Chuwei Zhang
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 Apr 2 comment Stopping times and hitting times for càdlàg processes But it is not true that the set $\{(r_n)_{n\in\mathbb{N}}:r_n\in\mathbb{Q}\forall n\}$, which is $\mathbb{Q}^{\mathbb{N}}$, is countable. Even its subset $\{0,1\}^{\mathbb{N}}$ is uncountable, for the binary expansion gives a surjection from $\{0,1\}^{\mathbb{N}}$ onto the interval $[0,1]$. Jan 26 comment Weak derivative zero implies constant function @avati91, yes, the weak limit of constant functions here is constant. Take a decreasing sequence $\epsilon_n$ that tends to zero. We know that $u\ast \rho_{\epsilon_n}$ converges to $u$ in $L^p_{\text{loc}}(U)$. But unless the numbers $c_{\epsilon_n}$ form a convergent sequence, this $L^p_{\text{loc}}(U)$ convergence cannot happen. Therefore, the constants $c_{\epsilon_n}$ must converge as numbers. Then it is easy to see that the limit of $u\ast \rho_{\epsilon_n}$ in $L^p_{\text{loc}}(U)$ is $\lim_n c_{\epsilon_n}$. Dec 6 awarded Popular Question Nov 19 awarded Self-Learner 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 Mar 19 awarded Yearling Mar 4 asked Weak convergence and norm convergnce along a subsequnece in $H^1(\Omega)$