Show that $F=\{f\in H^1(D):\|f\|\le 1\}$ is locally bounded family After a brief search I've found multiple answers regarding similar question, but with Bergman Space rather than the Hardy Space.

Define $H^1(D)$ as the set of analytic functions on open unit disk $D$, with the norm
$$\|f\|=\sup_{r<1} \frac{1}{2\pi} \int_0^{2\pi} |f(re^{i\theta})| \, d\theta < \infty$$
Show that $F=\{f\in H^1(D):\|f\|\le 1\}$ is a locally bounded family (ie uniformly bounded on compact subsets).

In Bergman Space case it's good to start with Cauchy formula an then integrate, apply CS inequality. But for this one similar strategy doesn't work since the integral appearing in the norm part is centered at $0$.
In particular I don't see how to use the bounded norm condition - it seems the upper bound $1$ is chosen arbitrary...

You have for $|z|<r<1$:
$$ |f(z)| \leq \oint_{|u|=r} \frac{|f(u)|}{|u-z|} \frac{|du|}{2\pi}\leq  \frac{r}{r-|z|} \int_0^{2\pi}|f(re^{i\theta})| \frac{d\theta}{2\pi}\leq \frac{r}{r-|z|} \|f\|_1
$$
Let $r\rightarrow 1$ to get $|f(z)|\leq \frac{1}{1-|z|} \|f\|_1$. So on the disk $D(0,a)$, $a<1$ we have 
$$\sup_{zz\in D(0,a)} |f(z)|\leq \frac{1}{1-a} \|f\|_1.$$
