TITLE: Does $|\Psi(x,y)| \le K|\langle x,y\rangle|$ hold?
QUESTION [2 upvotes]: This question follows my previous question. 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}$$
(2) $Q(x+y)+Q(x-y) = 2Q(x) + 2Q(y)$ for every $x,y \in H$ and
(3) $Q(\lambda x) = |\lambda|^{2}Q(x)$ for every $x \in H$ and $\lambda \in \mathbb{C}$.
For each $x,y \in H$, let:
\begin{eqnarray}
 \Psi(x,y) = \frac{1}{4}[Q(x+y)-Q(x-y)+iQ(x+iy)-iQ(x-iy)] \tag{1}\label{1}
\end{eqnarray}
In the previous post, I got an answer in which the following estimate is used:
$$|\Psi(x,e_{\alpha})| \le K|\langle x, e_{\alpha}\rangle|$$
where $\{e_{\alpha}\}_{\alpha \in I}$ is an orthonormal basis on $H$. And I cannot prove such inequality, since all estimates of $Q$ are given in terms of norms. What I can prove is the following:
\begin{eqnarray}
|\Psi(x,y)| \le K(||x||^{2}+||y||^{2}) \Rightarrow |\Psi(x,e_{\alpha})| \le K'(||x||^{2}) \tag{2}\label{2}
\end{eqnarray}
However, my intetion with (\ref{1}) is to prove the convergence of the series $\sum_{\alpha \in I}\Psi(x,e_{\alpha})e_{\alpha}$ in $H$, so that my estimate (\ref{2}) is not very useful. I believe that, if (\ref{1}) really holds, it might be the particular case of the following:
\begin{eqnarray}
|\Psi(x,y)| \le K|\langle x, y\rangle| \tag{3}\label{3}
\end{eqnarray}
Question: Does (\ref{3}) (or, at least (\ref{1})) really hold? And how to prove it?

REPLY [1 votes]: My ideas do not fit in the comment section, so I am posting them as an answer. But these are really only my thoughts on this exercise, to which I offered an answer. Alas, the problem is harder, at least for me, than I previously thought. My apologies. My reasoning was the following: Let's assume $H$ is real and $Q\ge 0$ to simplify the problem. Then, except for the triangle equality, $\sqrt Q$ is a norm that satisfies the parallelogram identity, so it induces an inner product on $H$, which is precisely the definition of $\Psi$ via a polarization identity. I then thought the estimate would fall out by comparing $\Psi$ to the polarization identity for $\|\cdot\|,$ as in my answer. For example, if one had $Q(x)=C\|x\|$ the result would be a triviality. Or if $\Psi$ preserved orthonomality. Or even if $Q(x)=Q(y)$  whenever $\|x\|=\|y\|.$ Note that as $Q$ is not given continuous, (but it is continuous at $x=0,)\ \Psi$ satisfies all the properties of inner product except that the equation  $\Psi(rx,z)=r\Psi(x,z)$ is known to hold only for rational numbers $r$. But we do have $\Psi(x+y,z)=\Psi(x,z)+\Psi(y,z).$ Now, let's suppose that $Q$ is continuous and set  $\Psi(e_i,e_j)=c_{ij}$. Then,
$|\Psi(x,e_j)|=|\Psi(\sum_i\langle x,e_i\rangle e_i,e_j)|=|\sum_i\Psi(\langle x,e_i\rangle e_i,e_j)|\le |\langle x,e_j\rangle| \sum_i|\Psi(e_i,e_j)|=|\langle x,e_j\rangle| \sum_i|c_{ij}|.$
Thus, it comes down to proving convergence of $\sum_i|c_{ij}|$, and so we need to consider $\Psi(e_i,e_j) = \frac{1}{4}(Q(e_i+e_j)-Q(e_i-e_j)$. Now, $\Psi(e_j,e_j)=Q(e_j)$, but if $i\neq j$, I can't see how the data gives us enough information to control the $c_{ij}$.
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