Lecture 17: Unbounded Operators

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tags: 224a

Densely Defined Operators, Extensions

An operator

$T:D(T)\to H$ is densely defined if

$D(T)$ is a dense subspace of

$H$. Many of the subtleties of dealing with such operators are related to the fact that there may be many reasonable choices of domain, and this choice matters a lot because e.g.

1. $ker(T)$ and
   $im(T)$ are now defined relative to
   $D(T)$, and this complicates crucial identities such as
   $ker(T^{\ast })=ran(T{)}^{\perp }$ since
   $D(T^{\ast })$ needn't be the same as
   $D(T)$.
2. For two such operators, simple operations such as
   $A+B$ only make sense on
   $D(A)\cap D(B)$.
3. The spectrum is highly sensitive to the choice of domain.

One thing that we will use often which works exactly the same way is equivalence under invertible transformations: if

$A,A^{-1}\in L(H)$ then

$D(AT{A}^{-1})=AD(T)$.

Our goal is to develop notions of spectrum, spectral measure, and functional calculus for selfadjoint unbounded operators. These are motivated by concrete applications in physics: the position and momentum operators

$M_x$ and

$-id/dx$ on various subspaces of

$\mathbb{R}$ are examples of such operators, and we need a way of associating spectral measures to them to make sense of experimental outcomes in quantum mechanics. Time evolution is given by the semigroup

$e^{itA}$ for various unbounded operators, so we must find a way (other than power series, which doesn't work here due to unboundedness) to make sense of this.

We will have three running examples:

Example 1.

$M_x$ on

$D(M_x):=\{f\in L^2(\mathbb{R}):M_{x}f\in L^2(\mathbb{R})\}$.
Example 2.

$A_1=-id/dx$ on

$C_c^1(\mathbb{R})$, the space of compactly supported continuous functions.
Example 3.

$A_2=-id/dx$ on

$C_c^1([0,1])\cap \{f:f(0)=f(1)=0\}$.

Closed Operators

We will focus on operators which have "maximal" domains in the following sense.

Definition.

$T:D(T)\to H$ is closed if whenever a sequence

$f_n\in D(T)$ satisfies

$f_n\to f\in H$ and

$T{f}_n\to g\in H$, one has

$f\in D(T)$ and

$Tf=g$. This is equivalent to saying that the graph

$$\mathrm{\Gamma }(T):=\{<f,g>:f\in D(T),g=Tf\}$$ is closed as a subspace of

$H\times H$ in the norm topology.

Example 1 is closed (see Dimock 1.1.4:ex1.1 for a proof), and the second example is not. For the second example, consider the sequence

$f_n(x)={\chi }_n(x)e^{-x^2/2}$ where

${\chi }_n(x)$ is a smooth cutoff supported on

$[-n,n]$. Then we have

$f_n\to f=e^{-x^2/2}$ in

$H$ and the images

$A_1{f}_n$ also converge in

$L^2(\mathbb{R})$ (via a dominated convergence argument), but

$f\notin D(T)$.

Spectrum

Spectra of closed operators are defined the same way as bounded operators (as complements of resolvent sets); see 1.2.2 of Dimock for details. The reason to insist on closed operators stems from the closed graph theorem, which implies that

$z-T$ is invertible for a closed

$T$ iff it is boundedly invertible, allowing the same classification into point, continuous, and residual spectra as before. The main difference from the bounded case is that the spectrum is no longer bounded.

If an operator

$T$ is not closed, define its closure

$\bar{T}$ to be the operator with graph

$\overline{\mathrm{\Gamma }(T)}$, provided this closure contains no points of type

$<0,g>,g\ne 0$ (this is necessary and sufficient for a graph to correspond to a linear operator). Then

$\sigma (T)$ refers to

$\sigma (\bar{T})$.

Adjoints, Symmetric and Self-adjoint Operators

If

$T:D(T)\to H$, the domain of the adjoint

$T^{\ast }$ is

$$D(T^{\ast })=\{g\in H:\mathrm{\exists }C,|(g,Tf)|\le C\Vert f\Vert \mathrm{\forall }f\in D(T)\},$$
which is the set of

$g$ such that the functional

$f\mapsto (g,Tf)$ is continuous on

$D(T)$.
The adjoint

$T^{\ast }$ is then defined via the Riesz theorem as the unique

$g^{\ast }$ such that

$(g^{\ast },f)=(g,Tf)$ for all

$f\in D(T)$. It is shown in Dimock 1.2.3 that the adjoint of a densely defined operator is always closed (proof: its graph can be written as the kernel of another operator).

