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Aug
20
answered does a simple random walk eventually hit every point?
Aug
8
awarded  Benefactor
Aug
7
accepted On every simply connected domain, there exists a holomorphic function with no analytic continuation.
Aug
2
awarded  Promoter
Aug
1
comment On every simply connected domain, there exists a holomorphic function with no analytic continuation.
Thanks for the added information! I have thought about what you suggested but I couldn't get it to work. Using the notation in your solution, as $N$ increases, $r$ needs to get closer and closer to $1$ in order that $|f(re^{2i\pi k/N})|>N$. So I can take a sequence $b_n$ in the disc that approaches the unit circle such that $b_n$ is not in the closure of $\{a_m:m\in\mathbb{N}\}$. Then there is a neighborhood of $\{b_n:n\in\mathbb{N}\}$, which is open and not relatively compact (since $|b_n|\rightarrow 1$), and which does not contain any $a_n$.
Jul
31
comment Proof or Counterexample:Is every open connected set $D \subset \mathbb C$ is a domain of holomorphy?
@MarianoSuárez-Alvarez The reason why $\sum_n z^{n!}$ works is that the function value goes off to infinity along any radial path towards a root of unity and the roots of unity are dense on the unit circle, so the function cannot even be continuously extended to any point on the boundary of the disc. This argument clearly relies on the geometry of the disc. But is there a topological reason that doesn't depend on the geometry of the disc for this?
Jul
31
comment On every simply connected domain, there exists a holomorphic function with no analytic continuation.
I still cannot work out what the $a_n$ should be. Could you tell me how to construct them?
Jul
31
comment On every simply connected domain, there exists a holomorphic function with no analytic continuation.
Thank you for the hint! But I cannot quite follow it. A subset $A$ of the unit disc not being relatively compact means that the closure of $A$ in the unit disc is not compact, so $A$ has a limit point on the unit circle. But if $A$ only has one limit point on the unit circle, why should the value of $f$ be unbounded on $A$?
Jul
31
comment On every simply connected domain, there exists a holomorphic function with no analytic continuation.
@Gary.That is not true. For example, the unit disc and the right half plane are conformally equivalent, but a circle is not homeomorphic to a straight line (by compactness).
Jul
31
asked On every simply connected domain, there exists a holomorphic function with no analytic continuation.
Jul
31
asked Is there a holomorphic function $f$ on the unit disc such that $|f(z)|\rightarrow\infty$ as $|z|\rightarrow 1$?
Apr
29
comment Estimating a power series for the order of an entire function
Thanks a lot! In fact I have been trying to prove the more general fact that you just stated: the order of $f$ equals $\alpha:=\limsup_n (n\log n)/\log(1/|a_n|)$. I have an estimate that for any $\epsilon>0$, $|a_n|\le n^{-n/(\alpha+\epsilon)}$ if $n$ is sufficiently large. But I could not estimate $\sum_{n=0}^{\infty}|a_n|r^n$ when $\alpha<1$. Do you have a reference to this result?
Apr
29
revised Estimating a power series for the order of an entire function
edited title
Apr
29
asked Estimating a power series for the order of an entire function
Apr
12
asked Decomposing continuous linear functionals on a locally convex space with 2 seminorms
Mar
19
awarded  Yearling
Mar
4
asked Weak convergence and norm convergnce along a subsequnece in $H^1(\Omega)$
Feb
15
accepted What is the Fourier transform of $e^{(-a+bi)x^2}$?
Feb
13
asked What is the Fourier transform of $e^{(-a+bi)x^2}$?
Dec
7
accepted Summable family in a normed linear space