# 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.

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.

• I might miss something, but.. why wouldn't $v\mapsto \langle u,v\rangle_X$ be continuous for each $u\in X$? – Berci Aug 30 '17 at 21:26
• Because the $Y$-induced topology on $X$ is weaker than the original topology. – Monstrous Moonshine Aug 30 '17 at 21:39
• If it's any help, this part of Lions and Magenes is here. – Keith McClary Sep 4 '17 at 4:56

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]

• Thanks for that. However, I still don't see why we should have that $D(\Lambda)=X$ where $\Lambda:=S^{1/2}$. Equivalently, why should we have that $$\sqrt{j j^*}:Y\to Y$$ has image $jX\subset Y$? – Monstrous Moonshine Sep 6 '17 at 17:38
• While I'm familiar with the spectral theorem for unbounded operators, I still don't quite grasp the step where you show that $\Lambda$ can be defined on all of $jX$, or why its domain should lie in $jX$. – Monstrous Moonshine Sep 9 '17 at 1:58
• For the first it's by closure, i.e. if $x_n\rightarrow x$ in $X$, then $jx_n \rightarrow j x$ (automatic) and $\Lambda j x_n \rightarrow y$ (by the above-mentioned identity, for some $y$) so $(jx,y)$ lies in the closure of the graph of $\Lambda$, whence $jx$ is in the domain. To see that all $y\in D(\Lambda)$ is of this form, note that $\Lambda j$ maps $X$ (which is complete) unitarily onto a dense set in $Y$ which must be complete whence equal $Y$. As $\Lambda$ must be injective on its domain, the domain cannot be larger than $jX$. – H. H. Rugh Sep 9 '17 at 5:55
• Thanks, I fully understand your answer now. However, I have one more question (which is nonessential to the proof): how do we know that for any positive self-adjoint unbound operator $S:Y\to Y$, there exists a unique power $S^\lambda$ (which is also self-adjoint and positive, and with certain continuity properties w.r.t. $\lambda$) for $\lambda\ge0$? The reference I used only gives existence, not uniqueness of such powers. – Monstrous Moonshine Sep 9 '17 at 18:06
• Not sure. For the square root of bounded positive operators you may look in Rudin, Functional Analysis, section 12.32 This is ok here as $S$ is strictly positive, so the inverse of $S$ is bounded and you conclude from the inverse. I don't have a reference for uniqueness in the general positive unbounded case, but it probably follows from the unique family of spectral measures associated with the operator and $S^\lambda$ must correspond to $t^\lambda$ on the spectrum. But there are clearly details to fill in (notably for the domain). Continuity properties are not obvious when unbounded. – H. H. Rugh Sep 9 '17 at 19:45