An operator

$T$ is symmetric if

$(g,Tf)=(Tg,f)$, and selfadjoint if

$T=T^{\ast }$ and

$D(T)=D(T^{\ast })$. Symmetry is a rather weak algebraic condition, and one can check by integration by parts that all three of our examples have it. It does, however immediately imply that

$D(T)\subset D(T^{\ast })$ since we trivially have

$|(g,Tf)|=|(Tg,f)|\le \Vert Tg\Vert \Vert f\Vert$ for every

$g\in D(T)$. Thus, every symmetric operator satisfies

$$T\subset T^{\ast }$$
which means

$T^{\ast }$ is an extension of

$T$.

We now check that example 1 is selfadjoint: suppose

$g\in D(M_x^{\ast })$. This means that there is a constant

$C$ such that

$|(g,M_{x}f){|}^2\le C\Vert f{\Vert }^2$ for all

$f\in D(M_x)$. Choose

$f=xg(x){\chi }_R(x)$ for a cutoff function on

$[-R,R]$. Plugging this into the above, we have

$$({\int }_{[-R,R]}{x}^2|g(x){|}^{2}dx{)}^2\le C{\int }_{[-R,R]}{x}^2|g(x){|}^{2}dx.$$
Letting

$R\to \mathrm{\infty }$, monotone convergence implies that

$\Vert M_{x}g{\Vert }_{L^2}\le C$, implying

$g\in D(M_x)$, as desired.

Example 2 is not selfadjoint. To see this, consider

$g(x)=e^{-x^2/2}$. A simple calculation shows that

$(g,Tf)$ is continuous in

$f$, so

$g\in D(A_1^{\ast })$ but clearly

$g\notin D(A_1)$.

For the third example, Dimock 1.3.1:ex1.3 shows that

$A_2^{\ast }g=ig$ for

$g=e^x$, so

$A_2$ has an eigenvalue of

$i$. We will show in the next section that this contradicts selfadjointness.

Spectrum of Selfadjoint operators

The spectrum of the first example is easily seen to be

$\mathbb{R}$. In general, if

$g:\mathbb{R}\to \mathbb{C}$ is any measurable function, the multiplication operator

$M_g$ defined on

$$D(M_g):=\{f\in L^2(\mathbb{R}):gf\in L^2(\mathbb{R})\}$$ has spectrum equal to the essential range of

$g$; the proof is the same as in the bounded case.

One can show that the closure of

$A_1$ is actually the operator

$\hat{A}$ from the previous lecture. Since

$\hat{A}$ is unitarily equivalent to

$M_x$ via the

$L^2$ Fourier transform, its spectrum is also

$\mathbb{R}$.

In general, an operator is called essentially selfadjoint if its closure is selfadjoint.

We now characterize the spectra of selfadjoint operators.
Theorem. If

$T:D(T)\to H$ is self-adjoint,

$\sigma (T)\subset \mathbb{R}$.
Proof. Dimock 1.3.2:thm1.8. The proof is identical to the selfadjoint case, which crucially uses the identity

$ker({\lambda }^{\ast }-T^{\ast })=ran(\lambda -T{)}^{\perp }$. This is where selfadjointness is essential: if

$D(T)\ne D(T^{\ast })$ then this identity doesn't hold.

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In particular, this shows that example

$A_2$ is not selfadjoint.

We end by stating a very concrete criterion for checking selfadjointness which how one checks this property in practice.
Theorem. If

$T:D(T)\to H$ is symmetric, the following are equivalent:

1. $T$ is self-adjoint.
2. $ran(T\pm i)=H$.
3. $T$ is closed and
   $ker(T\pm i)=\{0\}.$

Proof. Dimock 1.3.2:thm1.9 for parts 1 and 2. For the full theorem, see Reed and Simon VIII.3

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