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$\mathcal{H}$: Real hilbert space with inner product $(\cdot,\cdot)$. $D$ is a subspace of $\mathcal{H}$. Let $\mathcal{E}$ with domain $D$ be a positive definite (i.e. for all $u \in D $, $\mathcal{E}(u,u)\geq0$) bilinear form on $\mathcal{H}$.

For all $u,v$ in $D$, we set \begin{eqnarray*} \mathcal{\tilde{E}}(u,v)&=&\frac{1}{2}(\mathcal{E}(u,v)+\mathcal{E}(v,u))\\ \mathcal{\tilde{E_{1}}}(u,v)&=&\mathcal{\tilde{E}}(u,v)+(u,v) \end{eqnarray*} Then $\mathcal{\tilde{E_{1}}}(\cdot,\cdot)$ is the inner product on $D$. Therefore $(D,\mathcal{\tilde{E_{1}}}(\cdot,\cdot))$ is the pre-Hilbert space.

We denote its completion (w.r.t $\mathcal{\tilde{E_{1}}}^{1/2}$) by $\bar{D}$ equipped with (the unique extention of) $\mathcal{\tilde{E_{1}}}$.

$\underline{\rm{Claim1}}$

There is a unique continuous map $i:\bar{D}\to \mathcal{H}$ extending the inclusion map $D\subset \mathcal{H}$.

$\underline{\rm{Proof}}$

Since $\|u\|\leq \mathcal{\tilde{E_{1}}}(u,u) $ for all $u \in D$, the inclusion map $D\subset \mathcal{H}$ is uniform continuous on $D$.

$\underline{\rm{Claim2}}$

$(\mathcal{E},D)$ is $closable$* if and only if $i:\bar{D}\to \mathcal{H}$ is injective.

*Positive definite linear operator $(\mathcal{E},D)$ is called $closable$ (on $\mathcal{H}$) if for all $(u_{n})_{n=1}^{\infty}\subset D$, $\mathcal{E}(u_{n}-u_{m},u_{n}-u_{m})\to 0 $ as $n,m\to \infty$ and $u_{n}\to0$ in $\mathcal{H}$, it follows that $\mathcal{E}(u_{n},u_{n})\to 0$ as $n\to \infty$

I want to prove Claim2, but I don't know where to start.　Please give me a hint.

ko4
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1 Answers1

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The elements of $\overline{D}$ are equivalence classes of $\mathcal{\tilde E_1}$-Cauchy sequences $(x_n)$ with respect to $\mathcal{\tilde E_1}$. Note that any such sequence is Cauchy in $\mathcal H$, hence has a limit in $x\in\mathcal H$. However, it may happen that two $\mathcal{\tilde E_1}$-inequivalent sequences have the same limit, which means precisely when the inclusion $i:\overline{D}\to \mathcal H$ is not injective.

So, if $i:\overline{D}\to \mathcal H$ is not injective, there are two $\mathcal{\tilde E_1}$-Cauchy sequences that are $\mathcal{\tilde E_1}$-inequivalent but have the same limit in $\mathcal H$. Let $u_n$ be their difference: then $u_n\to 0$ in $\mathcal H$, the sequence is $\mathcal{\tilde E_1}$-Cauchy, but $\mathcal{E}(u_n,u_n)$ does not go to zero.

The converse is similar. If the operator is not closable, then we have $(u_n)$ as in the definition of closable, but with $\mathcal E(u_n,u_n)\not\to0$. This sequence determines a nonzero element of $\overline{D}$ that is mapped to $0$ by $i$.