# Lecture 17: Unbounded Operators
$\newcommand{\R}{\mathbb{R}}$
$\newcommand{\C}{\mathbb{C}}$
###### tags: `224a`
## Densely Defined Operators, Extensions
An operator $T:D(T)\rightarrow H$ 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 $ker(T^*)=ran(T)^\perp$ since $D(T^*)$ 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 $D(ATA^{-1})=AD(T)$.
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 $\R$ 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 $e^{it A}$ 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 $D(M_x):=\{f\in L^2(\R): M_xf\in L^2(\R)\}$.
**Example 2.** $A_1 = -id/dx$ on $C_c^1(\R)$, 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.** $T:D(T)\rightarrow H$ is *closed* if whenever a sequence $f_n\in D(T)$ satisfies $f_n\rightarrow f\in H$ and $Tf_n\rightarrow g\in H$, one has $f\in D(T)$ and $Tf=g$. This is equivalent to saying that the *graph* $$\Gamma(T):=\{<f,g>: f\in D(T), g=Tf\}$$ 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 $f_n\rightarrow f=e^{-x^2/2}$ in $H$ and the images $A_1f_n$ also converge in $L^2(\R)$ (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* $\overline{T}$ to be the operator with graph $\overline{\Gamma(T)}$, provided this closure contains no points of type $<0,g>, g\neq 0$ (this is necessary and sufficient for a graph to correspond to a linear operator). Then $\sigma(T)$ refers to $\sigma(\overline{T})$.
## Adjoints, Symmetric and Self-adjoint Operators
If $T:D(T)\rightarrow H$, the domain of the adjoint $T^*$ is
$$ D(T^*)=\{ g\in H: \exists C, |(g,Tf)|\le C\|f\|\forall f\in D(T)\},$$
which is the set of $g$ such that the functional $f\mapsto (g,Tf)$ is continuous on $D(T)$.
The *adjoint* $T^*$ is then defined via the Riesz theorem as the unique $g^*$ such that $(g^*,f)=(g,Tf)$ 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 $T=T^*$ and $D(T)=D(T^*)$. 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 $D(T)\subset D(T^*)$ since we trivially have $|(g,Tf)|=|(Tg,f)|\le \|Tg\|\|f\|$ for every $g\in D(T)$. Thus, every symmetric operator satisfies
$$ T\subset T^*$$
which means $T^*$ is an extension of $T$.
We now check that example 1 is selfadjoint: suppose $g\in D(M_x^*)$. This means that there is a constant $C$ such that $|(g,M_xf)|^2\le C\|f\|^2$ 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
$$(\int_{[-R,R]} x^2 |g(x)|^2dx)^2 \le C \int_{[-R,R]} x^2|g(x)|^2dx.$$
Letting $R\rightarrow\infty$, monotone convergence implies that $\|M_xg\|_{L^2}\le C$, 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 $g\in D(A_1^*)$ but clearly $g\notin D(A_1)$.
For the third example, Dimock 1.3.1:ex1.3 shows that $A_2^*g=ig$ 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 $\R$. In general, if $g:\R\rightarrow \C$ is any measurable function, the multiplication operator $M_g$ defined on $$D(M_g):=\{f\in L^2(\R): gf\in L^2(\R)\}$$ 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 $\R$.
In general, an operator is called *essentially selfadjoint* if its closure is selfadjoint.
We now characterize the spectra of selfadjoint operators.
**Theorem.** If $T:D(T)\rightarrow H$ is self-adjoint, $\sigma(T)\subset \R$.
*Proof.* Dimock 1.3.2:thm1.8. The proof is identical to the selfadjoint case, which crucially uses the identity $ker(\lambda^*-T^*)=ran(\lambda-T)^\perp$. This is where selfadjointness is essential: if $D(T)\neq D(T^*)$ then this identity doesn't hold. $\square$
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 $T:D(T)\rightarrow H$ 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$\square$

