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Nov 24, 2016 at 4:07 comment added T. Amdeberhan I made a mistake.
Nov 24, 2016 at 2:35 comment added Anthony Quas Yes! This does bother us! Why not say: here is a fact about sequences of complex numbers where the only proof I know goes through high level harmonic analysis (give reference). Does anyone see how to give a more elementary proof?
Nov 24, 2016 at 0:09 comment added T. Amdeberhan I thought it would be a distraction to know the complicated methods of proof while we seek for "a basic complex analysis proof". Look, if this bothers everyone then I apologize.
Nov 23, 2016 at 23:58 comment added nfdc23 If the answer to Gerry Myerson's question is "yes" then that is somehow not very professional.
Nov 23, 2016 at 23:23 comment added Gerry Myerson So, wait: you knew about the Boutet de Monvel & Gabber work, Amdeberhan, and deliberatley withheld this information when posting the question?
Nov 23, 2016 at 23:01 comment added T. Amdeberhan @ChristianRemling: I kind of agree with your assessment. However, we should probably get $H^{1/2}$ out of our minds and consider the problem/solution from a "novel" viewpoint. Maybe.
Nov 23, 2016 at 22:57 comment added Christian Remling @T.Amdeberhan: My feeling is the elementary part of this is what Anthony explained in his original comment. We're then still left with a problem about approximation in $H^{1/2}$, which doesn't feel elementary to me, though of course there could be a way around it.
Nov 23, 2016 at 22:18 comment added T. Amdeberhan Yes, I was aware of some "highbrow" proofs. I wish for an elementary argument (quite original it would be!) because after all the problem I stated in simple terms of infinite series.
Nov 23, 2016 at 21:18 comment added Christian Remling @AnthonyQuas: Thanks. It was detective work, but in the meantime I also received an e-mail from Richard Fournier on this who pointed out many interesting papers on these topics (some by himself).
Nov 23, 2016 at 21:12 comment added Anthony Quas This is a spectacular piece of detective work. I wonder if the OP already knew of the existence of this result? (somehow the fact that the question had exactly the right space and was asking for an elementary proof suggests this).
Nov 23, 2016 at 20:58 comment added Christian Remling @AnthonyQuas: You are right of course.
Nov 23, 2016 at 20:58 history undeleted Christian Remling
Nov 23, 2016 at 20:56 history edited Christian Remling CC BY-SA 3.0
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Nov 23, 2016 at 20:12 history deleted Christian Remling via Vote
Nov 23, 2016 at 6:25 comment added Anthony Quas Hang on a minute! I agree that $|u(y)-u(x)|$ is comparable to $|g(x)-g(y)|$ on every $A_n$, but the comparison gets worse and worse as $n$ increases. I don't think I believe that there is a $g$ with $|g(x)|<\pi$ that lies in $H^{1/2}$. For a specific example, what if $f(z)=z^2$. Then $g(x)$ should be $g(x)=4\pi x$, except this doesn't take values in $(-\pi,\pi)$. If you force it to take principal values, I think you don't satisfy the integrability condition any more.
Nov 23, 2016 at 4:32 comment added T. Amdeberhan I don't seem to follow where property (3) above has been shown for $h_k$ and they lead to property (3) for $f$.
Nov 23, 2016 at 0:43 history answered Christian Remling CC BY-SA 3.0