup$
    – fedja
    Commented May 13, 2019 at 4:41

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