Brownian Motion and Stochastic Calculus

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/Part 1: Stochastic Processes/Chapter 2: Diffusion Operators

2.1 Self-Adjoint Operators

Definition 2.1.1.label Let $H$ be a Hilbert space, $\varphi: H^{*} \to H$ be the canonical map, and $T: D(T) \to H$ be a densely defined linear operator. Let

\[D(T^{*}) = \bracsn{\psi \in H| \dpb{T\cdot, \psi}{H} \in H^*}\]

then the adjoint of $T$ is the mapping

\[D(T^{*}) \to H \quad \phi \mapsto \varphi(\dpb{T\cdot, \psi}{H})\]

Definition 2.1.2 ([X.2.2, Con85]).label Let $H$ be a Hilbert space and $T: D(T) \to H$ be densely defined linear map, then the following are equivalent:

  1. (1)

    $T^{*}$ is an extension of $T$.

  2. (2)

    $\angles{Tx, y}= \angles{x, Ty}$ for all $x, y \in D(T)$.

  3. (3)

    $\angles{Tx, x}\in \real$ for all $x \in D(T)$.

If the above holds, then $T$ is a symmetric operator.

If $T = T^{*}$, then $T$ is self-adjoint.

Proof. $(1) \Rightarrow (2)$: Let $x, y \in D(T)$, then

\[\angles{Tx, y}_{H} = \angles{x, T^*y}_{H} = \angles{x, Ty}_{H}\]

$(2) \Rightarrow (3)$: For any $x, y \in D(T)$,

\[\angles{Tx, x}_{H} = \overline{\angles{Tx, x}_H}\quad \angles{Tx, x}_{H} \in \real\]

$(3) \Rightarrow (1)$: By the polarisation identity, for any $x, y \in D(T)$,

\begin{align*}4\angles{Tx, y}_{H}&= \sum_{k = 0}^{3}i^{k}\angles{T(x + i^ky), x + i^ky}_{H} \\&= \sum_{k = 0}^{3}i^{k}\angles{x + i^ky, T(x + i^ky)}_{H} \\&= 4\angles{x, Ty}_{H}\end{align*}

Hence $T^{*}$ is an extension of $T$.$\square$

Definition 2.1.3.label Let $H$ be a Hilbert space and $T: D(T) \to H$ be a symmetric linear map, then $T$ is semipositive if $\dpb{x, Tx}H \ge 0$ for all $x \in H$, positive if $\dpb{x, Tx}H > 0$ for all $x \in H$, and coercive if there exists $\eps > 0$ such that $\dpb{x, Tx}H \ge \eps \norm{x}_{H}$ for all $x \in H$.

Lemma 2.1.4.label Let $H$ be a Hilbert space and $T: D(T) \to T(H)$ be an injective, self-adjoint operator with dense range, then $T^{-1}: T(H) \to D(T)$ is also self-adjoint.

Proof. Let $y \in D((T^{-1})^{*})$, then there exists $\phi \in H$ such that $\dpb{T^{-1}x, y}H = \dpb{x, \phi}H$ for all $x \in T(H)$. Under the substitution $x = Tz$, this implies that $\dpb{z, y}H = \dpb{Tz, \phi}H$ for all $x \in T(H)$. Hence $y = T^{*}(\phi) = T(\phi) \in T(H)$.$\square$

Theorem 2.1.5 (Friedrichs Extension, [Theorem 4.6, Bau14]).label Let $H$ be a Hilbert space and $T: D(T) \to H$ be a semipositive operator, then $T$ admits a self-adjoint extension.

Proof. For any $f, g \in D(T)$, let

\[\dpb{f, g}{E}= \dpb{f, Tg}{H}+ \dpb{f, g}{H}\]

then $\dpb{\cdot, \cdot}{E}$ is an inner product on $D(T)$. Let $E$ be the completion of $D(T)$ with respect to the norm induced by $\dpb{\cdot, \cdot}{E}$. Since $\norm{\cdot}_{E} \ge \norm{\cdot}_{H}$, the inclusion $\iota: D(T) \to H$ admits a continuous extension to $E$, which maps it into $H$ as a dense subspace.

To this end, let $\seq{x_n}\subset D(T)$ be a Cauchy sequence in $E$ such that $\norm{x_n}_{H} \to 0$ as $n \to \infty$, then

\begin{align*}\limv{n}\dpb{x_n, x_n}E&= \limv{n}\limv{m}\dpb{x_m, x_n}E \\&= \limv{n}\limv{m}\dpb{x_m, Tx_n}H \\&+ \limv{n}\limv{m}\dpb{x_m, x_n}H = 0\end{align*}

Hence $x_{n} \to 0$ in $E$, and the extension $\iota: E \to H$ is injective.

Since $\iota: E \to H$ is bounded and injective, it admits a bounded adjoint $\iota^{*}: H \to E$. Let $B = \iota \circ \iota^{*}: H \to H$, then $B$ is a bounded, injective, and self-adjoint.

Now, let $A = B^{-1}: B(H) \to H$, then $A$ is self-adjoint by Lemma 2.1.4. Moreover, for any $x, y \in B(H)$,

\begin{align*}\dpb{x, Ay}{H}&= \dpb{x, (\iota \circ \iota^*)^{-1}y}{H}\\&= \dpb{x, ({\iota^*})^{-1}\iota^{-1}y}H \\&= \dpb{\iota^{-1}x, \iota^{-1}y}E \\&= \dpb{x, (T + 1)y}H\end{align*}

So $(A - 1): B(H) \to H$ is the self-adjoint extension of $T$.$\square$

Definition 2.1.6 (Essentially Self-Adjoint).label Let $H$ be a Hilbert space and $T: D(T) \to H$ be a semipositive operator, then $T$ is essentially self-adjoint if the Friedrichs extension is its unique self-adjoint extension.

Lemma 2.1.7.label Let $H$ be a Hilbert space and $T: D(T) \to H$ be a semipositive. If there exists $\lambda > 0$ such that $(T^{*} + \lambda)$ is injective, then $T$ is essentially self-adjoint.

Proof. Let

\[\dpb{x, y}E = \dpb{x, (T^* + \lambda)y}H\]

be the inner product on $D(T^{*})$, then for any $y \in D(T^{*})$ with $y \perp D(T)$,

\[\dpb{x, (T^* + \lambda)y}H = 0 \quad \forall x \in D(T)\]

By density of $D(T)$ and injectivity of $T^{*} + \lambda$, $y = 0$. Hence $D(T)$ is dense in $D(T^{*})$ with respect to the inner product $\dpb{\cdot, \cdot}E$.

Let $S$ be a semipositive self-adjoint extension of $T$, then $D(S) \subset D(T^{*})$, so $D(T)$ is dense in $D(S)$ with respect to the inner product $\dpb{\cdot, \cdot}E$.

Let $E$ be the completion of $D(T^{*})$ with respect to the inner product $\dpb{\cdot, \cdot}E$ and $\iota: E \to H$ be the extension of the inclusion map, then for any $x, y \in D(T) \subset E$,

\[\dpb{x, (S + \lambda)y}H = \dpn{\iota^{-1}x, \iota^{-1}y}E\]

Hence $S$ and the Friedrichs extension coincide on $D(T)$. By density of $D(T)$ in $D(S)$, they also coincide on $D(S)$.$\square$

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Brownian Motion and Stochastic Calculus

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