Representation theorem for quadratic form on Hilbert space

Question

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.

Answer 1

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.

Answer 2

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 $H$.

Answer 3

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.