Lecture 11 - Riesz Representation Theorem and Adjoints

#math

Theorem (*)

If $H$ is a Hilbert space, and $T \in B (H)$ , then $\exists! T^{*} \in B (H)$ the adjoint of $T$ such that:

⟨ T u, v ⟩ = ⟨ u, T^{*} v ⟩

In addition, $(T^{*})^{*} = T$ and $| | T^{*} | | = | | T | | .$

For the proof we need Riesz Representation Theorem which establishes a connection between Hilbert space and it's continuous dual space.

Definition

For a topological vector space $V$ , its continuous dual space or topological dual space or just dual space $V^{*}$ is defined as the space of all continuous linear functional:

φ : V ⟶ R (or C)

Theorem

(Riesz Representation Theorem) Let $H$ be a Hilbert space, whose inner product $⟨ ., . ⟩$ is linear in its first argument and anti linear in its second argument. For every continuous linear functional $φ \in H^{*}$ , $\exists! f_{φ} \in H$ called the Riesz representation of $φ$ such that:

φ (x) = ⟨ x, f_{φ} ⟩, \forall x \in H

Furthermore $| | f_{φ} | |_{H} = | | φ | |_{H^{*}}$ , and $f_{φ}$ is the unique vector $f_{φ} \in K e r (φ)^{⊥}$ with,

φ (f_{φ}) = | | φ | |^{2}

Now we prove the first theorem $(*)$ .

Proof

Fix $v \in H$ and let:

l (u) = ⟨ T u, v ⟩

$l$ is a linear operator on $H$ , and $| | l (u) | | = | ⟨ T u, v ⟩ | \leq | | T u | | | | v | | \leq | | T | | | | u | | | | v | |$ . Therefore $l \in H^{*}$ with

| | l | | \leq | | T | | | | v | |

By the Riesz representation theorem, $\exists! v^{'} \in H$ such that

l (u) = ⟨ u, v^{'} ⟩

We define $T^{*} v = v^{'}$ so that:

T^{*} : H ⟶ H

and equivalently $⟨ T u, v ⟩ = ⟨ u, T^{*} v ⟩$ .

Proposition

$T^{*}$ is linear.

Proof

For $v_{1}, v_{2}, u \in H$ and $c_{1}, c_{2}$ scalars we have:

\begin{aligned} ⟨ u, T^{*} (c_{1} v_{1} + c_{2} v_{2}) ⟩ & = ⟨ T u, c_{1} v_{2} + c_{2} v_{2} ⟩ \\ = {\bar{c}}_{1} ⟨ T u, v_{1} ⟩ + {\bar{c}}_{2} ⟨ T u, v_{2} ⟩ \\ = {\bar{c}}_{1} ⟨ u, T^{*} v_{1} ⟩ + {\bar{c}}_{2} ⟨ u, T^{*} v_{2} ⟩ \\ = ⟨ u, c_{1} T^{*} v_{1} + c_{2} T^{*} v_{2} ⟩ \end{aligned}

Since $u$ is arbitrary we have shown that $T^{*}$ is linear.

Proposition

$T^{*}$ is bounded

Proof

$\begin{aligned} | | T^{*} | | & = | | l | | \leq | | T | | | | v | | \end{aligned}$

Proposition

$T^{*}$ is unique

Proof

Assume there exists $S \in B (H)$ such that:

⟨ u, T^{*} v ⟩ = ⟨ u, S v ⟩

Then we obtain that:

⟨ u, T^{*} v - S v ⟩ = 0, \forall u \in H

This implies $T^{*} v = S v$ for all $v \in H$ , thus $T^{*} = S$ .

Proposition

$(T^{*})^{*} = T$

Proof

As $T^{*} \in B (H)$ , it also has an adjoint:

T^{* *} := (T^{*})^{*}

satisfying

⟨ T^{*} u, v ⟩ = ⟨ u, T^{* *} v ⟩, \forall u, v H

But:

\begin{aligned} ⟨ T^{*} u, v ⟩ & = \overset{―}{⟨ v, T^{*} u ⟩} \\ = \overset{―}{⟨ T u, v ⟩} \\ = ⟨ v, T u ⟩ \end{aligned}

Thus we must have $T^{* *} = T$ . Moreover since:

| | T | | = | | T^{* *} | | \leq | | T^{*} | |

Thus $| | T | | = | | T^{*} | |$ .

Special Cases

Definition

If $T \in B (H)$ ,

If $T^{*} = T$ , we say that $T$ is self adjoint.
If $T^{*} = - T$ , we say that $T$ is skew adjoint.
If $T^{*} = T^{- 1}$ we say that $T$ is unitary.

Proposition

If $S, T \in B (H)$ , then $S T \in B (H)$ and:

(S T)^{*} = T^{*} S^{*}

If $T^{- 1} \in B (H)$ , then $(T^{*})^{- 1} \in B (H)$ and

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

Proof is left as an exercise to the readers.

Example

$H = C^{n}$ with $T u = A u$ where $A$ is an $n \times n$ matrix. We know $⟨ T u, v ⟩ = ⟨ A u, v ⟩ = ⟨ u, A^{*} v ⟩$ . $A^{*}$ is the conjugate transpose matrix of $A$ . By uniqueness $T^{*} v = A^{*} v$ . T is obviously self-adjoint if $A^{*} = A$ (if so $A$ is also said to be a Hermitian matrix, or symmetric if $A$ is real.)

Example

$T u (x) = a (x) u (x)$ on $L^{2} (Ω)$ where $a \in L^{\infty} (Ω)$ , then:

\begin{aligned} ⟨ T u, v ⟩ & = \int_{Ω} a (x) u (x) \overset{―}{v (x)} d x \\ = \int_{Ω} u (x) \overset{―}{\overset{―}{a (x)} v (x)} d x \end{aligned}

Thus $T^{*} v (x) = \overset{―}{a (x)} v (x)$
T is:

Self adjoint if $a$ is real valued.
Skew adjoint if $a$ is purely imaginary.
Unitary if $| a | = 1$ .

Example

$T u (x) = \int_{Ω} K (x, y) u (y) d y \in L^{2} (Ω)$

with $K \in L^{2} (Ω \times Ω)$ . $T$ is bounded.

\begin{aligned} ⟨ T u, v ⟩ & = \int_{Ω} (\int_{Ω} K (x, y) u (y) d y) \overset{―}{v (x)} d x \\ = \int_{Ω} u (y) (\overset{―}{\int_{Ω} \overset{―}{K (x, y)} v (x) d x}) d y \end{aligned}

Thus $T^{*} v = \int_{Ω} \overset{―}{K (x, y)} v (x) d x$ .
(Fubini is justified as $K (x, y) u (y) v (x) \in L^{1} (Ω \times Ω)$ )

$T$ will be self adjoint if $K$ is real-valued and $K (x, y) = K (y, x)$ .