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If I have a self-adjoint operator $U : \operatorname{dom}(U) \subset H \rightarrow H$ and $\lambda \in \rho(U)$, then I assume assume that it is correct that the operator $(U - \lambda I)^{-1} \in L(H)$ is also self-adjoint?(for $\lambda \in \mathbb{R})

I mean, if $U$ would have been bounded, then I guess this would have been clear, as $(U- \lambda I)$ would have been self-adjoint and $(T^*)^{-1}= (T^{-1})^*$ holds, but in this more general setting, I am not so sure anymore.

Edit: So one question, if I have a self-adjoint operator $T$ (unbounded) and I have $((T-\lambda)^{-1})^{*}$ where $\lambda \in \rho(T)$, then it is clear that $(T-\overline{\lambda})^{-1}$ is also a bounded operator. So if I now show that $(f,g) \in G(((T-\lambda)^*)^{-1}) \Leftrightarrow (g,f) \in G((T-\lambda)^*) \text{ and }\forall (h,k) \in G((T-\lambda)): \langle k,g \rangle = \langle h,f \rangle \Leftrightarrow (f,g) \in G(((T-\lambda)^{-1})^{*}) \text{ and } \forall (k,h) \in G((T-\lambda)^{-1}) : \langle k,g \rangle = \langle h,f\rangle$, then this is the full proof?

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  • $\begingroup$ If $U$ is selfadjoint then $(U-\lambda I)^{-1}$ has adjoint $(U-\overline{\lambda}I)^{-1}$. So the resolvent is normal, but is not selfadjoint unless $\lambda$ is real. And there may not be any real $\lambda$ in the resolvent if $U$ is unbounded. $\endgroup$ – DisintegratingByParts Nov 1 '14 at 18:57
  • $\begingroup$ @T.A.E. yes, I forgot that. Is it difficult to show this? $\endgroup$ – user159356 Nov 1 '14 at 19:33
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Adjoints of closed densely-defined linear operators on a Hilbert space $X$ are nice, once you get used to working in the graph space. In fact, the proofs are easier using these techniques for general closed densely-defined operators than the special-case proofs offered for the bounded case. John von Neumann introduced this way of working with densely-defined linear operators on a Hilbert space $X$. I'll explain his approach.

A Graph: The first thing to observe is that a subspace $\mathcal{M}\subseteq X\times X$ is the graph $\mathcal{G}(L)=\{ \langle x, Lx\rangle : x\in\mathcal{D}(L)\}$ of a linear operator $L : \mathcal{D}(L)\subseteq X\rightarrow X$ iff $\langle 0,y\rangle \in \mathcal{M}$ implies $y=0$. And $L$ is a closed linear operator iff its graph is a closed in the product space $X\times X$. If the subspace $\mathcal{M}$ is a graph, then the domain of $L$ becomes the set of all first coordinates in the subspace $\mathcal{M}$. It is easy to show that this unique correspondence between $x \in \mathcal{D}(L)$ and the second coordinate is linear because $\mathcal{M}$ is linear.

Closable: A linear operator $L : \mathcal{D}(L)\subseteq X\rightarrow X$ is closable iff the closure $\mathcal{G}(L)^{c}$ in $X\times X$ of its graph $\mathcal{G}(L)$ is a graph. Equivalently, if $\{ x_{n} \}\subseteq \mathcal{D}(L)$ converges to $0$ and $\{ Ax_{n} \}$ converges to some $y$, then $y=0$. That's the condition for a linear operator to have a closed extension, and is equivalent to the requirement that the closure of the graph of $L$ be the graph of a linear operator.

Inverses: If $L$ is a linear operator, then $L^{-1}$ exists iff the transpose of the graph of $L$ is a graph. Using the transpose map $\tau\langle x,y\rangle = \langle y,x\rangle$, $$ \tau \mathcal{G}(L)=\mathcal{G}(L^{-1}), $$ provided both exist. Notice that if $L^{-1}$ exists, then $L^{-1}$ is closed iff $L$ is closed because $\tau$ is a unitary map on $X\times X$ with $\tau^{2}=I$.

Adjoints: The real power of using graphs comes in defining the adjoint. If $L$ is a closed densely-defined linear operator, then $L$ has a closed densely-defined adjoint $L^{\star}$ whose graph is $$ \mathcal{G}(L^{\star})=J[\mathcal{G}(L)^{\perp}], $$ where the orthogonal complement is taken in $X\times X$ and $J$ is the unitary symplectic transpose $$ J\langle x, y\rangle = \langle y, -x\rangle. $$ Notice that $J^{2}=-I$. Because $\tau$ and $J$ are unitary, they commute with the action of taking orthogonal complement. So you may also write $$ \mathcal{G}(L^{\star})=J[\mathcal{G}(L)^{\perp}]=[J\mathcal{G}(L)]^{\perp}. $$ Of course the transpose operator $\tau$ also commutes with the action of taking the orthogonal complement.

Commutativity: Also note that $\tau J = - J\tau$, which means $\tau J\mathcal{M}=J\tau\mathcal{M}$ for subspaces $\mathcal{M}$ of $X\times X$. So, when you are considering the action of $J$, $\tau$ and $\perp$ on subspaces $\mathcal{M}\subseteq X\times X$, you can freely interchange these operations. That makes life very simple.

Your Example: Suppose that $L$ is a closed densely defined linear operator with a densely-defined inverse $L^{-1}$. Automatically $L^{-1}$ is closed because $L$ is closed (their graphs are transposes of each other.) That means that $L^{-1}$ will have a closed densely-defined adjoint $(L^{-1})^{\star}$. As you might guess, $(L^{-1})^{\star}=(L^{\star})^{-1}$.

