Defining interpolation spaces of Hilbert spaces using domains of unbounded operators

Question

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.

1. The operator $S$ is self-adjoint.
2. The operator $S$ is closed, and $\ker(S\pm i)=0$ as subspaces of $D(S)$.
3. 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.

Answer 1

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]