2.6 The Hille-Yosida Theorem

Definition 2.6.1.label Let $E$ be a Banach space and $\bracs{T_t|t \ge 0}$ be a strongly continuous contraction semigroup on $E$. Let

\[D = \bracs{x \in E \bigg | \lim_{t \to 0}\frac{T_{t}x - x}{t} \text{ exists}}\]

then:

(1)
For each $t \ge 0$, let
\[A_{t}: E \to E \quad x \mapsto \int_{0}^{t} T_{s}x ds\]

then for any $x \in E$, $A_{t}x \in D$ with
\[A(A_{t}x) = T_{t}x - x\]
(2)
$D$ is dense in $E$.
(3)
The operator
\[A: D(A) \to E \quad x \mapsto \lim_{t \to 0}\frac{T_{t}x - x}{t}\]

is closed.

and operator $A$ in $(2)$ is the generator of $\bracs{T_t|t \ge 0}$.

Proof. (1): For any $r \in (0, t)$,

\begin{align*}\frac{T_{r}A_{t}x - A_{t}x}{r}&= \frac{1}{r}\braks{T_r\int_0^tT_sx ds - \int_0^tT_sxds}\\&= \frac{1}{r}\braks{\int_r^{t+r}T_sx ds - \int_0^tT_sxds}\\&= \frac{1}{r}\braks{\int_t^{t+r}T_sx ds - \int_0^rT_sxds}\end{align*}

By the Fundamental Theorem of Calculus,

\[\lim_{r \downto 0}\frac{T_{r}A_{t}x - A_{t}x}{r}= T_{t}x - x\]

so for any $x \in E$ and $t > 0$, $A_{t}x \in D$.

(2): By continuity of $s \mapsto T_{s}x$ and the Fundamental Theorem of Calculus, $A_{t}x/t \to x$ strongly as $t \downto 0$. Therefore $D$ is dense in $E$.

(3): Let $\seq{x_n}\subset D$, $x \in E$, and $y \in E$ such that $x_{n} \to x$ and $Ax_{n} \to y$ as $n \to \infty$. By the Fundamental Theorem of Calculus,

\[T_{t}x_{n} - x_{n} = \int_{0}^{t} T_{s}Ax_{n} ds\]

By the Dominated Convergence Theorem,

\[T_{t}x - x = \int_{0}^{t}T_{s}yds\]

Using the Fundamental Theorem of Calculus again,

\[\lim_{t \downto 0}\frac{T_{t}x - x}{t}= y\]

Hence $y \in D$ with $Ax = y$.$\square$

Theorem 2.6.2 (Hille-Yosida).label Let $E$ be a Banach space and $A: D(A) \to E$ be a densely defined closed operator, then $A$ generates a strongly continuous contraction semigroup if and only if:

(1)
$\sigma(A) \subset (-\infty, 0]$.
(2)
$\normn{(\lambda - A)^{-1}}_{L(E; E)}\le \lambda^{-1}$ for all $\lambda > 0$.

Proof. First suppose that $A$ generates a strongly continuous contraction semigroup $\bracs{T_t|t \ge 0}$.

If the following integral does exist, then

\[\int_{0}^{\infty} e^{-t(\lambda - A)}dt = (\lambda - A)^{-1}\]

which suggests that

\[R_{\lambda} = \int_{0}^{\infty} e^{-\lambda t}T_{t}dt\]

is the inverse of $(\lambda - A)$. Since $t \mapsto T_{t}$ is continuous and $\norm{T_t}_{L(E; E)}\le 1$ for all $t \ge 0$, the integral converges absolutely as a Riemann integral with $\norm{R_\lambda}_{L(E; E)}\le \lambda^{-1}$.

