## 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.

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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