# Representation theorem for quadratic form on Hilbert space

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 (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 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.

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.

• Thanks for your answer! The only point not clear to me yet is the justification to use Riesz Representation Theorem to write $\Psi(x,y) = \langle A(x),y\rangle$. Don't you need $\Psi(x,y)$ to be linear in $y$ to each $x$? I haven't checked if this is the case, but it certainly doesn't seem to be the case from the properties of $Q$. Or the reason to use Riesz is something different? Aug 27, 2020 at 10:58
• The map $\Psi$ is sequilinear (linear in one argument, conjugate linear in the other), although it is probably a little painful to check. Aug 27, 2020 at 11:09
• Well, this must be it then. I'll try to prove it by myself. It is certainly a nice approach! Aug 27, 2020 at 11:24
• By the way, you also need sesquilinearity of $\Psi$ to see that the operator $A$ defined by your formula (2) is linear (to be precise, you only need linearity in the first argument, but conjugate linearity in the second argument then follows from the symmetry of formula (1)). Aug 27, 2020 at 13:09

I think this is a fine question for mathoverflow. There does indeed seem to be a convergence issue. However, it can be finessed by restricting to the span of some finite subset of the basis. Then we are working on a finite dimensional space and convergence is trivial. Next, use the uniqueness of $$A$$ to show that when we pass to a larger finite subset the values $$\langle Ax,x\rangle$$ do not change. We can also use (1) to get a uniform bound on the norms of the partial versions of $$A$$, so that they do ultimately yield a bounded operator on all of $$H$$.

• Hello Nik! Thanks for you answer again. I didn't comment it before because I didn't have time to elaborate the details on my own. But the idea is: take $\mathcal{E}_{k}:=\{e_{\alpha_{1}},...,e_{\alpha_{k}}\}$ a finite subset of the basis and consider $\mbox{span}\mathcal{E}_{k}$. Then, define $Ax$, as in my post, but only for those elements of $\mbox{span}\mathcal{E}_{k}$. Now, take $\mathcal{E}_{j} :=\{e_{\alpha_{1}},...,e_{\alpha_{j}}\}$ such that $\mathcal{E}_{k}\subseteq \mathcal{E}_{j}$. (continues) Aug 26, 2020 at 23:47
• Define $A$ on $\mathcal{E}_{j}$ as before. Then, if $x \in \mbox{span}\mathcal{E}_{k}$, say $x=\sum_{n=1}^{k}\alpha_{n}e_{\alpha_{n}}$, then $x \in \mbox{span}\mathcal{E}_{k}$, say $x = \sum_{n=0}^{j}\alpha_{n}e_{\alpha_{n}}$ where $\alpha_{n} = 0$ if $e_{\alpha_{n}} \not\in \mathcal{E}_{k}$. Then, I must show that the two definitions agree, using the representation of $x$ on $\mathcal{E}_{j}$. I think, so far, this is correct, right? Aug 26, 2020 at 23:53
• Next I shall get a uniform bound on $A$ and extend. This is the part where is not entirely clear to me yet. Aug 26, 2020 at 23:53
• I should probably take the bound $\sup_{x\in \mbox{span}\mathcal{E}_{k}}|\Psi(x,e_{\alpha})|$ to bound $A$ uniformly in each step? Aug 26, 2020 at 23:59
• Sure, no problem. Aug 27, 2020 at 2:29

Too long for a comment. Let: $$\tilde{\Psi}(x,y) := Q(x+y)-Q(x-y)$$

Fact 1: $$\tilde{\Psi}(x+z,y) = \tilde{\Psi}(x,y)+\tilde{\Psi}(z,y)$$, for every $$x,y,z \in H$$.

