Skip to main content
Search type Search syntax
Tags [tag]
Exact "words here"
Author user:1234
user:me (yours)
Score score:3 (3+)
score:0 (none)
Answers answers:3 (3+)
answers:0 (none)
isaccepted:yes
hasaccepted:no
inquestion:1234
Views views:250
Code code:"if (foo != bar)"
Sections title:apples
body:"apples oranges"
URL url:"*.example.com"
Saves in:saves
Status closed:yes
duplicate:no
migrated:no
wiki:no
Types is:question
is:answer
Exclude -[tag]
-apples
For more details on advanced search visit our help page
Results tagged with
Search options not deleted user 153260

Complex analysis, holomorphic functions, automorphic group actions and forms, pseudoconvexity, complex geometry, analytic spaces, analytic sheaves.

7 votes

Is $\frac{|t|}{e^{a|t|}-e^{-b|t|}}$ the Fourier transform of a positive function

No for $0<a<b$ your function has a global maximum at $\pm x_0 \neq 0$. Then, if it was $\hat{\varphi} = f $ for some positive function $\varphi$, $f(x-y)$ would be a positive semidefinite kernel. In p …
an_ordinary_mathematician's user avatar
5 votes
1 answer
266 views

A domination property for the Hardy space $H^1$

In the theory of Hardy spaces of the unit disc, a fact that is implicitely used quite often is that if $f\in H^p, 1<p<\infty$, then there exists a function $F\in H^p$ such that $|f(z)| \leq |F(z)|, \, …
an_ordinary_mathematician's user avatar
5 votes
Accepted

Is Toeplitz operator on the Bergman space bounded iff its symbol is bounded?

It is not neccesary in general that $\varphi \in L^\infty(\mathbb{D})$, but it is necessary and sufficient that in a certain sense $\varphi$ must be bounded ``on average in the hyperbolic sense''. The …
an_ordinary_mathematician's user avatar
4 votes
Accepted

A question on Bloch functions

As stated this property cannot be true. Consider $f(z)=z$. Clearly $f \in X_\frac12$. Let any other $g\in \mathcal{B}$ such that $\Vert g \Vert_\mathcal{B} < \varepsilon$. Then we have that $|f'(0)+g' …
an_ordinary_mathematician's user avatar
3 votes
0 answers
187 views

Beurling's theorem on invariant subspaces

Beurling's theorem characterize the closed subspaces $M\subset H^2$ of the Hardy space, which are invariant under the shift operator $Sf(z):=zf(z)$, as spaces of the form $\varphi H^2 $ where $\varphi …
an_ordinary_mathematician's user avatar
3 votes
0 answers
118 views

An open problem of Hardy and Littlewood on $p$-integral means

In Duren's book "Theory of $H^p$ spaces" (MSN) in the comment section after Section 4, it is mentioned that Littlewood and Hardy proved in Some properties of conjugate functions that if $u$ is a harmo …
an_ordinary_mathematician's user avatar
2 votes
Accepted

weakly separated sequences in RKHS are separated by Gleason metric

If a sequence is weakly separated, i.e. there exists a multiplier $\varphi_{ij}$ of multiplier norm at most one such that $\varphi_{ij}(\lambda_i)=\varepsilon, \varphi_{ij}(\lambda_j)=0$, then necessa …
an_ordinary_mathematician's user avatar
2 votes
Accepted

Criteria for Hardy space membership

This actually holds for all $p>0$. The function \begin{equation} G[f](\zeta):=\Big( \int_0^1(1-r)|f'(r\zeta)|^2 dr \Big)^{\frac 12}\end{equation} is sometimes called Paley Littlewood $g$-function. Th …
an_ordinary_mathematician's user avatar
2 votes

Examining the Hilbert transform of functions over the positive real line

The answer to the second question is negative as well. Take for example $g$ supported in $(-\infty,-1)$ and discontinuous in some point. If $f$ is supported in $\mathbb{R}_+$ and $y,z<-1$ it holds tha …
an_ordinary_mathematician's user avatar
1 vote

Pair of positive harmonic functions with negative inner product in Drury-Arveson space

I will try to prove that such functions do not exist. Suppose that $f,g$ are positive (pluri)harmonic functions in the pluri harmonic Drury Arveson space $\mathcal{H}DA_d$ such that $ \langle f ,g \ra …
an_ordinary_mathematician's user avatar
1 vote
Accepted

Equivalent condition for the Pick matrix being positive semidefinite

About your first question this is exactly the positivity of the determinant of the Pick matrix. That is because if $\lambda_1,\lambda_2 \in \mathbb{D}$ then \begin{align*} \det(P) & = \begin{bmatrix} …
an_ordinary_mathematician's user avatar
1 vote
0 answers
61 views

A question on a paper of B. S. Henriksen

I have been reading the article "A peak set of Hausdorff dimension $2n-1$ for the algebra $A(\mathcal{D})$ in the boundary of a domain $\mathcal{D}$ with $C^\infty$-boundary in $\mathbb{C}^n$" by B. S …
an_ordinary_mathematician's user avatar
1 vote
0 answers
28 views

About Carleson measures on the Hardy space on the bidisc

I have been reading the paper "Carleson Measures in Hardy and Weighted Bergman Spaces of Polydiscs" by F. Jafari and there are a few things that going on that I am not entirely convinced of. The paper …
an_ordinary_mathematician's user avatar
0 votes
Accepted

Isoperimetric inequality for analytic functions on an annulus

You can probably prove that $$ \Big( \int_\mathbb{A_r} |f(z)|^2 \frac{dxdy}{\pi(1-r^2)} \Big)^{1/2} \leq \int_{ \mathbb{T}} |f(e^{i\theta})| \frac{d\theta}{2\pi}+\int_{ \mathbb{T_r}} |f(re^{i\theta}) …
an_ordinary_mathematician's user avatar