For any $x \in E$ and $h > 0$,

\begin{align*}\frac{T_{h}R_{\lambda} x - R_{\lambda} x}{h}&= \int_{0}^{\infty} e^{-\lambda t}\frac{T_{h}T_{t}x - T_{t}x}{h}dt \\&= \frac{e^{\lambda h}}{h}\int_{h}^{\infty} e^{-\lambda s}(T_{s} - T_{s - h})xds \\&= \frac{e^{\lambda h}}{h}\braks{R_\lambda x - \int_0^h e^{-\lambda s}T_sxds}\\&- \frac{e^{\lambda h}}{h}\int_{h}^{\infty} e^{-\lambda s}T_{s - h}xds \\&= \frac{e^{\lambda h}}{h}\braks{R_\lambda x - \int_0^h e^{-\lambda s}T_sxds}\\&- \frac{1}{h}\int_{0}^{\infty} e^{-\lambda s}T_{s}xds \\&= \frac{e^{\lambda h}- 1}{h}R_{\lambda} x - \frac{e^{\lambda h}}{h}\int_{0}^{h}e^{-\lambda s}T_{s}xds\end{align*}

By the Fundamental Theorem of Calculus,

\[\lim_{h \downto 0}\frac{e^{\lambda h}}{h}\int_{0}^{h}e^{-\lambda s}= x\]

Therefore

\begin{align*}AR_{\lambda} x&= \lambda R_{\lambda} x - x \\ (\lambda - A)R_{\lambda} x&= x \\ (\lambda - A)R_{\lambda}&= \text{Id}\end{align*}

On the other hand, since $A$ is closed,

\[AR_{\lambda} x = \int_{0}^{\infty} e^{-\lambda t}T_{t}Axdt = R_{\lambda} Ax\]

Hence $R_{\lambda}(\lambda - A) = \text{Id}$ as well, so $R_{\lambda} = (\lambda - A)^{-1}$.

Now suppose that $A$ is a densely defined closed operator satisfying assumptions (1) and (2). For each $\lambda > 0$, let

\[A_{\lambda} = -\lambda + \lambda^{2}(\lambda - A)^{-1}\]

then for any $x \in D(A)$,

\begin{align*}A_{\lambda} x&= -\lambda(\lambda-A)(\lambda-A)^{-1}x + \lambda^{2}(\lambda-A)^{-1}x \\&= \lambda A(\lambda-A)^{-1}x = \lambda(\lambda-A)^{-1}Ax\end{align*}

By assumption (2), for any $y \in E$,

\begin{align*}\lambda(\lambda - A)^{-1}y - y&= \lambda(\lambda - A)^{-1}(y - \lambda^{-1}(\lambda-A)y) \\&= (\lambda - A)^{-1}(\lambda y - (\lambda-A)y) \\&= (\lambda-A)^{-1}Ay \\ \lim_{\lambda \to \infty}\lambda(\lambda - A)^{-1}y - y&= 0\end{align*}

Therefore for any $x \in D(A)$,

\[\limv{\lambda}A_{n}x = x\]

For each $\lambda > 0$, define

\[T^{\lambda}_{t} = \sum_{k = 0}^{\infty} \frac{t^{k}A_{\lambda}^{k}}{k!}\]

then $\bracsn{T^\lambda_t|t \ge 0}$ is a (very) strongly continuous semigroup with

\begin{align*}T_{t}^{\lambda}&= e^{-\lambda t}\sum_{k = 0}^{\infty} \frac{\lambda^{2k}t^{k}(\lambda - A)^{-k}}{k!}\\ \normn{T_t^\lambda}_{L(E; E)}&\le e^{-\lambda t}\sum_{k = 0}^{\infty} \frac{\lambda^{k} t^{k}}{k!}\le 1\end{align*}

so $\bracsn{T^\lambda_t|t \ge 0}$ is a contraction semigroup.