Proof: Let us evaluate the difference $$\tilde{\Psi}(x+z,y)-\tilde{\Psi}(x,y)-\tilde{\Psi}(z,y)$$. We have:

$$\tilde{\Psi}(x+z,y)-\tilde{\Psi}(x,y)-\tilde{\Psi}(z,y) = Q(x+z-y)-Q(x+z-y)-Q(x+y)+Q(x-y)-Q(z+y)+Q(z-y)$$

Now, note that: \begin{align} Q(x-y)-Q(z-y) &= \frac{1}{2}[Q(x-y+z-y)+Q(x-y-z+y)] \\ &= \frac{1}{2}[Q(x+z-2y)+Q(x-z)] \end{align} and also: \begin{align} Q(x+y)+Q(z+y) &= \frac{1}{2}[Q(x+y+z+y)+Q(x+y-z-y)] \\&= \frac{1}{2}[Q(x+z+2y)+Q(x-z)] \end{align}

Thus, we get: $$\tilde{\Psi}(x+z,y)-\tilde{\Psi}(x,y)-\tilde{\Psi}(z,y) = Q(x+z+y)-Q(x+z-y)+\frac{1}{2}Q(x+z-2y)-\frac{1}{2}Q(x+z+2y)$$

Moreover: \begin{align} Q(x+z+y)-\frac{1}{2}Q(\overbrace{x+z+2y}^{=x+z+y+y}) &= \frac{1}{2}Q(x+z+y-y)-Q(y) \\&= \frac{1}{2}Q(x+z)-Q(y) \end{align} and also: \begin{align} Q(x+z-y)-\frac{1}{2}Q(\overbrace{x+z-2y}^{=x+z-y-y}) &= \frac{1}{2}Q(x+z-y+y)-Q(y) \\&= \frac{1}{2}Q(x+z)-Q(y) \end{align} and this proves the result.

Fact 2: $$\tilde{\Psi}(-x,y) = -\tilde{\Psi}(x,y)$$

Fact 3: $$\tilde{\Psi}(y,x) = Q(y+x)-Q(y-x) = Q(x+y)-Q(x-y) = \tilde{\Psi}(x,y)$$

Fact 4: In particular, Fact 1 + Fact 2 lead to $$\tilde{\Psi}(kx,y) = k\tilde{\Psi}(x,y)$$ for every $$x,y \in H$$ and $$k \in \mathbb{Z}$$.

Fact 5: Let $$b \in \mathbb{Z}\setminus \{0\}$$. Then, $$\tilde{\Psi}(x,\frac{1}{b}y) = \frac{1}{b}\tilde{\Psi}(x,y)$$.

Proof: Note that: $$\tilde{\Psi}(x,\frac{1}{b}y) = Q(x+\frac{1}{b}y)-Q(x-\frac{1}{b}y) = \frac{1}{b^{2}}[Q(bx+y)-Q(bx-y)] = \frac{1}{b^{2}}\tilde{\Psi}(bx,y) = \frac{1}{b}\tilde{\Psi}(x,y)$$ where, in the last equality, I used fact 4.

Fact 6: $$\tilde{\Psi}(x,y+z) = \tilde{\Psi}(x,y)+\tilde{\Psi}(x,z)$$

Proof: By fact 3, $$\tilde{\Psi}(x,y+z) = \tilde{\Psi}(y+z,x) = \tilde{\Psi}(y,x)+\tilde{\Psi}(z,x) = \tilde{\Psi}(x,y)+\tilde{\Psi}(x,z)$$

Fact 7: Set $$\hat{\Psi}(x,y) := iQ(x+iy)-iQ(x-iy) = i\tilde{\Psi}(x,iy)$$. Then all the above facts also hold for $$\hat{\Psi}(x,y)$$.

Fact 8: $$|\Psi(x,y)|\le K(||x||^{2}+||y||^{2})$$ implies $$\Psi$$ is continuous in the product topology.

Now, according to MaoWao's answer, the result follows from the Riesz Representation Theorem if $$\Psi(x,y)$$ is sesquilinear. It is easy to see that $$-i\Psi(x,y) = \Psi(x,iy)$$. Finally, let $$\alpha = a+ib \in \mathbb{C}$$. Then, we have: $$\Psi(x,\alpha y) = \Psi(x,ay+iby) = \Psi(x,ay)-i\Psi(x,by)$$ Thus, find sequences $$\{a_{n}\}_{n\in \mathbb{N}}$$ and $$\{b_{n}\}_{n\in \mathbb{N}}$$ of rational numbers such that $$a_{n}\to a$$ and $$b_{n}\to b$$ and use the continuity of $$\Psi$$ to prove that it is anti-linear in the $$y$$ entry. The same reasoning leads us to the linearity in the $$x$$ entry. This, together with MaoWao's answer should be enough to prove the result.