If a function belongs to $H^p$ 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?
 A: 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}) $
$$g \in Hol(\Bbb D) \implies g_\rho\in Hol(D(0,\frac{1}{\rho})) \implies g_\rho ~~~\text{is bounded on $\Bbb D$}.$$
A: 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 $f\in H^\infty \implies g\in H^\infty.$
