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What should you be acquainted with? Measure Theory and some basics in Functional Analysis. For an in depth study we refer to E.B. Davies, One-Parameter Semigroups.

Continuous Contraction Semigroups

Definitions

For $t\in\R_0^+$ let $P_t:E\rar E$ be a family of bounded linear operators. This family is called a semigroup on $E$, if $P_0=1$ and for all $s,t\geq0$: $P_sP_t=P_{s+t}$. If $P_t$ is a semigroup such that for all $x\in E$ the map $t\mapsto P_tx$ is continuous, then the family is called a continuous semigroup. If $P_t$ is a semigroup such that for all $t$: $\norm{P_t}\leq1$, $P_t$ is called a contraction semigroup
In this chapter we study so called continuous contraction semigroups $P_t$.
If $E$ is finite dimensional, then for all linear operators $A:E\rar E$ the family $P_t\colon=e^{tA}$ is a continuous semigroup. 2. If $P_t$ is a contraction semigroup then for all eigen values $\l$ of $A$ we have: $\Re\l\leq0$. The converse is not necessarily true: take for example $A:\ell_2^2\rar\ell_2^2$, $Ae_1=0$, $Ae_2=e_1$, then $P_t=1+tA$, which is not a contraction.
Suppose $P_t$ is a contraction semigroup on $E$ and $F$ a dense subspace of $E$, such that for all $x\in F$: $$ \lim_{t\to0}\norm{P_tx-x}=0~. $$ Then $P_t$ is a continuous contraction semigroup on $E$.
$\proof$ By the semigroup property it suffices to prove continuity at $t=0$. For $x\in E$ choose $y\in F$, such that $\norm{x-y}\leq\e$ and choose $t$ such that $\norm{P_ty-y}\leq\e$. Then we get $$ \norm{P_tx-x} \leq\norm{P_t(x-y)}+\norm{x-y}+\norm{P_ty-y} \leq3\e~. $$ $\eofproof$
Let $X$ be a complete vector field on a manifold $M$ with flow $\theta$ and $\r$ a smooth density such that $\divergence(\r X)=0$. Then $P_tf\colon=f\circ\theta_t$ is a continuous contraction semigroup on $L_p(\mu)$ for all $1\leq p < \infty$.
By Liouville's Theorem $\mu$ is $\theta_t$-invariant and thus $P_t:L_p(\mu)\rar L_p(\mu)$ is an isometry.
We check continuity of $t\mapsto P_tf$ for all $f\in C_c^\infty(M)$: Let $K$ be a compact subset of $S$ such that for all $|t| < 1$: $f\circ\theta_t|K^c=0$, this is possible since $\theta$ is continuous and $f\in C_c(M)$. Since $(t,x)\mapsto d(\theta_t(x),x))$ is continuous on $\R\times K$, the function $h:t\mapsto\sup\{d(\theta_t(x),x)):x\in K\}$ is continuous (cf. e.g. exam). Thus if $C$ is the Lipschitz constant of $f$, then $$ \sup_{x\in K}|f(\theta_t(x))-f(x)|\leq Ch(t), $$ which implies that for all $1\leq p\leq\infty$ the map $t\mapsto P_tf$ is continuous in $L_p(\mu)$. Since $C_c^\infty(M)$ is dense in $L_p(\mu)$ for $p < \infty$, we are done by lemma.
Suppose $P_t$ is a contraction semigroup of self-adjoint operators on a Hilbert space $E$. Let $F$ be a dense subspace of $E$ such that for all $x\in F$: $$ \lim_{t\to0}\la P_tx,x\ra=\Vert x\Vert^2~. $$ Then $P_t$ is a continuous contraction semigroup on $E$. Suggested solution.
Let $P_1,\ldots,P_m$ be commuting projections on a Banach space $E$ such that $\norm{P_j}\leq1$. Then $$ P_t\colon=\prod_{j=1}^m(P_j+e^{-t}(1-P_j)) $$ is a continuous contraction semigroup on $E$. Solution by T. Speckhofer.
