# Does $|\Psi(x,y)| \le K|\langle x,y\rangle|$ hold?

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?

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 $$c_{ij}$$.