Sesquilinear Forms: Hamiltonian (II)

Question

Given a Hilbert space $\mathcal{H}$.

Consider a positive form: $$s:\mathcal{D}\to\mathcal{H}:\quad s(\varphi,\varphi)\geq0$$

Introduce its form space: $$\mathcal{H}_s:=\mathcal{D}:\quad\langle\varphi,\psi\rangle_s:=s(\varphi,\psi)+\langle\varphi,\psi\rangle$$

Then it is closable: $$s\subseteq\hat{s}:\iff\mathcal{H}_s\subseteq\mathcal{H}_\hat{s}:=\hat{\mathcal{H}}_s\subseteq\mathcal{H}$$

Is there a sophisticated way?

Answer 1

Suppose the form is given by: $$s(\varphi,\psi)=\langle A\varphi,\psi\rangle=\langle\varphi,A\psi\rangle$$

Regard the embedding: $$\iota:\mathcal{H}_s\to\mathcal{H}:\varphi\mapsto\varphi$$

Clearly it is bounded: $$\|\iota(\varphi)\|=\|\varphi\|\leq\|\varphi\|_s$$

So it extends to: $$\hat{\iota}:\hat{\mathcal{H}_s}\to\mathcal{H}:\quad\|\hat{\iota}\|<\infty$$

By continuity one has: $$\hat{\iota}(\varphi)=0:\quad\varphi_n\to\varphi\implies\hat{\iota}(\varphi_n)\to0$$

By density one can choose: $$\varphi_n\in\mathcal{H}_s:\quad\hat{\iota}(\varphi_n)=\iota(\varphi_n)=\varphi_n$$

So it must vanish: $$\|\varphi\|_s\leftarrow\langle\varphi,\varphi_n\rangle_s=\lim_m\{\langle \varphi_m,A\varphi_n\rangle+\langle\varphi_m,\varphi_n\rangle\}=0\to0$$

Thus it embeds.

Answer 2

ATTENTION: THIS PROOF HAS A FLAW!!!

Failure

It is not guaranteed: $$\|\varphi_m-\varphi_n\|_s\to0\nRightarrow\|\varphi-\varphi_n\|_\hat{s}\to0$$ So the norm is ill-defined.

Proof

It holds the estimate: $$\|\varphi\|^2\leq s(\varphi,\varphi)+\|\varphi\|^2=\|\varphi\|_s$$

That offers explicitely: $$\|\varphi_m-\varphi_n\|_s\to0\implies\|\varphi_m-\varphi_n\|\to0\implies\|\varphi-\varphi_n\|\to0$$

So complete it by: $$\mathcal{H}_\hat{s}:\quad\langle\varphi,\psi\rangle_\hat{s}:=\lim_n\langle\varphi_n,\psi_n\rangle_s$$

That limit exists since: $$0\leq|\langle\varphi_m,\psi_m\rangle_s-\langle\varphi_n,\psi_n\rangle_s|\leq\|\varphi_m-\varphi_n\|_s\|\psi_m\|_s+\|\varphi_n\|_s\|\psi_m-\psi_n\|_s\to0$$

And is well-defined since: $$0\leq|\langle\varphi_n,\psi_n\rangle_s-\langle\tilde{\varphi}_n,\tilde{\psi}_n\rangle_s|\leq\|\varphi_n-\tilde{\varphi}_n\|_s\|\psi_n\|_s+\|\tilde{\varphi}_n\|_s\|\psi_n-\tilde{\psi}_n\|_s\to0$$

Noting the estimate: $$0\leq\|\varphi_n-\tilde{\varphi}_n\|_s\leq\|\varphi-\varphi_n\|_s+\|\tilde{\varphi}-\tilde{\varphi}_n\|_s\to0$$

But algebraic properties remain: $$\hat{s}(\varphi,\psi):=\langle\varphi,\psi\rangle_\hat{s}-\langle\varphi,\psi\rangle:\quad\hat{s}(\varphi,\varphi)=\|\varphi\|_\hat{s}^2-\|\varphi\|^2\geq0$$

So one obtains the desired form.

Answer 3

Given the Hilbert space $\mathcal{L}^2(\mathbb{R})$.

Consider the positive form: $$s:\mathcal{C}^\infty_0(\mathbb{R})\times\mathcal{C}^\infty_0(\mathbb{R})\to\mathcal{L}^2(\mathbb{R}):\quad s(\varphi,\psi):=\overline{\varphi(0)}\psi(0)$$

Regard a sequence of bumps: $$\varphi_n\in\mathcal{C}^\infty_0(\mathbb{R}):\quad\varphi_n\left(|x|\geq\tfrac{2}{n}\right)\equiv0\leq\varphi_n\left(\tfrac{1}{n}\leq|x|\leq\frac{2}{n}\right)\leq1\equiv\varphi_n\left(|x|\leq\tfrac{1}{n}\right)$$

Contradictory one has: $$\|\varphi_m-\varphi_n\|_s\to0\quad\|0-\varphi_n\|_s\not\to0$$

So the form is not closable!