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.

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.

• Thanks for the answer! So, if $x = \sum_{\alpha \in I}x_{\alpha}e_{\alpha}$, I know that this sum is actually a countable sum because only countable many $\langle x, e_{\alpha}\rangle$ are nonzero. But I'm confused about the $Ax$ term. I need to prove that the sum $\sum_{\alpha \in I}\Psi(x,e_{\alpha})e_{\alpha}$ converges to define it as an element $z = Ax$ in $H$, right? But, what's the apropriate notion of convergence here? I mean...at first, the sum need not to be countable, right? It is countable if you know $Ax \in H$ but you can only garantee that if you prove it converges Commented Aug 25, 2020 at 4:08
• Yes, and that is the gist of my sketch of the proof. The convergence is in $H$. But you know that $\sum_{\alpha \in I}|\langle x_{\alpha},e_{\alpha}\rangle |^2$ converges (Bessel) so it's basically the comparison test in the scalar field (bounding $|\Psi(x,e_{\alpha})|$, as above). btw: your second question is a slog through some substitutions, but you really only have three equations to work with. Commented Aug 25, 2020 at 4:29
• Yes, this is something I was about to ask you too. Is my first sketch correct. I mean, should I start with $\langle Ax, y\rangle = \sum_{\alpha \in I}\Psi(x,e_{\alpha})y_{\alpha}$ to get the result? Commented Aug 25, 2020 at 4:42
• Here is a primer on Hilbert Spaces. For your question, start by writing $\langle Ax, y\rangle = \langle\sum_{\alpha\in I}\Psi(x,e_{\alpha})e_{\alpha},y_{\alpha}\rangle=\langle\sum_{\alpha\in I}\Psi(x,e_{\alpha})e_{\alpha},\sum_{\beta\in I}\langle y_{\beta},e_{\beta}\rangle e_{\beta}\rangle=\sum_{\alpha\in I}\Psi(x,e_{\alpha})\langle y_{\alpha},e_{\alpha}\rangle$. Commented Aug 25, 2020 at 11:34
• Right! Furthermore, I just realized: because $|\Psi(x,e_{\alpha})| \le C'|\langle x, e_{\alpha}\rangle|$, only finitely many $\Psi(x,e_{\alpha})$ are nonzero, so that $Ax$ is defined by means of a countable sum as well. Commented Aug 25, 2020 at 14:47

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 $$$$\tag{1} \Psi_1(-x,y) = -\Psi_1(x,y)$$$$ $$$$\tag{2} \Psi_1(x+z,y) = \Psi_1(x,y) + \Psi_1(z,y)$$$$

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 $$A$$.