Proving the existence of a bounded linear operator $A$ such that $Q(x) = \langle Ax, x\rangle$ I'm following this lecture notes on functional analysis where Lemma 12.2.7 (page 7) states the following:
Lemma: Let $H$ be a complex Hilbert space and $Q: H \to \mathbb{C}$ be a function. Then, the following are equivalent:
(i) There exists exactly one operator $A \in \mathcal{L}(\mathcal{H})$ such that $Q(x) = \langle Ax, x\rangle$ for all $x \in H$
(ii) There is a constant $C>0$ such that $|Q(x)| \le C||x||^{2}$, $Q(x+y)+Q(x-y)=2Q(x)+2Q(y)$ and $Q(\lambda x) = |\lambda|^{2}Q(x)$ for all $x,y \in H$ and $\lambda \in \mathbb{C}$.
Now, I'm interested in the implication (ii) $\Rightarrow$ (i). The reasoning is sketched in the notes but I don't seem to understand the proof. The first step is to consider $\Psi(x,y)$, defined by:
$$\Psi(x,y) = \frac{1}{4}[Q(x+y)-Q(x-y)+iQ(x+iy)-iQ(x-iy)]$$
and then define $A$ by setting:
\begin{eqnarray}
Ax := \sum_{\alpha \in I}\Psi(x,e_{\alpha})e_{\alpha} \tag{1}\label{1}
\end{eqnarray}
where $\{e_{\alpha}\}_{\alpha \in I}$ is a Hilbert basis.
Question 1: What does the sum (\ref{1}) mean? I mean, what is the notion of convergence behind it? Is it well-defined (i.e. does this sum always converge?)
Question 2: Supposedly, one must have $\langle Ax,y\rangle = \Psi(x,y)$, but I don't know how this follows. I think the approach here would be: take $y \in H$ and write $y = \sum_{\alpha \in I}y_{\alpha}e_{\alpha}$ (this sum is actually a countable one). Then, I think the idea is to write:
$$\langle Ax, y\rangle = \sum_{\alpha\in I}\Psi(x,e_{\alpha})y_{\alpha}$$
but how come does the latter become $\Psi(x,y)$?
Note: I added the tag 'alternative-proof' because I'd be happy to see an alternative (and easier) proof, if somebody knows one.
 A: I am assuming that $\{e_{\alpha}\}_{\alpha}$ is orthonormal and that $x=\sum_{\alpha\in I}\langle x,e_{\alpha}\rangle e_{\alpha}.$ The sum has only countably many non-zero terms (why?) so we may take it to be countable.
If you recognize that the definition of $\Psi$ is through an analogue of the polarization identity,, then the way forward falls out. The key fact is that
$\displaystyle \langle x,\ y\rangle ={\frac {1}{4}}\left(\|x+y\|^{2}-\|x-y\|^{2}+i\|x-iy\|^{2}-i\|x+iy\|^{2}\right)$,
Now we have
$\Psi(x,y) = \frac{1}{4}|[Q(x+y)-Q(x-y)+iQ(x+iy)-iQ(x-iy)]$
Using the conditions on $Q$, you can show that $|\Psi(x,e_{\alpha})|\le C'|\langle x,e_{\alpha}\rangle|$ for some constant $C'$.
It follows that $\sum_{\alpha \in I}\Psi(x,e_{\alpha})e_{\alpha}$ converges. The rest of the exercise is a calculation.
A: I will try to prove $(ii) \implies (i)$ with an additional assumption that $Q$ is continuous. The condition $(i)$ implies continuity of $Q$, so it must hold anyways, but I have not managed to prove it from the conditions of $(ii)$.
WLOG we can assume that $Q(x) \in \mathbb{R}$ for each $x \in H$. Otherwise consider $Q_1 = \operatorname{Re} Q$ and $Q_2 = \operatorname{Im} Q$, find $A_i$ such that $Q_i(x) = \langle A_i x,x \rangle$. Then $$Q(x) = Q_1(x) + iQ_2(x) = \langle A_1 x,x \rangle + i \langle A_2 x,x \rangle = \langle (A_1 + i A_2) x,x \rangle.$$
We will prove that $(x,y) \mapsto \Psi(x,y)$ is bounded, linear in $x$ and conjugate linear in $y$ and proceed with the Lax-Milgram theorem.
Consider $\Psi_1(x,y) = \frac{1}{4} (Q(x+y) - Q(x-y))$. Then
\begin{equation}\tag{1}
\Psi_1(-x,y) =  -\Psi_1(x,y)
\end{equation}
\begin{equation}\tag{2}
\Psi_1(x+z,y) = \Psi_1(x,y) + \Psi_1(z,y)
\end{equation}
By $(1)$, $(2)$ and mathematical induction, $\Psi_1$ is $\mathbb{Q}$-linear in $x$. Analogically $\Psi_1$ is $\mathbb{Q}$-linear in $y$. By continuity of $Q$ it is also $\mathbb{R}$-linear in $x$ and $y$.
Now consider $\Psi(x,y) = \Psi_1(x,y) + i \Psi_1(x,iy)$, which is the $\Psi$ defined in the question. Then $\Psi$ is $\mathbb{R}$-linear in $x$ and $y$. To show it is also ($\mathbb{C}$-)linear in $x$ and conjugate linear in $y$, we just need to show that $\Psi(ix,y) = i \Psi(x,y)$ and $\Psi(x,iy) = -i\Psi(x,y)$, which holds by straightforward computation.
By $(ii)$ you also have $|\Psi(x,y)| \leq K (||x||^2 + ||y||^2)$.
To sum up, $\Psi$ is bounded, linear in $x$ and conjugate linear in $y$. By Lax-Milgram theorem there is $A \in L(H)$, such that $\Psi(x,y) = \langle Ax,y \rangle$ for each $x,y 
 \in H$. Now what is left is just to prove the equality $Q(x) = \Psi(x,x)$.
If you are not familiar with Lax-Milgram theorem, it is basically the procedure you used in your previous question on Borel calculus. Just use the Riesz reprezentation theorem twice to get the operator $A$.
