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Mar 27, 2010 at 1:47 answer added Jonas Meyer timeline score: 9
Jan 5, 2010 at 21:11 answer added engelbrekt timeline score: 13
Jan 5, 2010 at 19:42 answer added Jorge Vitório Pereira timeline score: 10
Jan 5, 2010 at 19:41 answer added Harald Hanche-Olsen timeline score: 34
Jan 5, 2010 at 19:38 vote accept Johan
Jan 5, 2010 at 19:31 comment added Harald Hanche-Olsen Write up the complex Fourier series for the function on the boundary and throw away all the Fourier coefficients with negative index. What you're left with extends to the analytic function proposed in the answer, but that is useless.
Jan 5, 2010 at 19:14 comment added Johan That probably works but there is one small point I wonder about: if I take an arbitrary continuous function $f$ on the unit circle there is not always an analytic function $g$ on the unit disk that has $f$ as its boundary value. (There exists such a function only if the Dirichlet problem for $f$ happens to have an analytic solution as I see it.) Thus when I extend by the Cauchy formula and then go back to the unit circle I am not sure I end up with the same function as I started with nor that the function suddenly has not acquired derivatives everywhere.
Jan 5, 2010 at 19:10 answer added David E Speyer timeline score: 20
Jan 5, 2010 at 19:10 comment added Anweshi That does give an analytic function in the interior. But what about its behavior on the boundary? You don't know what happens.
Jan 5, 2010 at 18:52 comment added Mariano Suárez-Álvarez Take a continuous function on the unit circle which does not have derivatives at a dense set, and use Cauchy's formula to extend it to an analytic function in the interior.
Jan 5, 2010 at 18:39 history asked Johan CC BY-SA 2.5