Now, for any $x \in D(A)$, $t \ge 0$, and $\lambda, \mu > 0$, by the Fundamental Theorem of Calculus,

\begin{align*}{(T_t^\lambda - T_t^\mu) x}&= \int_{0}^{t} \frac{d}{dx}(T_{s}^{\lambda} T_{t - s}^{\mu} x)ds \\&= \int_{0}^{t} T_{s}^{\lambda} T_{t - s}^{\mu} (A_{\lambda} - A_{\mu})xds \\ \normn{(T_t^\lambda - T_t^\mu) x}_{E}&\le t\norm{(A_\lambda - A_\mu)x}_{E}\end{align*}

This allows defining

\[T_{t}x = \limv{n}T_{t}^{n}x\]

for all $x \in D(A)$, and extend to $E$ by density. By continuity, $\bracs{T_t|t \ge 0}$ is a strongly continuous contraction semigroup.

Finally, let $B$ be the generator of $\bracs{T_t|t \ge 0}$, then for any $t \ge 0$, $x \in D(A)$, and $\lambda > 0$, by the Dominated Convergence Theorem,

\begin{align*}T_{t}^{\lambda} x&= x + \int_{0}^{t} T_{s}^{\lambda} Axds \\ T_{t}x&= x + \int_{0}^{t}T_{s} Axds\end{align*}

Hence $B$ is an extension of $A$. On the other hand, $(\lambda - A)$ and $(\lambda - B)$ are both bijective. Therefore $A$ is an extension of $B$, and $A= B$.$\square$

Definition 2.6.3.label Let $E$ be a Banach space, $A: E \to E$ be a densely defined, closed operator, then $A$ is dissipative if for every $x \in E$, there exists $\phi \in E^{*}$ such that:

(1)
$\norm{\phi}_{E^*}= \norm{x}_{E}$.
(2)
$\dpb{x, \phi}{E}= \norm{x}_{E}^{2}$.
(3)
$\phi(Ax) \le 0$.

Proposition 2.6.4.label Let $E$ be a Banach space and $A: E \to E$ be a densely defined, closed operator, then $A$ generates a strongly continuous contraction semigroup if and only if:

(1)
$A$ is dissipative.
(2)
For each $\lambda > 0$, $(\lambda - A)$ is surjective.

Proof. First suppose that $A$ generates a strongly continuous contraction semigroup $\bracs{T_t|t \ge 0}$. By the Hahn-Banach theorem, there exists $\phi \in E^{*}$ with $\norm{\phi}_{E^*}= \norm{x}$, $\dpb{x, \phi}{E}= \norm{x}_{E}^{2}$.

Since $\bracs{T_t|t \ge 0}$ is a contraction semigroup,

\[\dpb{T_tx, \phi}{E}\le \norm{\phi}_{E^*}\norm{x}_{E} \le \norm{x}^{2} = \dpb{x, \phi}{E}\]

At $t = 0$,

\[0 \ge \frac{d}{dt}\dpb{T_tx, \phi}{E}= \dpb{Ax, \phi}{E}\]

Now suppose that $A$ is a dissipative operator with $(\lambda - A)$ being surjective for all $\lambda > 0$. For any $x \in D(A)$, let $\phi \in E^{*}$ with $\norm{\phi}_{E^*}= \norm{x}_{E}$, $\dpb{x, \phi}{E}= \norm{x}_{E}^{2}$, and $\dpb{Ax, \phi}{E}\le 0$, then

\begin{align*}\lambda \norm{x}_{E}^{2}&\le \lambda \dpb{x, \phi}{E}- \dpb{Ax, \phi}{E}\\&= \dpb{(\lambda - A)x, \phi}{E}\le \norm{(\lambda - A)x}_{E}\norm{\phi}_{E^*}\end{align*}

so $(\lambda - A)$ is coercive with $\normn{(\lambda - A)^{-1}}_{L(E; E)}\le \lambda^{-1}$.$\square$

Brownian Motion and Stochastic Calculus

2.6 The Hille-Yosida Theorem

Brownian Motion and Stochastic Calculus