Verify that the following are continuous contraction semigroups on $C_0(\R^+)$. \begin{eqnarray*} P_tf(x)&\colon=&e^{-t/x}f(x)+\int_x^\infty\frac{te^{-t/y}f(y)}{y^2}\,dy\quad\mbox{and}\\ Q_tf(x)&\colon=&\frac{xf(x)}{x+t}+\int_x^\infty\frac{tf(t+y)}{(t+y)^2}\,dy \end{eqnarray*}
For any continuous contraction semigroup $P_t:E\rar E$ we put $$ \dom L\colon=\Big\{x\in E:\,\lim_{t\dar0}\frac{P_tx-x}t\mbox{ exists}\Big\} $$ and for all $x\in\dom L$ let us denote this limit by $Lx$. The linear operator $L$ is called the generator of the continuous contraction semigroup $P_t$ with domain $\dom L$.
For instance, by exam the domain of the generator $-\D$ of the heat semigroup on $L_2(\R^d)$ contains the space $H^2(\R^d)$. Obviously, the generator $-\D$ of the heat semigroup on $C_0(\R^d)$ has a different domain!
For all $y\in\R^d$ the operators $P_tf(x)\colon=e^{it\la x,y\ra}f(x)$ form a continuous contraction semigroup on $L_2(\R^d)$ with generator $Lf(x)=i\la x,y\ra f(x)$.
Find the domain of the generator of the continuous contraction group $P_tf(x)=f(x+t)$ on $C_0(\R)$ and on $L_2(\R)$.
Let $\g$ denote the standard gaussian measure on $\R^d$. The semigroup $$ P_tf(x) \colon=\int f(e^{-t}x+\sqrt{1-e^{-2t}}\,y)\,\g(dy) =\Big(\frac{1-e^{-2t}}{2\pi}\Big)^{d/2} \int f(y)e^{-\frac{\Vert e^{-t}x-y\Vert^2}{2(1-e^{-2t})}}\,dy $$ is called the Ornstein-Uhlenbeck semigroup on $L_2(\g)$. $P_t$ is self-adjoint on $L_2(\g)$ and its generator is the diffusion operator $$ -Hf(x)=\sum\pa_j^2f(x)-\sum x_j\pa_jf(x)~. $$ Hence $\g$ is the speed measure of $H$ and by lemma: $$ \int Hf.g\,d\g =\int\la\nabla f,\nabla g\ra\,d\g~. $$
$\proof$ $\eofproof$
Prove the following assertions for the Ornstein-Uhlenbeck semigroup $P_t$: 1. $P_t$ is a continuous contraction semigroup on $L_1(\g)$. 2. For all $f\in C_c^\infty(\R^d)$: $\Vert\nabla P_tf(x)\Vert\leq e^{-t}\Vert\nabla f(x)\Vert$.
Verify that the following is a continuous contraction semigroups on $C(\TT)$. $$ P_tf(x)\colon=\frac1{2\pi}\int_0^{2\pi} P(e^{-t},x-y)f(y)\,dy \quad\mbox{where}\quad P(r,x)\colon=\frac{1-r^2}{1-2r\cos x+r^2} =\Re\Big(\frac{1+re^{ix}}{1-re^{ix}}\Big)~. $$ This is called the Poisson semigroup on the torus $\TT$. 2. Show that $P_te_m=e^{-t|m|}e_m$ for $e_m(x)=e^{imx}$, $m\in\Z$. 3. Compute the generator of $P_t$. Remark: Use the fact that for $f\in C(\TT)$ and $z=e^{-t}e^{ix}$, then function $z\mapsto P_tf(x)$ is the unique harmonic extension of the function $e^{ix}\mapsto f(x)$ defined on the boundary $S^1=\pa D$ to the unit disc $D\colon=[|z| < 1]$. Suggested solution.