The official schedule is posted at https://www.jointmathematicsmeetings.org/meetings/national/jmm2023/2270_program_ss94.html#title The Zoom link for virtual attendance/virtual talks is: https://berkeley.zoom.us/j/94982941411?pwd=RWRXeVdoZUNVTmdqK1FIMksrZFllZz09 Friday, January 6 (101 Hynes Convention Center) 1-2pm: Matt Colbrook, The foundations of infinite-dimensional spectral computations (Zoom) 2-3pm: JosuĂ© Tonelli-Cueto, Condition-based Low-Degree Approximation of Real Polynomial Systems

1/3/2023Lecture 1: Propositional Logic A proposition is a declarative sentence that is either true or false but not both. A proposition has a truth value which is either $T$ or $F$. A letter used to denote a proposition is called a propositional variable. If $p$ is a proposition, the negation of $p$, denoted $\lnot p$, is the proposition "it is not the case that $p$". The truth value of $\lnot p$ is the opposite of that of $p$. If $p$ and $q$ are propositions, the conjunction of $p$ and $q$ (denoted $p\land q$) is the proposition "p and q." Its truth value is $T$ when both $p$ and $q$ are $T$, and it is $F$ otherwise. If $p$ and $q$ are propositions,he disjunction of $p$ and $q$ is the proposition "p or q". Its truth value is $T$ when at least one of $p$ or $q$ is $T$, and $F$ if both $p$ and $q$ are $F$. If $p$ and $q$ are propositions, the conditional $p\rightarrow q$ is the proposition "if $p$ then $q$". Its truth values are given by the following truth table (which lists its truth values in terms of those of $p$ and $q$): [see Table 5 in Rosen] The converse of $p\rightarrow q$ is $q\rightarrow p$. A compound proposition consists of logical operations (the four listed above) applied to propositions or propositional variables, possibly with parentheses. The truth value of a compound proposition can be mechanically determined given the truth values of its constituents.

12/12/2022Lecture 14 Basic Terminology, Handshaking Theorem A graph is a pair $G=(V,E)$ where $V$ is a finite set of vertices and $E$ is a finite multiset of $2-$element subsets of $V$, called edges. If $E$ has repeated elements $G$ is called a multigraph, otherwise it is called a simple graph. We will not consider directed graphs or graphs with loops (edges from a vertex to itself). Two vertices $x,y\in V$ are adjacent if ${x,y}\in E$. An edge $e={x,y}$ is said to be incident with $x$ and $y$. The degree of a vertex $x$ is the number of edges incident with it. Theorem. In any graph $G=(V,E)$, $$ \sum_{v\in V} deg(v) = 2|E|.$$ Proof. Let $I={(e,v):e\in E,v\in V,\textrm{$e$ is incident with $v$ in $G$}}$ be the set of edge-vertex incidences in $G$. We will count the number of incidences in two ways: (i) Observe that every edge participates in exactly two incidences, and no incidence participates in more than one edge. Thus, the total number of incidences is $|I|=2|E|$.

12/12/2022Prove that every simple graph with $n$ vertices and $k$ edges has at least $n − k$ connected components (hint: induction, or contradiction). Let $p\ge 3$ be a prime. Consider the graph $G=(V,E)$ with $V={0,1,2,\ldots, (p-1)}$ and $$E={{x,y}:x-y\equiv 2(mod p)\lor x-y\equiv -2 (mod p)}$$ Show that $G$ is connected. Does $G$ have an Euler circuit? Prove your answer. Let $n\ge 1$ be an integer and let $k\le n/2$. Consider the graph $G=(V,E)$ with $V={S\subseteq {1,2,\ldots,n}: |S|=k}$ and $$E = {{S,T}:|(S-T)\cup (T-S)|=1}.$$ What are the degrees of the vertices in $G$? Is $G$ connected? For which values of $n$ and $k$ does $G$ have an Eulerian circuit? For which values of $n$ and $k$ is $G$ $2-$colorable? Prove your answers.

11/8/2022
Published on ** HackMD**

or

By clicking below, you agree to our terms of service.

Sign in via Facebook
Sign in via Twitter
Sign in via GitHub
Sign in via Dropbox
Sign in with Wallet

Wallet
(
)

Connect another wallet
New to HackMD? Sign up