Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $H^2$ denote the Hardy space on the strip $S:=\{z\in{\mathbb C}\,:\,0<\Im z <1\}$ (or the upper half plane), i.e. $H^2$ consists of all holomorphic functions $f:S\to\mathbb C$ such that for each $0< y< 1$, the function $x\mapsto f_y(x):= f(x+iy)$ lies in $L^2({\mathbb R},dx)$, and $\|f\| :=\sup_{0< y < 1}\|f_y\|_2<\infty$.

In this setting, it is known that $f$ has $L^2$-boundary values at the two boundaries of $S$, and that there holds the pointwise bound $|f(x+iy)|\leq c \|f\| d(y)^{-1/2}$, where $d(y)$ denotes the distance of $iy$ from the boundary of $S$, and $c$ is some constant.

I am interested in the following question: Suppose you have $f\in H^2$ with "regular" boundary behaviour -- for example, $f$ extends continuously to the closure of $S$, or $f$ continues holomorphically to a neighbourhood of $S$, or even to an entire function. Is a better pointwise bound than the one I wrote above known in such cases (for example, uniformly bounded)?

share|improve this question
Regularity won't help. Just take a bad function in the unrestricted setting, draw a nice curve close to the boundary that approaches the boundary quickly at infinity and map the srunk domain back to the strip with bounded distortion. –  fedja May 30 '13 at 2:32
For this argument to work completely one would need an example function $f \in H^2$ which behaves badly (in the sense of growth) at infinity at the boundary, because local divergencies like a pole at the boundary are ruled out by requiring continuous extension to the closure of $S$. You probably have such an example in mind? I currently don't know one, but it would be nice to have, because then one could also check if stronger regularity assumptions like holomorphic extension to a wider strip are still compatible with it. –  Gandalf Lechner May 30 '13 at 11:35
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.