The higher dimensional analogue is a bit more intricate:
Let $\s$ denote the normalized surface measure on the sphere $S^{d-1}$. The Poisson semigroup on $S^{d-1}$ is given by $$ P_tf(x) =\int_{S^{d-1}}\frac{(1-e^{-2t})}{\norm{e^{-t}x-y}^d}\,f(y)\,\s(dy)~. $$ The generator of this semigroup is $$ -H=\tfrac12(d-2)-\sqrt{\tfrac14(d-2)^2+\D_{S^{d-1}}} $$ where $\D_{S^{d-1}}$ is the Laplacian on $S^{d-1}$; $-H$ is not a diffusion operator!
$\proof$ $\eofproof$
In fact both the Poisson- and Ornstein-Uhlenbeck semgroups are special cases of a more general construction: Given an orthogonal decomposition $E_n$, $n\in\N_0$ of a Hilbert space $E$; let $Q_n:E\rar E_n$ be the sequence of orthogonal projections and put for all $z\in D\colon=[|z| < 1]$: $$ T(z)\colon=\sum_{n\geq0}z^nQ_n~. $$ Obviously $T(z)$ is a linear operator and for all normed $x\in E$ we have: $\norm{T(z)x}^2\leq\sum\norm{Q_nx}^2=1$; for $w,z\in D$ we have: $T(z)T(w)=T(zw)$, proving that the family $P_t\colon=T(e^{-t})$, $t\geq0$, is a contraction semigroup on $E$. Moreover it's a continuous semigroup: this clearly holds for $x\in E_n$ and thus for $x\in\bigoplus E_n$, which by definition is a dense subspace of $E$. Finally the generator $-H$ is easily computed on the space $E_n$: for all $x\in E_n$. $-Hx=-nx$, i.e. $$ \dom(H)=\{x\in E:\sum n^2\norm{Q_nx}^2 < \infty\} \quad\mbox{and}\quad Hx=\sum nQ_nx\quad \ker H=E_0~. $$ The Poisson semigroup is generated by taking for $E_n$ the space of spherical harmonics of degree $n$ and the Ornstein-Uhlenbeck semigroup is obtained by taking for $E_n$ the space of Hermite polynomials of degree $n$.
But there is still another semigroup, actually a continuous group: $U_t\colon=T(e^{-it})$
Verify that $-iH$ is the generator of $U_t$, $t\in\R$. Make an educated guess for an integral formula for $U_t$ in case of the Poisson and the Ornstein-Uhlenbeck semigroup. Suggested solution.

Integration in Banach Spaces

Let $E$ be a separable Banach space, $I=(a,b]$ an interval (possibly unbounded) and $f:I\rar E$ a simple integrable function, i.e. there are finitely many pairwise disjoint measurable subsets $A_j$ of $I$ and $x_j\in E$, such that $$ f=\sum x_jI_{A_j} \quad\mbox{and}\quad \l(A_j) < \infty $$ In this case we define $$ \int_a^b f(t)\,dt \colon=\int_I f\,d\l \colon=\sum_j x_j\l(A_j)~. $$ By the triangle inequality we get the widely-used inequality: $$ \Big\Vert\int f\,d\l\Big\Vert\leq\int\norm{f(t)}\,dt~. $$ Now we extend this definition to integrable functions: So let $f:I\rar E$ be measurable, i.e. for all $x^*\in E^*$ the function $t\mapsto x^*(f(t))$ is measurable. Then $\norm f$ is measurable: Let $x_n$ be dense in $E$ and choose by Hahn-Banach $x_n^*\in E^*$, $\norm{x_n^*}=1$ such that $\norm{x_n}=x_n^*(x_n)$ - such a sequence $x_n^*$ is called a norming sequence, then for all $x\in E$: $\Vert x\Vert=\sup|x_n^*(x)|$ and thus $$ \norm{f(t)}=\sup_n|x_n^*(f(t))|~. $$ Suppose $f$ is measurable and $\int_a^b\norm{f(t)}\,dt < \infty$ - such a function is called integrable. In this case we can find a sequence of simple integrable functions $f_n$, such that $$ \norm{f_n(t)}\leq2\norm{f(t)} \quad\mbox{and}\quad \lim_n\int_a^b\norm{f(t)-f_n(t)}\,dt=0~. $$ This is not at all obvious, we simply take it for granted (cf. e.g. N. Dunford and J.T. Schwartz, Linear Operators I). More generally, we call a sequence $f_n$ of simple functions a determining sequence for $f$, if there exists a sequence of simple integrable functions $f_n:I\rar E$ such that $$ \lim_n\int_a^b\norm{f(t)-f_n(t)}\,dt=0~. $$ It can be checked readily that $\int f_n\,d\l$ is a Cauchy sequence and therefore we define: $$ \int f\,d\l\colon=\lim_n\int f_n\,d\l~. $$ Again, it's easily verified that this limit doesn't depend on the particular determining sequence choosen.
