"Define $D_\alpha$ as the open disc in $\mathbb{C}$ with centre $0$ and radius $\alpha$
We can think of holomorphic functions on $D_\alpha$ as power series $\sum_{n=0}^{\infty}c_n z^n$ which are convergent on $D_\alpha$
Define $H_\alpha$ as the vector space of holo. functions on $D_\alpha$ with the property:
Given $z=x+yi$, $$\iint_{D_\alpha}\lvert f(z)\rvert^2 \mathrm{d} x\,\mathrm{d} y \lt \infty$$
1) Given the inner product, $$\langle f,g \rangle_{\alpha} = \iint_{D_\alpha}f(z)\overline{g(z)} \mathrm{d} x\,\mathrm{d} y $$ Show that $H_\alpha$ is an inner product space and write the associated norm
2) Define $p_n(z)=z^n$ for integers $n\geq 0$.
Prove that $\langle p_m,p_n \rangle_{\alpha} = 0$ whenever $m \neq n$ and compute $\lvert \lvert p_n \rvert \rvert_{\alpha}$
3) Let $f(z)=\sum_{n=0}^{\infty}c_n z^n$ be an element of $H_\alpha$. Prove $$\lvert \lvert f \rvert \rvert_{\alpha}^2 = \sum_{n=0}^{\infty}\lvert c_n\rvert^2\lvert \lvert p_n \rvert \rvert_{\alpha}^2$$ when:
i) We assume the power series converges uniformly in $D_\alpha$
ii) We do not assume the power series converges uniformly in $D_\alpha$"
So part 1 is very simple, we are given that $H_\alpha$ is a vector space, so we just have to show that the inner product properties hold for elements of $H_\alpha$ and we know the associated norm is $\lvert \lvert f \rvert \rvert_{\alpha} = \sqrt{\langle f,f \rangle_{\alpha}}$
I'm really struggling with parts 2 and 3. In part 2, I'm struggling with the integral we get. EDIT: So thinking some more, I've solved part 2 (same technique as user284331)
Where do I start with part 3? So far, I get:
$$\lvert \lvert f \rvert \rvert_{\alpha}^2 = \langle f,f \rangle_{\alpha} = \iint_{D_\alpha}\lvert f\rvert^2 \mathrm{d} x\,\mathrm{d} y = \iint_{D_\alpha}\lvert \sum_{n=0}^{\infty}c_n z^n\rvert^2 \mathrm{d} x\,\mathrm{d} y $$
EDIT: I have a proof of part 3i that uses the properties of the inner product:
$$\lvert \lvert f \rvert \rvert_{\alpha}^2 = \langle f,f \rangle_{\alpha} = \langle \sum_{n=0}^{\infty}c_n z^n , \sum_{n=0}^{\infty}c_n z^n \rangle_{\alpha}$$
$$ = \overline{c_0}\langle \sum_{n=0}^{\infty}c_n z^n , 1 \rangle_{\alpha} + \overline{c_1}\langle \sum_{n=0}^{\infty}c_n z^n , z \rangle_{\alpha} + ...$$
$$ = \overline{c_0}{c_0}\langle 1 , 1 \rangle_{\alpha} + \overline{c_0}{c_1}\langle z, 1 \rangle_{\alpha} + ...$$
But we know polynomials of different powers are orthogonal, $\langle z^m, z^n \rangle = 0$ when $n \neq m$, so
$$ = \overline{c_0}{c_0}\langle 1 , 1 \rangle_{\alpha} + \overline{c_1}{c_1}\langle z, z \rangle_{\alpha} + ...$$
$$ = \sum_{n=0}^{\infty}\overline{c_n}{c_n}\langle z^n , z^n \rangle_{\alpha}$$
$$ = \sum_{n=0}^{\infty}\lvert c_n\rvert^2\lvert \lvert p_n \rvert \rvert_{\alpha}^2$$
This completes my proof, and I don't think it uses uniform convergence anywhere.
But I don't understand how to use the assumption of uniform convergence to prove it. Any suggestions?
EDIT 2: So I've realised where my proof uses uniform convergence, and I have a hint towards part 3ii
Hint: You can assume that $\lvert \lvert f \rvert \rvert_{\alpha}$ is the limit of $\lvert \lvert f \rvert \rvert_{\alpha^{'}}$ as $\alpha^{'}$ increases towards $\alpha$
So my thought process is:
Find $D_{\alpha^{'}} \subset D_{\alpha}$ where the power series converges uniformly (but not in $D_{\alpha}$). That is, the radius of convergence of the power series is $\alpha^{'}$
Therefore we can use our result of part 3i for $\lvert \lvert f \rvert \rvert_{\alpha^{'}}$ then take the limit as $\alpha^{'}$ increases towards $\alpha$ of both sides, giving us:
$$\lim_{\alpha^{'} \to \alpha} \lvert \lvert f \rvert \rvert_{\alpha^{'}}^{2} = \lim_{\alpha^{'} \to \alpha} \sum_{n=0}^{\infty}\lvert c_n\rvert^2\lvert \lvert p_n \rvert \rvert_{\alpha^{'}}^2 = \lim_{\alpha^{'} \to \alpha} \sum_{n=0}^{\infty}\frac{\lvert c_n\rvert^2 \pi}{n+1}(\alpha^{'})^{2(n+1)} $$
That last equality comes from the result of part 2. Now the left hand side becomes what we are looking for, $\lvert \lvert f \rvert \rvert_{\alpha}^2$, but how do I calculate the limit of the sum on the right hand side. If I can do that, the proof is finished, I think.