To show $(L^{-1})^{\star}=(L^{\star})^{-1}$: First, you must show that $L^{\star}$ has an inverse, which comes down to showing $\tau\mathcal{G}(L^{\star})$ is a graph: $$ \tau\mathcal{G}(L^{\star})=\tau[J\mathcal{G}(L)^{\perp}]= [J\tau\mathcal{G}(L)]^{\perp}= [J\mathcal{G}(L^{-1})]^{\perp}=\mathcal{G}((L^{-1})^{\star}). $$ Obviously the subspace on the far right is a graph. So $L^{\star}$ has an inverse and $(L^{\star})^{-1}=(L^{-1})^{\star}$. This proves the following:

Lemma: Let $H$ be a Hilbert space and $L$ a densely-defined closed linear operator on $H$. If $L$ has a densely-defined inverse $L^{-1}$, then $L^{\star}$ has a densely-defined inverse, and $(L^{\star})^{-1}=(L^{-1})^{\star}$.

Note: If $L^{-1}$ is defined everywhere then it is bounded by the closed graph theorem. In that case $(L^{\star})^{-1}$ is also defined everywhere and is bounded because of the graph equation stated in the lemma. This is the case in your problem for $L=U-\lambda I$ because resolvents are, by definition, defined everywhere and are bounded.

Added in Response to your Addition: Another big fact. If $A$ is closed and densely-defined then $A^{\star\star}=A$. This is because $(\mathcal{M}^{\perp})^{\perp}=\mathcal{M}$ for a closed subspace of a Hilbert space such as $H\times H$. This extends the previous lemma.

Lemma: Let $L$ be a closed densely-defined linear operator on a Hilbert space $H$ with adjoint $L^{\star}$. Then $L^{-1}$ exists as a densely-defined linear operator iff $(L^{\star})^{-1}$ exists as a densely-defined linear operator and, in either case, $(L^{-1})^{\star}=(L^{\star})^{-1}$.

Proof: I showed you that if $L$ has a densely-defined inverse, then $L^{\star}$ has a densely defined inverse and $(L^{-1})^{\star}=(L^{\star})^{-1}$. Conversely, if $L^{\star}$ has a densely-defined inverse, then $(L^{\star})^{\star}=L$ has a densely-defined inverse and $((L^{\star})^{-1})^{\star}=L^{-1} \implies (L^{\star})^{-1}=(L^{-1})^{\star}$. $\;\;\Box$

As a corollary: If $L$ is a closed densely-defined linear operator on a Hilbert space, then $L-\lambda I$ has a densely-defined inverse iff $L^{\star}-\overline{\lambda}I$ has a densely-defined inverse and, in either case, $$ ((L-\lambda I)^{-1})^{\star}=(L^{\star}-\overline{\lambda}I)^{-1}. $$ In particular, $\lambda \in\rho(L)$ iff $\overline{\lambda}\in\rho(L^{\star})$ and the resolvents satisfy $R_{L}(\lambda)^{\star}=R_{L^{\star}}(\overline{\lambda})$. This last equation holds very generally in the sense that one exists iff the other does and, in that case, the two always equal. So, if $L=L^{\star}$ then $\lambda\in\rho(L)$ iff $\overline{\lambda}\in\rho(L)$ and, in that case, $R_{L}(\lambda)^{\star}=R_{L}(\overline{\lambda})$.

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  • $\begingroup$ ah yes, i read about it. these things are also called linear relations, right? $\endgroup$ – user159356 Nov 2 '14 at 0:08
  • $\begingroup$ I added a question about the proof to my initial question, cause it became to long for a comment. somehow I feel surprised about the shortness of this proof and since I am not very familiar with this concept, I wanted to be sure that I understood this. $\endgroup$ – user159356 Nov 5 '14 at 0:06
  • $\begingroup$ Indeed, I forgot switching elements in one place.Well, I think what I do is not so different from what you do, but I just wanted to stick to the elements. So if I have an inverse, I change components and if I have an adjoint, I look at $\langle k,f \rangle - \langle g,h \rangle = 0$. So $T^{-1}:=\{(f,g); (g,f) \in T\}$ and $T^*:=\{(f,g) ; (k,h) \in T: \langle h,f \rangle = \langle g,k \rangle\}$. Does it now make sense to you? $\endgroup$ – user159356 Nov 5 '14 at 8:19
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    $\begingroup$ @TobiasHurth : The reason I don't like your definition of $T^{\star}$ is that it suggests the set of elements is a graph already. If $T$ is closed and densely-defined, then your definition is okay, but it hides those details of closed and densely-defined. A more practical definition would start with the domain consisting of all $y$ for which there exists $z$ such that $(Tx,y)=(x,z)$ for all $x\in\mathcal{D}(T)$. Equivalently, there exists a constant $m$ such that $|(Tx,y)|\le m\|x\|$ for all $x\in\mathcal{D}(T)$. For a given $y$, such $z$ is unique because $\mathcal{D}(T)$ is dense. $\endgroup$ – DisintegratingByParts Nov 5 '14 at 14:41
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    $\begingroup$ @TobiasHurth : Yes, your argument is sound. Side issue: the second definition of domain as the set of $y$ such that $|(Tx,y)|\le m\|x\|$, $x\in\mathcal{D}(T)$ is a very useful alternative that deals with $y$ without the second element $z$, and this is an advantage of an element approach that uses Riesz representation. $\endgroup$ – DisintegratingByParts Nov 5 '14 at 14:57

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