If $f:I\rar E$ is continuous and $I=[a,b]$ compact, then $f$ is integrable and $$ \int_a^b f(t)\,dt =\lim_{n\to\infty}\frac{b-a}{2^n}\sum_{j=0}^{2^n-1} f(t_j), $$ where $t_j$ may be any point in the inverval $[a+(b-a)j2^{-n},a+(b-a)(j+1)2^{-n}]$. Thus for continuous functions on compact intervals the integral coincides with the Riemann integral.
If $f:I\rar E$ is integrable and $c\in I$, then $F(t)\colon=\int_c^t f(s)\,ds$ is continuous. Solution by T. Speckhofer.
Two properties are checked easily: 1. the triangle inequality: $$ \Big\Vert\int f\,d\l\Big\Vert \leq\int\norm f\,d\l $$ and 2. for all $x^*\in E^*$: $$ x^*\Big(\int f(t)\,dt\Big) =\int x^*(f(t))\,dt~. $$ This makes it particularly easy to evalute integrals in function spaces like $C_b(S)$ or $C_0(S)$:
$f:[0,1]\rar C[0,1]$, $f(t)(x)\colon=x^t$ is integrable and $$ \Big(\int_0^1f(t)\,dt\Big)(x) =\frac{x-1}{\log x}~. $$
Since evaluation at $x$ is a continuous linear functional on $C[0,1]$, we get $$ \Big(\int_0^1f(t)\,dt\Big)(x) =\int_0^1f(t)(x)\,dt =\frac{x-1}{\log x}~. $$ 3. If $f:I\rar E$ happens to be continuous, then for any $c\in I^\circ$ we put $$ F(t)\colon=\int_c^t f(s)\,ds~. $$ We have the so called fundamental theorem of calculus: $F$ is differentiable and $$ \forall t\in I^\circ:\quad F^\prime(t)=f(t)~. $$ As in the real case it's this theorem that usually allows to calculate integrals explicitely!
Prove the fundamental theorem of calculus for continuous functions $f:I\rar E$.
Suppose $f:[a,b]\rar E$, $\vp:[a,b]\rar\C$ are $C^1$-functions on a compact interval $[a,b]$, then $$ \int_a^b\vp^\prime(t)f(t)\,dt =\vp(b)f(b)-\vp(a)f(a)-\int_a^b\vp(t)f^\prime(t)\,dt $$
Contrary to the case of functions with values in spaces like $C_b(S)$, evaluation is not at all defined on spaces like $L_1(\mu)$. The following will usually suffice for our purposes: Suppose $f:I\rar L_1(S,\mu)$ is integrable, i.e. $\int\int_S|f(t)(x)|\mu(dx)\,dt < \infty$. Then it can be proved (cf. exam) that there is a measurable mapping $F:I\times S\rar\R$, such that
  1. For $\l$ almost all $t\in\R$: $F(t,.)=f(t)$.
  2. For $\mu$ almost all $x\in S$: $F(.,x)\in L_1(I)$ and for $\mu$ almost all $x\in S$: $$ \int_I F(t,x)\,dt =\Big(\int_I f(t)\,dt\Big)(x)~. $$
Thus we can essentially integrate pointwise and the situation doesn't really differ from integration in $C_b(S)$ spaces.
If $f:[0,1]\rar E$ is integrable, then for $\mu$ almost all $x$ the function $t\mapsto\Big(\int_0^t f(s)\,ds\Big)(x)$ is continuous.
