bio | website | math.sunysb.edu/~myounsi |
---|---|---|
location | Stony Brook, NY, USA | |
age | ||
visits | member for | 5 years, 10 months |
seen | 2 hours ago | |
stats | profile views | 2,038 |
I am currently a NSERC postdoctoral fellow at Stony Brook University.
Aug
21 |
comment |
Absolute continuity and the Luzin N-Property for functions of two variables
Hi Trevor. I don't know about analogs of absolute continuity in several variables, but a very interesting family of continuous functions in $\mathbb{R}^n$ which preserve sets of zero $n$-dimensional Lebesgue measure are the so-called quasiregular mappings. A standard reference for this is the book of Rickman, Quasiregular mappings. You could have a look and figure out precisely what properties are needed in order to preserve zero-measure sets. |
Aug
19 |
comment |
Is there a reference for “computing $\pi$” using external rays of the Mandelbrot set?
This doesn't answer your very interesting question (and surely you are already aware of what follows), but a good reference for this $\pi$ phenomenon is the paper " $\pi$ in the Mandelbrot set" by Klebanoff. In the conclusion, Klebanoff conjectures that there are infinitely many routes at each of the infinitely many pinches of the Mandelbrot set which lead to $\pi$ in this way. |
Aug
13 |
awarded | Yearling |
Aug
13 |
revised |
A Generalization of the Ahlfors function to have varying degrees?
added 1766 characters in body |
Aug
13 |
answered | A Generalization of the Ahlfors function to have varying degrees? |
Aug
13 |
comment |
A Generalization of the Ahlfors function to have varying degrees?
Even if all $d_k$'s are equal to one, the branched cover is not unique. Indeed, if $\Omega$ is $n$-connected and say, contains $\infty$, then the Ahlfors function is not unique as a degree $n$ branched cover of $\Omega$ onto $\mathbb{D}$ which vanishes at $\infty$. The family of such degree $n$ branched covers depends on $n$ parameters. |
Aug
12 |
comment |
A Generalization of the Ahlfors function to have varying degrees?
A key step is showing that a certain linear system has a unique solution, in order to "remove the periods" of the harmonic function. I don't know if this works with arbitrary degrees though... See The structure of the semi-group of proper holomorphic mappings of a planar domain to the unit disk by Bell and Kaleem. |
Aug
12 |
comment |
A Generalization of the Ahlfors function to have varying degrees?
Note that one can obtain holomorphic branched covers of degree $n$ of $\Omega$ onto $\mathbb{D}$ without using Ahlfors functions. A theorem of Bieberbach states that for every $(b_1,\dots,b_n)$ where $b_j$ belongs to the curve $\gamma_j$, there is a degree $n$ branched cover $f:\Omega \to \mathbb{D}$ which maps each $b_j$ onto the point $1$. The proof involves taking linear combinations of Poisson kernels, and then showing that the coefficients can be chosen such that the resulting harmonic map has a well-defined conjugate. |
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? |