Defining interpolation spaces of Hilbert spaces using domains of unbounded operators This comes from the book Non-Homogeneous Boundary Value Problems and Applications I by Lions and Magenes, section 2.1.

Let $X\subset Y$ be a dense continuous injection of separable complex Hilbert spaces. We will define a strictly positive self-adjoint densely-defined unbounded operator $S$ in $Y$ as follows.
Let $D(S)$ denote those $x\in X\subset Y$ such that $$X\to\Bbb C,\quad v\mapsto\langle u,v\rangle_X$$ is continuous w.r.t. the topology on $X$ induced by $Y$. Then we may define an operator $S:D(S)\to Y$ by setting $$\langle u,v\rangle_X=\langle Su,v\rangle_Y$$ for all $v\in X$, which uniquely defines $Su$ by density.

Now, the authors state the following (without proof or explanation):
Proposition For $S$ defined as such, $D(S)$ is dense in $Y$, and furthermore that $S$ is self-adjoint. Using the spectral theorem for unbounded self-adjoint operators, if we set $\Lambda=S^{1/2}$, then $D(\Lambda)=X$. 
However, I haven't managed to figure out the proof of any of these claims.
For  general background, we have the following equivalence.

We say that a densely-defined unbounded operator $S:D(S)\to Y$ on $Y$ is symmetric if it is closable and $\langle Su,v\rangle_Y=\langle u,Sv\rangle_Y$ for all $u,v\in D(S)$. Then for any symmetric operator $S$, the following are equivalent.
  
  
*
  
*The operator $S$ is self-adjoint.
  
*The operator $S$ is closed, and $\ker(S\pm i)=0$ as subspaces of $D(S)$.
  
*We have $\operatorname{im}(S\pm i)=Y$ where the image is of $D(S)$.
  
  
  Equivalently, we may replace $i,-i$ above with $\lambda,\bar\lambda$ for any strictly complex $\lambda$.

Now, assuming that $D(S)\subset Y$ is dense and $S$ is closable, I see that $S$ is symmetric. Furthermore, for any $v\in D(S)$, we have that $$\langle Sv,v\rangle_Y=\lVert v\rVert_X\ge C\lVert v\rVert_Y$$ for some fixed $C>0$. This tells us that $S$ is injective, and furthermore that is $S$ if closed, then it is self-adjoint, since for any $\lambda\in\Bbb C\setminus\Bbb R$ with $|\lambda|<C$, if $Sv=\lambda v$ then $$0=|\langle Sv,v\rangle_Y-\langle\lambda v,v\rangle_Y|\ge C\lVert c\rVert_Y-|\lambda|\lVert v\rVert_Y$$ so that $v=0$. However, showing that $S$ is closed or even closable is beyond me.
Any help with approaching a proof of the proposition would be greatly appreciated.
 A: Lions and Magenes could  have made life easier by adding some details.
A  crucial point is that $S: D(S) \rightarrow Y$ is 
 a bijection onto $Y$ and as is often the case, 
 more easily studied by considering its inverse.
The issue becomes more clear if you introduce explicitly
an operator
$j:X \rightarrow Y$ for the continuous, dense injection.
Then the adjoint $j^* : Y \rightarrow X$ verifies:
$$ \langle y, jx \rangle_Y = \langle j^* y, x \rangle_X , \;  \forall
   x\in X, y\in Y .$$
For completeness recall the construction of the adjoint: For $y\in Y$, the subspace 
$$ N(y)  = \{ z\in X : \langle y, j z\rangle_Y = 0 \}$$
 is closed (j is continuous) and proper iff $y\neq 0$ (j has dense image). 
 It follows that there is a unique element $u=j^*(y)  \in N(y)^\perp$ for which:
 $$ \langle u,u \rangle_X = \langle y,ju \rangle_Y$$
As every $x\in X$ may be written: $x=tu + z$, $t\in {\Bbb C}$, $z\in N(y)$ you obtain:
 $$ \langle u,x \rangle_X =\langle u, tu \rangle_X
 = \langle y, j(tu)\rangle_Y = \langle y, jx \rangle_Y, \forall x\in X $$
This property also characterizes
$u$ as the unique element verifying the above. 
From the 
characterization we see that $j^*$
is linear.
 Furthermore, it is injective because $j$ has dense image and it has dense image 
 because $j$ is injective:
 To see e.g. the latter note that $v\in (j^*(Y))^\perp$ implies $jv \in Y^\perp$
 so $jv$ (whence also $v$) must be the zero-vector.
As already mentioned $j^*(Y)$ is dense in $X$ and mapping this back again into $Y$
we see that
$D = D(S) = j j^* (Y)$ is dense in $Y$. The map:
$S =(jj^*)^{-1}: D(S) \rightarrow  Y$ 
is injective and maps $D(S)$ onto $Y$.
The fact that $S$ is closed comes almost for free:
 For a sequence $(v_n)_n$ in $D(S)\subset Y$:
  $$ v_n  \rightarrow v\in Y, \; y_n = S v_n \rightarrow y\in Y$$
is equivalent to 
  $$ j j^* y_n = v_n \rightarrow v, \; y_n \rightarrow y$$
But then $j j^*y_n \rightarrow j j^*y$ by continuity and $ jj^*y=v$
(whence $y=Sv$ by injectivity) so $S$ is closed. 
Also for $u_1,u_2\in D(S)$ we have $u_1=jj^*y_1, u_2=jj^* y_2$ for some $y_1,y_2\in Y$
and then symmetry of $S$ follows from:
$$
\langle Su_1,u_2 \rangle_Y =
\langle y_1,jj^* y_2 \rangle_Y =
\langle j^*y_1,j^* y_2 \rangle_X =
\langle jj^*y_1,y_2 \rangle_Y =
\langle u_1,S u_2 \rangle_Y . $$
Finally, as $S$ maps $D(S)$ onto $Y$ it
is selfadjoint.
For $x=j^*y$, $y\in Y$ and $u\in X$ we have:
$$ \langle S jx,ju \rangle_Y = \langle S jj^* y,ju\rangle_Y=
\langle y,ju\rangle_Y=\langle j^*y,u\rangle_X = \langle x,u\rangle_X$$
Thus, $\langle S jx,jx \rangle_Y = \|x\|_X^2 \geq C^2 \|jx\|_Y^2$. This implies that $S$ is strictly positive, whence has a square-root by the spectral theorem (I don't think there is any shortcut to this). So there is a unique self-adjoint operator $\Lambda=S^{1/2}$ with a domain of definition $\Omega\subset Y$ that consists of precisely those $y$ for which $\|\Lambda y\|_Y^2 = \langle Sy,y\rangle_Y$ is finite. The domain contains in particular $D(S)$ and the previous identity shows that for $x=j^*y$, $y\in Y$ we have:
$$ \langle S jx,jx\rangle_Y^{1/2} = \|\Lambda jx\|_Y=\|x\|_X \geq C \|jx\|_Y$$
Since $j^* Y$ is dense in $X$ this identity extends by continuity to all of $x\in X$. As $S$ has dense image, so does $\Lambda$ (on $D(S)$) and finally $\Lambda^{-1}$ extends to an isomorphism of $Y$ onto $jX$, which is therefore the domain of $\Lambda$. [For this last non-trivial part you should check with the spectral theorem for unbounded operators]
