I think my question is more suitable for Mathematics Stack Exchange than to MathOverflow but I've already posted two related questions there and I got even more confused, so maybe I can clarify things here. I'm studying spectral theory by myself as part of my research activity and the following question arose.


Let $H$ be a Hilbert space and $Q: \mathcal{H}\to \mathbb{C}$ a function such that:


(1) There exists $C>0$ such that $|Q(x)| \le C||x||^{2}$ for every $x\in H$


(2) $Q(x+y)+Q(x-y) = 2Q(x) + 2Q(y)$ for every $x,y \in H$ and


(3) $Q(\lambda x) = |\lambda|^{2}Q(x)$ for every $x \in H$ and $\lambda \in \mathbb{C}$.


**Question:** Is there some bounded linear operator $A \in H$ such that $Q(x) = \langle Ax, x\rangle$?


The answer to this question seems to be affirmative and a sketch of a possible approach is given [here](http://msekce.karlin.mff.cuni.cz/%7Evybiral/FAII/FAII_Skript.pdf) (page 7, Lemma 12.2.7). The idea is to define:
\begin{eqnarray}
\Psi(x,y) = \frac{1}{4}[Q(x+y)-Q(x-y)+iQ(x+iy)-iQ(x-iy)] \tag{1}\label{1}
\end{eqnarray}
where $\{e\_{\alpha}\}\_{\alpha \in I}$ is an orthonormal basis of $H$ and then define $A$ by means of the rule:
\begin{eqnarray}
Ax = \sum\_{\alpha \in I}\Psi(x,e\_{\alpha})e\_{\alpha}\tag{2}\label{2}
\end{eqnarray}
But I'm puzzled with this approach since I was not able to prove that $\sum\_{\alpha \in I}\Psi(x,e\_{\alpha})e\_{\alpha}$ converges in the first place. All I could prove was $|\Psi(x,y)| \le K(||x||^{2}+||y||^{2})$ for some $K > 0$. As you can see [in my previous post](https://math.stackexchange.com/questions/3803175/does-psix-y-le-k-langle-x-y-rangle-hold) on math stack, it seems that the convergence problem is a bit tricky indeed.


In summary: I don't know how to prove that (\ref{2}) converges and, thus, I don't quite understand the proof of the result. However, I believe it's possible to find a more direct proof, maybe using Riesz Representation Theorem ideas (although $Q$ here is not linear) or something like that. I'd appreciate any help on either way.



Representation theorem for quadratic form on Hilbert space
There is indeed a simple proof using the Riesz representation theorem. First note that replacing $x$ by $\lambda^{-1}x$ and $y$ by $\lambda y$ in $\lvert \Psi(x,y)\rvert\leq K(\lVert x\rVert^2+\lVert y\rVert^2)$, you get $\lVert \Psi(x,y)\rvert\leq K(\lambda^{-2}\lVert x\rVert^2+\lambda^2\lVert y\rVert^2)$. With $\lambda=\lVert x\rVert^{1/2}\lVert y\rVert^{-1/2}$ this gives
$$
\lvert \Psi(x,y)\rvert\leq 2K\lVert x\rVert \lVert y\rVert.
$$
Thus for every $x\in H$ there exists $A(x)\in H$ such that $\Psi(x,y)=\langle A(x),y\rangle$ for $y\in H$ by the Riesz representation theorem. Since $\Psi$ is sesquilinear, the map $x\mapsto A(x)$ is linear, and moreover,
$$
\lVert A(x)\rVert=\sup\_{\lVert y\rVert=1}\lvert \Psi(x,y)\rvert\leq 2K\lVert x\rVert,
$$
so that $A$ is also bounded.


