# Pointwise bounds on Hardy space functions with regular boundary behaviour

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)?

• 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. May 30, 2013 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. May 30, 2013 at 11:35

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 ).
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.