7
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I might miss something, but.. why wouldn't $v\mapsto \langle u,v\rangle_X$ be continuous for each $u\in X$? $\endgroup$ – Berci Aug 30 '17 at 21:26
  • $\begingroup$ Because the $Y$-induced topology on $X$ is weaker than the original topology. $\endgroup$ – Monstrous Moonshine Aug 30 '17 at 21:39
  • $\begingroup$ If it's any help, this part of Lions and Magenes is here. $\endgroup$ – Keith McClary Sep 4 '17 at 4:56
1
+50
$\begingroup$

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]

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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 $jX\subset Y$? $\endgroup$ – Monstrous Moonshine Sep 6 '17 at 17:38
  • $\begingroup$ 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 $jX$. $\endgroup$ – Monstrous Moonshine Sep 9 '17 at 1:58
  • $\begingroup$ 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 $jX$. $\endgroup$ – H. H. Rugh Sep 9 '17 at 5:55
  • $\begingroup$ 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. $\endgroup$ – Monstrous Moonshine Sep 9 '17 at 18:06
  • $\begingroup$ 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. $\endgroup$ – H. H. Rugh Sep 9 '17 at 19:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.