Unbounded Linear Operators

Hellinger-Toeplitz

If a linear operator

T

is defined on every element of a complex hilbert space

H

and

T

is self-adjoint then

T

is bounded

Suppose for the sake of contradiction that $T$ is unbounded, therefore for each $N \in ℕ_{1}$ we obtain an element of $x_{N} \in dom (f)$ such that $‖ T x_{N} ‖ > N ‖ x_{N} ‖$ , therefore we can construct a sequence $(y_{n})$ such that $‖ y_{n} ‖ = 1$ and $‖ T y_{n} ‖ \to \infty$

We now construct a sequence of functionals of the form $f_{n} (x) = ⟨ T x, y_{n} ⟩ = ⟨ x, T y_{n} ⟩$ where we note that each such $f_{n}$ is defined on all of $H$ and is linear and moreover for each $n$ by the Schwarz inequality we have that: $| f_{n} (x) | = | ⟨ x, T y_{n} ⟩ | \leq ‖ T y_{n} ‖ ‖ x ‖$ which shows that the functional $f_{n}$ is bounded as a linear operator.

Additionally for any $x \in H$ the real sequence $(f_{n} (x))$ is also bounded because we have: $| f_{n} (x) | = | ⟨ T x, y_{n} ⟩ | \leq ‖ T x ‖$ therefore there exists some $c \in ℝ^{+}$ such that ${‖ f_{n} ‖}_{op} \leq c$ for every $n \in ℕ_{1}$ therefore for each $x \in H$ we have that $| f_{n} (x) | \leq {‖ f_{n} ‖}_{op} ‖ x ‖ \leq c ‖ x ‖$ thus taking $x = T y_{n}$ we have that ${‖ T y_{n} ‖}^{2} = ⟨ T y_{n}, T y_{n} ⟩ = | f_{n} (T y_{n}) | \leq c ‖ T y_{n} ‖$ therefore $‖ T y_{n} ‖ \leq c$ which is a contradiction because we assumed that $‖ T y_{n} ‖ \to \infty$

The above implies that given any unbounded linear operator it will never be defined on all elements of a hilbert space.

Extension of a Linear Operator

Suppose that

S, T

are linear operators, then we say that

T

is an extension of

S

$dom (S) \subseteq dom (T)$
$S = T ↾_{dom (S)}$

Densely Defined Linear Operator

We say that a linear operator

T

is densely defined in

H

dom (T)

is dense in

H

Hilbert Adjoint Operator

Let

T : X \to H

be a linear opeartor (possibly unbounded) denslely defined linear operator in a complex hilbert where

H

is a complex hilbert space, then the Hilbert adjoint operator

T^{*} : Y \to H

T

where

Y = {y \in H : \exists y^{*} \in H st \forall x \in H, ⟨ T x, y ⟩ = ⟨ x, y^{*} ⟩}

then

T^{*} (y)

is defined to be equal to

y^{*}

Adjoint Operators Respect the Subset Relation

Let

S : X \to H

and

T : Y \to Y

be densely defined linear operators in a complex hilbert space

H

then if

S \subseteq T

we have

T^{*} \subseteq S^{*}

By the definition of $T^{*}$ we have that for all $a \in Y, b \in dom (T^{*})$ that $α : ⟨ T a, b ⟩ = ⟨ a, T^{*} b ⟩$ and since $S \subseteq T$ then also that for every $s \in dom (S)$ we have $β : ⟨ S s, b ⟩ = ⟨ s, T^{*} b ⟩$

Now we want to prove that $dom (T^{*}) \subseteq dom (S^{*})$ . But recall by definition $dom (S^{*}) = {y \in H : \exists y^{*} \in H st \forall x \in H, ⟨ S x, y ⟩ = ⟨ x, y^{*} ⟩}$ Now the $b$ 's such that $β$ holds must be a subset of $dom (S)$ because it is an even more specific representation, ie $y^{*}$ must exist and be of the form $T^{*} y$ which is more restrictive, so we have that $dom (T^{*}) \subseteq dom (S^{*})$ .

