Completion, injection $\mathcal{H}$: 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.
 A: 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 $i$.