Does it make sense to do this? Is there an easier way to prove 3 without the assumption of uniform convergence?
A trick is to switch to polar coordinates:
Let $z=re^{i\theta}$, then \begin{align*} (p_{m},p_{n})&=\iint_{D_{\alpha}}z^{m}\overline{z^{n}}dxdy\\ &=\int_{0}^{\alpha}\int_{0}^{2\pi}r^{m}r^{n}e^{im\theta}e^{-in\theta}d\theta rdr\\ &=\int_{0}^{\alpha}r^{m+n+1}\int_{0}^{2\pi}e^{i(n-m)}d\theta dr. \end{align*} We note that \begin{align*} \int_{0}^{2\pi}e^{i(n-m)}=0,~~~~n\ne m,~~~~\int_{0}^{2\pi}e^{i(n-m)}=2\pi,~~~~n=m. \end{align*} So the result would be \begin{align*} \|p_{n}\|^{2}=(p_{n},p_{n})=\dfrac{\pi}{n+1}\alpha^{2(n+1)}. \end{align*}
Edit:
The third part:
It is all about interchanging limit with integral, and infinite sum with integral and all that. The best way to deal with them is to implement Lebesgue Dominated Convergence Theorem, as the following shown.
We know that $f$ is holomorphic on $D_{\alpha}$, the open disk with radius $\alpha$, the power series expansion of $f$ might not be uniformly convergent on the whole $D_{\alpha}$, but by a standard theory of complex analysis, it must be uniformly convergent on each compact subset of $D_{\alpha}$. We denote by $\overline{D}_{\alpha'}$ the closed disk with radius $\alpha'<\alpha$, so $\overline{D}_{\alpha'}$ is a compact subset of $D_{\alpha}$.
For each closed disk $\overline{D}_{\alpha'}$, there exists a constant $M_{\alpha'}$ such that for $f(z)=\displaystyle\sum_{n=0}^{\infty}c_{n}z^{n}$ \begin{align*} \sum_{n=0}^{\infty}|c_{n}||z|^{n}\leq M_{\alpha'},~~~~z\in\overline{D}_{\alpha'}. \end{align*} By Monotone Convergence Theorem, we have \begin{align*} \|f\|_{\alpha}^{2}=\lim_{\alpha'\rightarrow\alpha}\iint_{\overline{D}_{\alpha'}}|f(z)|^{2}dxdy. \end{align*} Pluggin the power series expansion of $f$, we obtain that \begin{align*} \|f\|_{\alpha}^{2}=\lim_{\alpha'\rightarrow\alpha}\iint_{\overline{D}_{\alpha'}}\sum_{n=0}^{\infty}c_{n}z^{n}\overline{f(z)}dxdy. \end{align*} Note that \begin{align*} \sum_{n=0}^{\infty}\left|c_{n}z^{n}\overline{f(z)}\right|=\sum_{n=0}^{\infty}|c_{n}||z|^{n}|f(z)|\leq M_{\alpha}^{2},~~~~z\in\overline{D}_{\alpha'}, \end{align*} by Lebesgue Dominated Convergence Theorem, we get \begin{align*} \iint_{\overline{D}_{\alpha'}}\sum_{n=0}^{\infty}c_{n}z^{n}\overline{f(z)}dxdy=\sum_{n=0}^{\infty}c_{\alpha}\iint_{\overline{D}_{\alpha'}}z^{n}\overline{f(z)}dxdy. \end{align*} Plugging the power series expansion of $f$ again, we obtain that \begin{align*} \sum_{n=0}^{\infty}c_{n}\iint_{\overline{D}_{\alpha'}}z^{n}\overline{f(z)}dxdy=\sum_{n=0}^{\infty}c_{n}\iint_{\overline{D}_{\alpha'}}z^{n}\sum_{m=0}^{\infty}\overline{c_{m}}\cdot\overline{z^{m}}dxdy. \end{align*} Also note that \begin{align*} \sum_{m=0}^{\infty}|z|^{n}|c_{m}||z|^{m}\leq|\alpha'|^{n}M_{\alpha},~~~~z\in\overline{D}_{\alpha'}, \end{align*} by Lebesgue Dominated Convergence Theorem again, we get \begin{align*} \sum_{n=0}^{\infty}c_{n}\iint_{\overline{D}_{\alpha'}}z^{n}\sum_{m=0}^{\infty}\overline{c_{m}}\cdot\overline{z^{m}}dxdy&=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}c_{n}\overline{c_{m}}\iint_{\overline{D}_{\alpha'}}z^{n}\overline{z^{m}}dxdy\\ &=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}c_{n}\overline{c_{m}}(p_{n},p_{m})_{\alpha'}\\ &=\sum_{n=0}^{\infty}|c_{n}|^{2}\|p_{n}\|_{\alpha'}^{2}. \end{align*} We are almost done, but now we get that \begin{align*} \|f\|_{\alpha}^{2}=\lim_{\alpha'\rightarrow\alpha}\sum_{n=0}^{\infty}|c_{n}|^{2}\|p_{n}\|_{\alpha'}^{2}. \end{align*} The issue is to interchange the infinite sum with the limit. The trick is to use Monotone Convergence Theorem. First of all, the limit can be realized as \begin{align*} \|f\|_{\alpha}^{2}=\lim_{N\rightarrow\infty}\sum_{n=0}^{\infty}|c_{n}|^{2}\|p_{n}\|_{\alpha-1/N}^{2}, \end{align*} whereas \begin{align*} \|p_{n}\|_{\alpha-1/N}^{2}=\dfrac{\pi}{n+1}\left(\alpha-\dfrac{1}{N}\right)^{2(n+1)}, \end{align*} so the sequence $N\rightarrow\|p_{n}\|_{\alpha-1/N}^{2}$ is increasing, and we use Monotone Convergence Theorem to deduce that \begin{align*} \|f\|_{\alpha}^{2}=\lim_{N\rightarrow\infty}\sum_{n=0}^{\infty}|c_{n}|^{2}\|p_{n}\|_{\alpha-1/N}^{2}&=\sum_{n=0}^{\infty}|c_{n}|^{2}\lim_{N\rightarrow\infty}\|p_{n}\|_{\alpha-1/N}^{2}\\ &=\sum_{n=0}^{\infty}|c_{n}|^{2}\lim_{N\rightarrow\infty}\dfrac{\pi}{n+1}\left(\alpha-\dfrac{1}{N}\right)^{2(n+1)}\\ &=\sum_{n=0}^{\infty}|c_{n}|^{2}\dfrac{\pi}{n+1}\alpha^{2(n+1)}\\ &=\sum_{n=0}^{\infty}|c_{n}|^{2}\|p_{n}\|_{\alpha}^{2}, \end{align*} we are done.
Note that the uniform convergence of the power series expansion of $f$ on the closed disk $\overline{D}_{\alpha'}$ is not strong enough to guarantee the exchange of the infinite sum and the integral. But we make no use of this fact indeed, and we go through by Lebesgue Dominated Convergence Theorem.
Edit:
Uniform convergence of the power series expansion of $f$ on the closed disk $\overline{D}_{\alpha'}$ still helps in the interchange of the infinite sum with the integral, as the following shown:
Denote $s_{n}(z)=\displaystyle\sum_{k=0}^{n}c_{k}z^{k}$, the partial sums of the power series. Being uniform convergent simply means that for every $\epsilon>0$, some $N$ is such that \begin{align*} |s_{n}(z)-f(z)|<\epsilon,~~~~z\in\overline{D}_{\alpha'},~~~~n\geq N, \end{align*} for all such $z$ and $n$, it follows that \begin{align*} \left|s_{n}(z)\overline{f(z)}-f(z)\overline{f(z)}\right|=|f(z)||s_{n}(z)-f(z)|\leq M_{\alpha'}\epsilon, \end{align*} this shows that $s_{n}(z)\overline{f(z)}\rightarrow f(z)\overline{f(z)}$ uniformly, so \begin{align*} \iint_{\overline{D}_{\alpha'}}\sum_{n=0}^{\infty}c_{n}z^{n}\overline{f(z)}dxdy=\sum_{n=0}^{\infty}c_{\alpha}\iint_{\overline{D}_{\alpha'}}z^{n}\overline{f(z)}dxdy \end{align*} is allowed by the elementary calculus theorem regarding the uniform convergence.
Similarly, the series $z^{n}\displaystyle\sum_{m=0}^{\infty}\overline{c_{m}}\cdot\overline{z^{m}}$ is also uniformly convergent, once again the following is allowed: \begin{align*} \iint_{\overline{D}_{\alpha'}}z^{n}\sum_{m=0}^{\infty}\overline{c_{m}}\cdot\overline{z^{m}}dxdy=\sum_{m=0}^{\infty}\overline{c_{m}}\iint_{\overline{D}_{\alpha'}}z^{n}\overline{z^{m}}dxdy. \end{align*}