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Let $\mathbb D$ denote the open unit disc in $\mathbb{C}$ . Let $Hol(\mathbb {D})$ denote the space of holomorphic functions on $\mathbb D$. The Hardy spaces on $\mathbb D$ are defined as follows.

$$H^p=\{f\in Hol(\mathbb {D}):\sup_{r<1}\int_{0}^{2\pi} |f(re^{i\theta}|^pd\theta<\infty\}\;\;\;(0<p<\infty)$$

$$H^\infty=\{f\in Hol(\mathbb {D}):\sup_{z\in\mathbb D}|f(z)|<\infty\}$$

Now let $w\in \mathbb {D}$ and $f\in H^p$ . Consider the new function $g$ on $\mathbb {D}$ defined as $$g(z)=\frac{f(z)-f(w)}{z-w}\;\;\;(z\in \mathbb {D})$$ It is said the $g\in H^p.$ Can anyone tell how? Does $g\in H^\infty$ also?

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  • $\begingroup$ Good exercise I like it $\endgroup$ – Guy Fsone Oct 12 '17 at 14:58
  • $\begingroup$ Are you asking if $f\in H^\infty \implies g\in H^\infty?$ $\endgroup$ – zhw. Oct 12 '17 at 19:55
  • $\begingroup$ @zhw No. I am asking if $f\in H^p$ then does $g\in H^\infty$? $\endgroup$ – user429197 Oct 12 '17 at 21:40
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Hint: By Analiticty,

$$f\in Hol(\Bbb D) \implies f(z) = \sum_{n =0}^{\infty}a_n(z-w)^n, ~~z\in D(w,\rho)$$ for som $\rho>$ such that $D(w,\rho)\subset \Bbb D.$

Remark $$a_n =\frac{f^{(n)}(w)}{n!}$$

Hence,

  • For $z\in D(w,\rho)$

$$ g(z) = \frac{f(z) - f(w)}{z-w} = \sum_{n =1}^{\infty}a_n(z-w)^{n-1} ~~$$

$\bar{D}(w,\rho)\ni\mapsto g(z)$ is continuous with $~~g(w) = f'(w).$ This implies that, there exists some constant $C_\rho>0$ $$|g(z)| \le C_\rho,~~z\in \bar{D}(w,\rho)$$

  • For $z\in \Bbb D\setminus D(w,\rho)$

$$ |g(z)|\le \rho^{-1} (|f(z)|+|f(w)|).$$

  • Altogether give

$$ |g(z)|\le C_\rho+ \rho^{-1} (|f(z)|+|f(w)|)~~~~\forall ~~z\in\Bbb D.$$

Then it easy to see that, $g\in H^p$

Now if $g\in H^\infty \implies f\in H^\infty $ this will mean that $H^p\subset H^\infty$ which is not true!!

in fact

$$|f(z)|\le |f(z)-f(w)| +|f(w)| = |g(z)(z-w)|+|f(w)|\le 2\|g\|_\infty+|f(w)|. $$

Hence there exists $f\in H^p$ such that $g\not \in H^\infty$.

Better Rescaling on $g$ If you set $$g_\rho (z) = g(\rho z)~~~\text{for fixed $0<\rho<1$} $$ then $$g_\rho\in H^\infty$$

Since $\bar{\Bbb D}\subset D(0,\frac{1}{\rho}) $

$$g \in Hol(\Bbb D) \implies g_\rho\in Hol(D(0,\frac{1}{\rho})) \implies g_\rho ~~~\text{is bounded on $\Bbb D$}.$$

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  • $\begingroup$ FsoneThank you for your answer. Can you tell if there is any way that I can scale $g$ so that it belongs to $H^\infty$? Or any factor which I multiply with $g $ and then it belongs to $H^\infty$? $\endgroup$ – user429197 Oct 13 '17 at 5:03
  • $\begingroup$ I think this question on rescaling is too broad and vague . since it depends on which kind of singularity that $f$ has and the type of resacaling you aim to do use. But any if you see the new edit this may be ok $\endgroup$ – Guy Fsone Oct 13 '17 at 7:43
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The isolated sigularity at $w$ is removable, so $g$ is holomorphic on $\mathbb D.$ Here's the proof for $1\le p < \infty$: Choose $r_0 \in (|w|,1).$ Then for $r_0<r<1,$

$$\int_0^{2\pi} |g(re^{it})|^p\, dt \le \int_0^{2\pi} \frac{(|(f((re^{it})|+|f(w)|)^p}{(r_0-|w|)^p}\, dt$$ $$ \le \frac{1}{(r_0-|w|)^p}\int_0^{2\pi} 2^{p-1}(|f((re^{it})|^p+|f(w)|^p)\, dt,$$

which shows $g\in H^p.$ The proof for $0<p<1$ is very similar. Try your hand at proving $f\in H^\infty \implies g\in H^\infty.$

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