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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)?

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    $\begingroup$ 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. $\endgroup$
    – fedja
    May 30, 2013 at 2:32
  • $\begingroup$ 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. $\endgroup$ May 30, 2013 at 11:35

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The construction you described fits in to the framework of the sinc methods. Thus one can get some insight into the properties of such functions by first approximating it via Sinc series and then use the properties of the series to get other (including various norm) estimates.
This approach should work at least for $H^p$, $p=1,2,\infty$, since the approximation theorems and errors estimates are readily available for such cases ( see Thm. 3.1.3 F. Stenger, Numerical Methods Based on Sinc and Analytic Functions ).

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Here's a way to make the approximation suggest by fedja more precise. By Paley-Wiener theorem, the Fourier transform is an isometry from $L^2(\mathbb R^+)$ to $H^2$. Suppose we have a bound $f(x+iy)\le C(y)\|f\|_{H^2}$ for entire $f\in H^2$. Then for any $f_0\in H^2$, let $f_n=\cal F(1_{[0,n]}\cal F^{-1}f_0)$. Then $f_n$ is entire and we have $f_n(x+iy)\le C(y)\|f\|_{H^2}$. Note that $\cal F^{-1}f_n\to\cal F^{-1}f_0$ in $L^2(\mathbb R^+)$, so $f_n\to f$ in $H^2$. Also, by the mean value theorem,

$$ f_n(x+iy)=\frac1{\pi y^2}\int_{B(x+iy,y)} f_n(z) =\frac1{\pi y^2}\int_0^{2y} \int_{x-\sqrt{y^2-y'^2}}^{x+\sqrt{y^2-y'^2}} f_n(x'+iy')dx'dy' \le\frac1{\pi y^2}\int_0^{2y} \sqrt y \|f_n(\cdot+iy')\|_{L^2}dy' \le\frac2{\pi\sqrt y}\|f_n\|_{H^2}. $$ Replacing $f_n$ by $f_n-f_0$ shows that $f_n(x+iy)\to f_0(x+iy)$. Taking the limit we have $f(x+iy)\le C(y)\|f\|_{H^2}$, so the best constant in the full case is the same as in the entire case.

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