753 reputation
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bio website math.sunysb.edu/~myounsi
location Stony Brook, NY, USA
age
visits member for 5 years, 9 months
seen yesterday

I am currently a NSERC postdoctoral fellow at Stony Brook University.


Jul
13
comment Do the bounded isophase lines of a complex polynomial $f$ through the zeroes of $f’$ define a spanning tree?
If I understand correctly, the questions are about isochromatic lines connecting a pair of zeros, not a pair of critical points.
Jul
10
comment Conformal welding of annuli
Yes. I'm not sure I understand Question 2, but usually $\phi_X$ is only assumed to be a homeomorphism on the circle. In that case, even if you assume real analyticity, there might be a problem in the glueing if $\phi_X'$ vanishes somewhere. Maybe this is why some people require that the maps are conformal on a neighborhood of the circle.
Jul
9
comment Conformal welding of annuli
In classical conformal welding, a sufficient condition for the existence of conformal welding is that the homeomorphism is quasisymmetric. In that case, the curve obtained is a quasicircle, and you get uniqueness (up to Mobius transformation). The uniqueness follows from the fact that quasicircles are removable for homeomorphisms of the sphere that are conformal on the exterior. In your case, I suppose a sufficient condition to get existence and uniqueness of welding is that the homeomorphisms have a quasiconformal extension.
Jul
5
comment A question of Erdos on entire functions
A proof of this can be found in Proofs from the Book by Aigner and Ziegler. See Theorem 5, p.103.
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?
added 282 characters in body
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
added 803 characters in body
Jan
12
revised Analytic diffeomorphisms of the circle from complex domains
added 803 characters in body