L2 inequality implies completeness (a) Let $$\|f\|=\sqrt{\int_{|z|<1}|f(z)|^2dxdy}$$ for all holomorphic $f:\{|z|<1\}\to\mathbb{C}$ and $$L^2=\left\{\text{holomorphic }f:\{|z|<1\}\to\mathbb{C},\|f\|<\infty\right\}$$ Show that there exists $C>0$ with $$|f(w)|\le\dfrac{C}{1-|w|}\|f\|$$ for all $f\in L^2$ and $|w|<1$.
(b) Show that $L^2$ is complete under $\|\cdot\|:L^2\to\mathbb{R}_{\ge 0}$.
For (a), my idea is $$|f(w)|=\dfrac{1}{1-|w|}|f(w)|(1-|w|)\le\dfrac{1}{1-|w|}\int_{|z|<1}|f(z)||g(z)|dxdy\le\dfrac{1}{1-|w|}\|f\|\|g\|$$ by Cauchy, where $g:\{|z|<1\}\to\mathbb{C}$ is independent of $f$. The difficulty is finding $g$ and relating $|f(w)|$ to an integral over $\{|z|<1\}$ involving $|f(z)|$.
For (b), does the following picture work? It gets $\sup_{|z|<e}|f(z)|\le C_e\|f\|$ whereas I get $\sup_{|z|<e}|f(z)|(1-|z|)\le C_e\|f\|$, and I don't see how to fix it. It's from page 13 of http://www.mat.unimi.it/users/peloso/Matematica/har-ber2a.pdf
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 A: As for point (a).
Let us examine the compex integral
$$
 I = \int_{D(0,r)}h(z)\,dxdy,
$$
where $h$ is a holomorphic function in a neighborhood of this disc. Then
$$
I = \int_0^r\int_0^{2\pi} h(se^{i\theta})s\,dsd\theta = \int_0^r2\pi sh(0)\,ds = \pi r^2h(0),
$$
where second equality follows from Cauchy's formula. I hope it is clear that the disc is centered at zero only for convinience and this caluculation holds for any disc.
The answer to your point (a) should be clear now. Let $D(w,r) \subset \{|z|<1\}$, then
$$
|f(w)| = \frac{1}{\pi r^2}\left|\,\int_{D(w,r)}f(z)\,dxdy\;\right| \leq \frac{1}{\pi r^2}\int_{D(w,r)}|f(z)|\,dxdy \\
\leq \frac{1}{\pi r^2}\|f\|\sqrt{\int_{D(w,r)}1\;dxdy} = \frac{\|f\|}{r\sqrt{\pi}}.
$$
The last inequality follows from H$\ddot{\text{o}}\,$lder. The only question left is the maximal value for $r$ and clearly it is the distance from $w$ to unit circle, which is equal $1-|w|$.
As for point (b) - which part specifically are you having a problem with?
EDIT:
Regarding point (b): Let us take the notation from your comment, so $f$ is the holomorphic limit under locally uniform convergence and $g$ is the $L^2$ limit. The crucial part is noticing that $f$ is in $L^2$. To see this notice that for any disc $D(0,1-\delta)$ with $\delta$ small, if we estimate the norm restricted to this disc, we get
$$
 \|f\||_{D(0,1-\delta)} \leq \|f-f_n\||_{D(0,1-\delta)} + \|f_n\||_{D(0,1-\delta)}
$$
The first part goes to zero with $n$ by uniform convergence and the second is estimated as follows:
$$
 \|f_n\||_{D(0,1-\delta)} \leq \sup_{n}\|f_n\||_{D(0,1-\delta)} \leq \sup_{n}\|f_n\||_{D(0,1)} = M < \infty.
$$
There must be a unifrom, finite estimate on this supremum since $\{f_n\}$ is a Cauchy sequence in Hilbert space. But this gives a uniform estimate of the norm of $f$, independent of $\delta$, thus by monotone convergence theorem $f$ must lie in $L^2$.
Now, since $f$ and $g$ both are in $L^2$, so must be their difference. Let us fix a small $\epsilon$. Using the monotone convergence again we get that there must be a small positive $\delta_0$, such that for every $\delta < \delta_0$ we have
$$
 \|f-g\||_{D(0,1-\delta)^C} < \epsilon/3.
$$
Let us take such $\delta$. Now, by the uniform convergence and the $L^2$ convergence we can take such an $n_0$, that for every $n > n_0$ we have that
$$
 \|f-f_n\||_{D(0,1-\delta)} < \epsilon/3 \quad \text{and(!)} \quad \|f_n-g\||_{D(0,1-\delta)^C} < \epsilon/3.
$$
Now the last thing to do is to estimate the norm:
$$
 \|f-f_n\||_{D(0,1)} \leq \|f-f_n\||_{D(0,1-\delta)} + \|f-f_n\||_{D(0,1-\delta)^C} \leq \epsilon/3 + \|f-g\||_{D(0,1-\delta)^C} + \|g-f_n\||_{D(0,1-\delta)^C} < \epsilon
$$
and we are done. Second inequality follows from triangle inequality, and the first one from obvious fact that $\sqrt{x+y} \leq \sqrt{x} + \sqrt{y}$, for $x,y \geq 0$.
