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Complex analysis, holomorphic functions, automorphic group actions and forms, pseudoconvexity, complex geometry, analytic spaces, analytic sheaves.
3
votes
2
answers
205
views
Bergman norm on a bigger domain
Let $D$ be a unit disc (I am actually interested in a much more general setting, but let's start with explicit examples). Let $E$ be an open subset of $D$. Consider the functional on the space of all …
1
vote
How to prove that weighted Bergman space is separable.
Let $B\subset D$ be dense and countable. Let $B'\subset A^{2}(D,e^{-\varphi})^*$ be the set of point evaluations on $A^{2}(D,e^{-\varphi})$ at the points of $B$. Any $f\in A^{2}(D,e^{-\varphi})$ is co …
1
vote
0
answers
86
views
Does there exist a bounded analytic function majorated by a given one?
Let $f$ be an element of the Hardy space $H^2$, i.e. $f$ is an analytic function on the unit disk such that $\sum|a_n|^2<\infty$, where $f(z)=\sum a_n z^n$. Assume also that $f\not\equiv 0$.
Is th …
4
votes
2
answers
248
views
Holomorphic union of sets
Let $\mathbb{D}$ be the unit disk on the plane and let $U,V\subset \mathbb{D}$ be open and such that $U\cup V=\mathbb{D}$.
Is there a holomorphic map $\varphi:\mathbb{D}\times \mathbb{D}\to \mathb …
1
vote
0
answers
75
views
Criterion of holomorphy
I have posted a similar question on MSE few days ago, but it received no attention.
Let $G$ be a homogeneous (or even symmetric) domain in $\mathbb{C}^{n}$ and $u:G\to \mathbb{C}\backslash\{0\}$.
Q1 …
4
votes
1
answer
130
views
Criterion for homogeneity
Let $\Omega$ be a bounded domain in $\mathbb{C}^{n}$ and let $G=Aut(\Omega)$ be the full group of self-biholomorphisms of $\Omega$. Assume that there is $z\in \Omega$, such that the orbit of $z$ is so …
1
vote
1
answer
80
views
How many points of a sequence can we catch with an analytic disc?
Let $X\subset \mathbb{C}^{n}$ be a domain. You can assume that it is nice (e.g. bounded convex balanced ). Let $\{x_n\}$ be a sequence of points that does not have a limit point in $X$.
Let $D$ be the …
3
votes
0
answers
58
views
Convergence of sesqui-holomorphic kernels on the diagonal
Let $X\subset \mathbb{C}^d$ be a domain.
A function (kernel) $K:X\times X\to \mathbb{C}$ is called sesqui-holomorphic if it is holomorphic in the first variable, and anti-holomorphic in the second va …
4
votes
0
answers
100
views
Generating $H^{\infty}(X)$
Let $X$ be a domain in $\mathbb{C}^d$ and let $\mathbb{D}$ be the open unit disk in $\mathbb{C}$. Consider the Banach algebra $H^{\infty}(X)$ consisting of bounded holomorphic functions on $X$ with th …
3
votes
0
answers
72
views
A holomorphic shrinking of a domain into a compact subset
This question is related to these two.
Let $X\subset \mathbb{C}^{n}$ be a bounded domain. I am interested in the following property: there is a sequence of continuous maps $\varphi_n:\overline{X}\to X …
5
votes
1
answer
393
views
Holomorphic Sard's theorem 2
My previous question on this topic had a negative answer, but Tom Goodwillie in the comments suggested a statement, which may be true, and even a strategy of how to prove it. I haven't been able to ne …
2
votes
2
answers
248
views
Measures, orthogonal to holomorphic functions
Let $G$ be a domain in $\mathbb{C}^{d}$ and let $H\left(G\right)$ be the space of all holomorphic functions on $G$.
My question is how to characterize all such (Radon) measures $\mu$ on $G$, that $\i …
2
votes
1
answer
112
views
Norm of vector-valued holomorphic functions
Let $G$ be a connected simply connected domain in $\mathbb{C}^{n}$, let $H$ be a Hilbert space.
Q1. Which functions $F:G\to(0,+\infty)$ are such that there is a holomorphic $f:G\to H\backslas …
4
votes
1
answer
289
views
Supremum over which sets makes $H^{\infty}$ non-separable?
It is known that the space $H^{\infty}$ of bounded holomorphic functions on the unit disk $D$ is non-separable with respect to the supremum norm $\|\cdot\|_{\infty}^{D}$. Let $E\subset D$ be connected …
6
votes
Accepted
Supremum over which sets makes $H^{\infty}$ non-separable?
Jochen Wengenroth suggested to look at Carleson's interpolation theorem, and it seems like it completely answers my question. Namely, the following is true.
Let $E$ be a subset of $D$. Then $H^\infty …