Finally we note that for any $y \in dom (T^{*})$ we have that $S^{*} y = T^{*} y$ by chaining $α$ and $β$ therefore we have that $T^{*} \subseteq S^{*}$

Double Adjoint Is a Superset

Suppose that

T : X \to H

is a linear operator which is densely defined in a complex hilbert space

H

then if

dom (T^{*})

is dense in

H

then

T \subseteq T^{* *}

We know that for all

x \in X

and

y \in dom (T^{*})

that we have

α : ⟨ T x, y ⟩ = ⟨ x, T^{*} y ⟩

therefore by taking complex conjugates that is:

β : ⟨ T^{*} x, y ⟩ = ⟨ y, T x ⟩

since

dom (T^{*})

was dense then we know that

T^{* *}

exists and by definition we have that for all

a \in dom (T^{*})

and all

b \in dom (T^{* *})

thus using

α

and

β

in the same way as the above prove we conclude that

dom (T) \subseteq dom (T^{* *})

and that

T^{* *} x = T x

for any

x \in dom (T)

, that is

T \subseteq T^{* *}

If a Linear Operator Is Injective and Has Dense Image Then the Adjoint Has an Inverse

Suppose that

T : X \to H

is a densely defined linear operator in a dense hilbert space

H

, if

T

is injective and

im (T)

is dense in

H

then

T^{*}

is injective and

{(T^{*})}^{- 1} = {(T^{- 1})}^{*}

We know that $T^{*}$ exists because $T$ is densely defined in $H$ , also that $T^{- 1}$ exists because $T$ is injective. Also $im (T) = dom (T^{- 1})$ is dense in $H$ therefore ${(T^{- 1})}^{*}$ exists.

We must now show that ${(T^{*})}^{- 1}$ exists and satisfies the equation as stated in the outset. So let $y \in dom (T^{*})$ , then for every $x \in dom (T^{- 1})$ we have that $T^{- 1} (x) \in dom (T)$ satisfying $α : ⟨ T^{- 1} x, T^{*} y ⟩ = ⟨ T T^{- 1} x, y ⟩ = ⟨ x, y ⟩$ then by the definition of the hilbert adjoint operator of $T^{- 1}$ we have for every $x \in dom (T^{- 1})$ $⟨ T^{- 1} x, T^{*} y ⟩ = ⟨ x, {(T^{- 1})}^{*} T^{*} y ⟩$ so we conlcude that $T^{*} y \in dom ({(T^{- 1})}^{*})$ combined with $α$ we conclude that $β : {(T^{- 1})}^{*} T^{*} y = y$ when $T^{*} y = 0$ we have that $y = 0$ thus ${(T^{*})}^{- 1} : im (T^{*}) \to dom (T^{*})$ exists. Since ${(T^{*})}^{- 1} T^{*}$ is the identity operator on $dom (T^{*})$ then $β$ implies that ${(T^{*})}^{- 1} \subseteq {(T^{- 1})}^{*}$

To complete the proof we must now show that ${(T^{*})}^{- 1} \supseteq {(T^{- 1})}^{*}$ , so note that for any $x \in dom (T)$ and $y \in dom ({(T^{- 1})}^{*})$ , then $T x \in im (T) = dom (T^{- 1})$ and $⟨ T x, {(T^{- 1})}^{*} y ⟩ = ⟨ T^{- 1} T x, y ⟩ = ⟨ x, y ⟩$ but also by the definition of the hilbert adjoint operator of $T$ we note for each $x \in dom (T)$ $⟨ T x, {(T^{- 1})}^{*} y ⟩ = ⟨ x, T^{*} (T^{- 1}) y ⟩$ so we have ${(T^{- 1})}^{*} y \in dom (T^{*})$ and that $γ : T^{*} {(T^{- 1})}^{*} y = y$ for every $y \in dom ({(T^{- 1})}^{*})$ .

Now we know that $T^{*} {(T^{*})}^{- 1}$ is the identity operator on $dom ({(T^{*})}^{- 1}) = im (T^{*})$ and ${(T^{*})}^{- 1} : im (T^{*}) \to dom (T^{*})$ is surjective, so by $γ$ we have that $dom ({(T^{*})}^{- 1}) \supseteq dom ({(T^{- 1})}^{*})$ to conclude that ${(T^{*})}^{- 1} \supseteq {(T^{- 1})}^{*}$ so we are done.