Let $f:\R\rar X$ be integrable. If for all $y\in\R$: $\int_\R f(t)e^{-iyt}\,dt=0$, then $f=0$.
Choose a norming sequence $x_n^*\in X^*$, then $\norm{f(t)}=\sup_n\{|x_n^*\circ f(t)|\}$ and thus the Fourier transforms of the functions $f_n\colon=x_n^*\circ f$ vanish, i.e. $f_n=0$ a.e. Hence $\norm f=\sup_n|f_n|=0$ a.e.
Let $f:\R^+\rar X$ be integrable. If for all $y\in\R^+$: $\int_0^\infty f(t)e^{-yt}\,dt=0$, then $f=0$.

Convergence

As in the real case every sequence $f_n$ converging in measure has a subsequence $f_{k(n)}$ converging almost uniformly and therefore $f_{k(n)}$ converges a.e. If $\l(I) < \infty$, then every sequence converging a.e. converges almost uniformly (Egorov's Theorem) and thus it converges in measure.
If $f_n:I\rar E$ converges a.e. or in measure to $f$ and if there exists an integrable function $g:I\rar\R_0^+$ such that for all $n\in\N$: $\norm{f_n}\leq g$, then $$ \lim_n\int f_n\,d\l =\int f\,d\l~. $$
$\proof$ Suppose $f_n\to f$ a.e. or in measure, then we can find a subsequence $f_{k(n)}$, converging a.e. to $f$. Hence $\norm f\leq g$ and $\norm{f-f_n}\leq 2g$. Thus $\norm f$ is integrable and by the usual dominated convergence theorem: $$ \norm{\int f\,d\l-\int f_n\,d\l} \leq\int\norm{f-f_n}\,d\l\rar0. $$ $\eofproof$
Now let $F$ be another separable Banach space, $A:E\rar F$ a bounded linear operator and $f:I\rar E$ integrable, then $A(f)$ is integrable as well and $$ \int A(f)\,d\l =A\Big(\int f\,d\l\Big)~. $$ In particular, if $X:I\rar E\times F$ is integrable, then so are $f\colon=\Prn_E X$ and $g=\Prn_F X$ and we have $$ \int X\,d\l =\Big(\int f\,d\l,\int g\,d\l\Big)~. $$
Let $E,F$ be Banach spaces and $A:\dom A(\sbe E)\rar F$ a closed (linear) operator, i.e. $\G(A)\colon=\{(x,Ax):x\in\dom A\}$ is a closed subspace of $E\times F$ and $\dom A$ is dense. Suppose both $f:I\rar\dom A$ and $t\mapsto Af(t)$ are integrable, then: $$ \int f\,d\l\in\dom A \quad\mbox{and}\quad A\Big(\int f\,d\l\Big) =\int Af\,d\l~. $$
$\proof$ Since $\G(A)$ is closed $\G(A)$ is a Banach space. Moreover $f$ and $A(f)$ are integrable and thus $X\colon=(f,A(f))$ is integrable and $$ \Big(\int f\,d\l,\int A(f)\,d\l\Big) =\int X\,d\l \in\G(A)~. $$ $\eofproof$
If $L$ is closed then the preimage of every compact set is closed. In particular $\ker L$ is a closed subspace of $E$. Suggested solution.
$f:[0,1]\rar L_p[0,1]$, $1\leq p < \infty$, $f(t)(x)\colon=x^t$ is integrable and $$ \Big(\int_0^1f(t)\,dt\Big)(x) =\frac{x-1}{\log x}~. $$
The canonical injection $J:C[0,1]\rar L_p[0,1]$ is a contraction and thus the assertion follows from exam and proposition
If $A_0:D(\sbe E)\rar E$ is a symmetric linear operator on a dense subspace $D$ of a Hilbert space $E$, then the closure $\cl{\G(A_0)}$ defines a closed, symmetric linear operator $A:\dom A(\sbe E)\rar E$ with $D\sbe\dom A$. 2. Find $\dom A$ in case $E=L_2(\R)$, $A_0f=-f^\dprime$ and $D=C_c^\infty(\R)$.
