Unbounded operators (1)

Bounded linear operators are the key ingredient studied in basic functional analysis. However, some important linear operators in analysis and physics are not bounded. Here I give an elementary introduction to unbounded operators, which may interest the readers especially analysts, hopefully.

1 CLOSED OPERATORS

Unless specified, always let X and Y be two Banach spaces. Endow the norm \|\cdot\|_{X}+\|\cdot\|_{Y} to the product space X\times Y. Denote X^* and Y^* to be their continuous duals. Also let H be a Hilbert space.

Definition 1.1

Given a linear operator T, which its domain D(T) is a subspace of X and its range space R(T) is a subspace of Y. Define the graph of T to be

G(T)=\{(x,Tx):x\in D(T)\}, which is a subspace of X\times Y.

T is said to be closed if G(T) is closed in X\times Y.

Definition 1.2    (Extension and closability)

Given T_1,T_2 linear. If G(T_1)\subset G(T_2), then we call T_2 to be an extension of T_1, denoted by T_1\subset T_2.

A linear operator T is said to be closable if there exists an extension S\supset T such that \overline{G(T)}=G(S). S is the closure of T, denoted by S=\overline{T}.

Not all linear operators are closable, we give a condition to determine whether \overline{G(T)} is the graph of some linear operator.

Proposition 1.3 Given T linear. The following are equivalent.

(1) T is closable.

(2) If (0,y)\in \overline{G(T)}, then y=0.

Proof.

The following are the basic properties about closed operators.

Proposition 1.4

(1) If T is an injective closed operator, then T^{-1}:R(T)\rightarrow D(T) is also closed.

(2) T is closed implies the null space N(T) is closed in X.

(3) If T is closable and S is closed, then \overline{T}\subset S. This means the closure of T is its smallest closed extension.

(4) If T is closed and D(T)=X, then T is bounded.

Proof. (1)(2)(3) follow by direct check. (4) follows by closed graph theorem.

By Propostion 1.4(4), what we interest in closed unbounded operators are those T with D(T)\neq X. What we like to study is

Definition 1.5

A linear operator T is said to be densely defined if D(T) is dense in X.

Consider the following “adjoint” T^*.

Definition 1.6

Given a densely defined linear T. Denote

D(T^*)=\{y^*\in Y^*:\text{there exists }x^*\in X^*\text{ such that }y^*(Tx)=x^*(x)\text{ for all }x\in D(T)\}.

Define T^*:Y^*\rightarrow X^* by y^*\mapsto x^*. T^* is called the adjoint operator of T, and D(T^*) to be the domain of T^*.

T is densely defined guarantees T^*(y^*) is uniquely determined, and it can be easily checked that T^* is also linear.

If H is a Hilbert space and X=Y=H, then can rewrite

D(T^*)=\{y\in H:\text{there exists }M_{y}>0\text{ such that }|\langle y,Tx \rangle|\leq M_{y}\|x\|\}.

Theorem 1.7

T^* defined above must be closed. Also, T_1\subset T_2 imples T_2^*\subset T_1^*.

Proof.

Theorem 1.8

(1)  T is a densely defined linear operator with X=Y=H. Then T is closable if and only if T^* is densely defined. In either case \overline{T}=T^{**}.

(2) If T is a densely defined linear operator and Y is reflexive, then the conclusion as above still holds, but in either case \overline{T}=\Phi_{Y}^{-1}T^{**}\Phi_{X}, where \Phi_{X}:X\rightarrow X^{**} and \Phi_{Y}:Y\rightarrow Y^{**} are canonical embeddings.

Proof.

Definition 1.9 (Symmetry, self-adjointness, essentially self-adjointness)

Given T densely defined linear operator from H to H. T is said to be symmetric if T\subset T^*. T is said to be self-adjoint if T=T^*. T is said to be essentially self-adjoint if T is closable and \overline{T} is self-adjoint.

Remark 1.10

(1)  T is symmetric if and only if for all x\in D(T), \langle Tx,y \rangle =\langle x,Ty \rangle.

(2) Symmetric operators are closable. Indeed, if T is symmetric, then T\subset T^* and hence D(T)\subset D(T^*). This implies T^* is densely defined. Thus T is closable by Theorem 1.8(1).

(3) If T\in B(H), symmetry is equivalent to self-adjointness. In general, T is self-adjoint if and only if T is symmetry and D(T)=D(T^*).

(4) If T is self-adjoint, S is symmetric and T\subset S, then T=S. Indeed, since T\subset S\subset S^*\subset T^* (Theorem 1.7 is used), T=T^*=S^*=S. This means, self-adjoint operator is the maximal symmetric extension of itself.

If T is essentially self-adjoint, then \overline{T} is its unique closed self-adjoint extension, because if S is another one, then \overline{T}\subset S implying \overline{T}=S.

Examples.

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