743 reputation
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bio website math.sunysb.edu/~myounsi
location Stony Brook, NY, USA
age
visits member for 5 years, 8 months
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I am currently a NSERC postdoctoral fellow at Stony Brook University.


Jun
30
comment Is Every Holomorphic Near an Entire?
@DavidSpeyer : To see that the condition implies analytic extension of $f$, consider for simplicity the case where $K$ is an annulus centered at the origin. Then we can use Cauchy's integral formula to write $f(z)=\sum_{n=-\infty}^{\infty} a_n z^n$ in $K$. Now let $k$ be any negative integer and apply the condition with $g(z)=z^{-k-1}$ to obtain that $a_k=0$. This shows that the Taylor coefficients with negative indices all vanish, thus $f$ extends analytically to the hole of $K$.
Jun
30
comment Is Every Holomorphic Near an Entire?
@DavidSpeyer : With the condition "for any Jordan curve $\gamma$ contained in the interior of $K$ and any function $g$ holomorphic in the interior of $K$, we have $\int_\gamma f(z)g(z) \, dz = 0$", I think the statement is true, at least for sufficiently nice compact sets $K$, e.g. if $K$ has finitely many holes. The point here is that this condition ensures that $f$ extends analytically to each hole, so that we can apply Mergelyan's theorem to uniformly approximate $f$ by polynomials.
Jun
30
comment Is Every Holomorphic Near an Entire?
@DavidSpeyer : Interesting question. Note that you should replace "circle" by "Jordan curve" in your condition, for otherwise one can consider $K$ a topological annulus around $0$ not containing any circle surrounding its hole and then the $1/z^2$ counterexample works.
Jun
30
awarded  Enlightened
Jun
30
awarded  Nice Answer
Jun
30
comment Is Every Holomorphic Near an Entire?
@AndréHenriques I assumed that $K$ was compact, thanks for the remarks. As Steven Gubkin mentioned, if $K$ is only assumed to be closed, it follows from Arakelian's approximation theorem. I have edited my answer accordingly.
Jun
30
revised Is Every Holomorphic Near an Entire?
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Jun
30
answered Is Every Holomorphic Near an Entire?
Jun
3
comment Mapping theorem in higher dimensions
The result that you mention in the first sentence is due to Riemann, and the map is not unique, unless normalized correctly.
May
23
comment Conformal map and Jordan curve
I think it's pretty clear. +1
May
21
comment Conformal map and Jordan curve
Well what if $\Omega$ is the whole plane and $\gamma$ is not a circle? In this case, such an $f$ would be linear, so that $f(\gamma)$ cannot be a circle.
May
21
comment Conformal map and Jordan curve
Not in general, of course, just because the curve $\gamma$ needs to be analytic if there is such a conformal map $f$.
Apr
19
comment If $f$ is separately holomorphic on $\Omega$ then $f\in\mathcal{C}^0(\bar\Omega)\Leftrightarrow f\in L^1(\Omega)$
What do you mean by "Hartog's theorem is not allowed"?
Jan
12
comment This inequality why can't solve it by now (Only four variables inequality)?
@AdamP.Goucher : But in your book "Mathematical Olympiad Dark Arts", you say "Hence [by Delzell's algorithm] it is theoretically possible to prove any inequality involving rational functions simply by reducing it to the sum of squares inequality. However, this approach is similar in its impracticality to building an automobile using Stone Age tools." As for the present problem, I would be quite interested to see the automobile built more efficiently..!
Jan
12
revised Analytic diffeomorphisms of the circle from complex domains
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Jan
12
revised Analytic diffeomorphisms of the circle from complex domains
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Jan
12
answered Analytic diffeomorphisms of the circle from complex domains
Dec
12
comment Bound areas of disks with respect to a quadratic differential
@DylanThurston : I'm glad I could help.
Dec
12
revised Bound areas of disks with respect to a quadratic differential
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Dec
12
revised Bound areas of disks with respect to a quadratic differential
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