Extend the definitions of measurable and integrable function $f:I\rar E$ to an arbitrary $\s$-finite measure space $(\O,\F,\mu)$ and $f:\O\rar E$.
Suppose $f:I\times J\rar E$ is integrable, then: $$ \int_{I\times J} f\,d\l =\int_I\int_Jf(s,t)\,dt\,ds =\int_J\int_If(s,t)\,ds\,dt~. $$
Suppose $F\in L_p(S_1\times S_2,\mu_1\otimes\mu_2)$. Then the mapping $f:S_1\rar:L_p(\mu_2)$, $f(x_1)\colon=F_{x_1}$ is measurable and if $p=1$ $f:S_1\rar L_1(S_2,\mu_2)$ is integrable and for $\mu_2$ almost all $x_2\in S_2$: $$ \Big(\int_{S_1} f\,d\mu_1\Big)(x_2) =\int_{S_1} F(x_1,x_2)\,\mu_1(dx_1)~. $$ If $f:S_1\rar L_p(S_2,\mu_2)$ is integrable, then this relation also holds in $L_p(S_2,\mu_2)$. Suggested solution.
If $F:I\times S\rar[0,\infty]$ is measurable, then for all $1\leq p < \infty$: $$ \Big(\int_S\Big(\int_I F(t,x)\,dt\Big)^p\,d\mu\Big)^{1/p} \leq\int_I\Big(\int_S F(t,x)^p\,d\mu\Big)^{1/p}\,dt $$ This is called Minkowski's inequality. Verify that it's just the triangle inequality for $f:I\rar L_p(S,\mu)$, $f(t)=F_t$.
The following result generalizes a statement alluded to above:
Suppose $(S_i,\F_i,\mu_i)$, $i=1,2$, are measure spaces. If $f:S_1\rar L_1(S_2,\mu_2)$ is integrable, then there is some function $F\in L_1(S_1\times S_2,\mu_1\otimes\mu_2)$, such that
  1. For $\mu_1$ almost all $x_1\in S_1$: $F_{x_1}=f(x_1)$.
  2. For $\mu_2$ almost all $x_2\in S_2$: $F_{x_2}\in L_1(S_1,\mu_1)$ for $\mu_2$ almost all $x_2\in S_2$: $$ \int_{S_1} F(.,x_2)\,d\mu_1 =\Big(\int_{S_1} f(x_1)\,\mu_1(dx_1)\Big)(x_2)~. $$

Integrating the heat equation

Lots of operators are defined by integration of semigroups. Just for motivation let us take a positive self-adjoint linear operator $H$ on a finite dimensional Hilbert space $E$ ($\dim E=d$): $-H$ is the generator of the continuous contraction semigroup $P_t\colon=e^{-tH}$. For any $\l > 0$ and $\Re z > 0$ we have: $$ \forall x\in E:\quad (\l^2+H)^{-z}x=\frac1{\G(z)}\int_0^\infty t^{z-1} e^{-\l^2t}P_tx\,dt~. $$ Indeed by the spectral theorem there is an ONB $e_1,\ldots,e_n$ and positive numbers $0\leq\l_1\leq\cdots\leq\l_d$, such that $He_n=\l_ne_n$. It follows that: \begin{eqnarray*} \int_0^\infty t^{z-1}e^{-\l^2t}P_te_k\,dt &=&\int_0^\infty t^{z-1} e^{-(\l^2+\l_k)t}\,dt\,e_k\\ &=&\G(z)(\l^2+\l_k)^{-z}e_k =\G(z)(\l^2+H)^{-z}e_k~. \end{eqnarray*} Now let's look at the heat semigroup $P_t=e^{-t\D}$ on $\R^d$: For all $f\in C_0(\R^d)$: $$ P_tf(x) =\frac1{(4\pi t)^{d/2}}\int f(y)e^{-\frac{\norm{x-y}^2}{4t}}\,dy~. $$ For $\l > 0$ and $\Re z > 0$ we put: $$ U_z^{\l^2}f \colon=(\l^2+\D)^{-z}f \colon=\frac1{\G(z)}\int_0^\infty t^{z-1}e^{-\l^2t}P_tf\,dt $$ This family of operators is called the family of Bessel's potential operators - we will see that $U_\l^1$ is just the resolvent of $-\D$. In case this definition makes sense for $\l=0$, the family of operators $G_z\colon=U_z^0$ is called the family of Riesz's potential operators.
