Is a de-zeroed Hardy function still Hardy?

Question

Suppose a Hardy function $f(z)$ on the upper half complex plane or $f\in H^{2+}$ (Chapter II, p.45 of Fulvio Ricci, Hardy Spaces in One Complex Variable) has a zero of order $m$ at $\omega$ with $\mathbf{Im}(\omega)>0$. A Hardy $H^{2+}$ function $f(z=x+iy)$ on the upper half complex plane is a holomorphic function on the that plane where $f_y(x):=f(x+iy)\in L^2(\mathbf R)$ with norm $\|f_y\|_2$ with respect to $x$ for any given $y$, and $\sup_y\|f_y\|_2<\infty$. Is $\displaystyle\frac{f(z)-f(\omega)}{(z-\omega)^n}\in H^{2+},\, \forall n\le m$?

Answer 1

Had I read further into Fulvio Ricci, Hardy Spaces in One Complex Variable, Chapter II, Theorem 5.4 provides the answer. This theorem also appears as Theorem 3.10 in Section 3, Chapter 20, of John B. Conway, Functions of One Complex Variable II, and Theorem 11.6 of Peter L. Duren, Theory of $H^p$ Spaces regarding the Blaschke product factorization.

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Edit: Actually, it is not necessary to use the explicit factorization as referred to above.

The answer is affirmative

Proof: Consider $n=1$. Larger $n\le m$ works the same. $g(z):=\frac{f(z)-f(\omega)}{z-\omega}$ is holomorphic on the closed upper half complex plane. Let $D(\omega;R)$ be the closed disk centered at $\omega$ and radius $0<R<\mathbf{Im}(\omega)$. $|g(z)|$ has a maximum and is bounded on the compact $D(\omega,R)$.

For an arbitrary $y\ge0$, let $\Omega_y:=\{x+iy: x\in\mathbf R\}\setminus D(\omega;R)$. We have $$|g(z)|\le \frac{|f(z)|}R+\frac{|f(\omega)|}{|z-\omega|},\ \forall |z-\omega|\ge R.$$ $|z-\omega|^2\ge(R^2+x^2)$, where $x$ is a real number such that $x+x_0=\mathbf{Re}(z-\omega)$ and that $|x_0+i\mathbf{Im}(z-\omega)|=R$.

By the Minkowski inequality $$\Big(\int_{\Omega_y} |g(x+iy)|^2 dx\Big)^\frac12 \le \frac1R\Big(\int_{\Omega_y} |f(x+iy)|^2 dx\Big)^\frac12+|g(\omega)|\Big(2\int_0^\infty\frac1{R^2+x^2}dx\Big)^\frac12$$ The first integral on the right hand side is uniformly bounded over $y$ by the definition of $H^{2+}$. $\displaystyle 2\int_0^\infty\frac1{R^2+x^2}\,dx=\frac\pi R$. We are done. $\quad\square$