For $0 < \Re z < d/2$ and $f\in C_c(\R^d)$ we have: $$ G_zf \colon=\frac1{\G(z)}\int_0^\infty t^{z-1}P_tf\,dt =\frac{\G(\frac d2-z)}{2^{2z}\pi^{\frac d2}\G(z)} \int\norm{x-y}^{2z-d}f(y)\,dy~. $$
For all $a\in\R$ we have: $$ \frac1{\G(z)} \int_0^\infty t^{z-1}(4\pi t)^{-\frac d2} e^{-\frac{a^2}{4t}}\,dt =\frac{\G(\frac{d-2z}2)|a|^{2z-d}} {2^{2z}\pi^{\frac d2}\G(z)}~. $$ Assume w.l.o.g. that $0\leq f\leq1$ and $\int f=1$, then $P_tf\geq0$ and $\norm{P_tf}\leq\max\{1,(4\pi t)^{-d/2}\}$. This shows that $t\mapsto t^{z-1}\norm{P_tf}$ is integrable in e.g. $C_0(\R^d)$ and since $f\mapsto f(x)$ is a continuous linear map we get by proposition, Fubini and the preceeding caculation: \begin{eqnarray*} G_zf(x) &=&\frac1{\G(z)}\int_0^\infty t^{z-1}P_tf(x)\,dt\\ &=&\frac1{\G(z)}\int\int_0^\infty t^{z-1}f(y)e^{-\frac{\norm{x-y}^2}{4t}}\,dy\,dt =\frac{\G(\frac d2-z)}{2^{2z}\pi^{\frac d2}\G(z)} \int\norm{x-y}^{2z-d}f(y)\,dy~. \end{eqnarray*}
For $\l,r > 0$ and $\nu\in\C$: $$ K(\l,\nu,r) \colon=\int_0^\infty t^{-\nu-1}e^{-\l^2 t-r^2/4t}\,dt $$ Prove that Bessel's potential operators are given by $$ U_z^{\l^2}f(x) =\frac1{(4\pi)^{d/2}\G(z)}\int_{\R^d}K(\l,d/2-z,\norm{y-x})f(y)\,dy~. $$ Moreover we have (suggested solution):
  1. $K(\l,\nu,r)=\l^{2\nu}K(1,\nu,\l r)$.
  2. $\pa_\l K(\l,\nu,r)=-2\l K(\l,\nu-1,r)$.
  3. $\pa_rK(\l,\nu,r)=-\frac12rK(\l,\nu+1,r)$.
  4. $r\pa_r^2K+(2\nu+1)\pa_rK-\l^2 rK=0$.
  5. $K(\l,1/2,r)=(4\pi)^{1/2}e^{-\l r}/r$ und $K(\l,-1/2,r)=(4\pi)^{1/2}e^{-\l r}/\l$.
Verify the Bessel potentials for the Laplacian on $\TT^d$ applied to $f\in C(\TT^d)$ are just the Bessel potentials of the Laplacian on $\TT^d$ applied to the periodic extension $F$ of $f$. The definition of Riesz potentials doesn't make sense in this case!
If $A$ is a positive self-adjoint linear operator on a finite dimensional Hilbert space $E$, then for all $0 < \a < 1$: $$ (1+A)^{-\a}=c_\a\int_0^\infty t^{-\a}(1+t+A)^{-1}\,dt \quad\mbox{where}\quad c_\a=\int_0^\infty t^{-\a}(1+t)^{-1}\,dt=\frac{\pi}{\sin(\pi\a)} $$

The Cauchy Problem

 The Cauchy problem is about solutions $f:\R_0^+\rar E$ of the equation $f^\prime(t)=Lf(t)$ given $f(0)$; we again remark that the derivative on the left hand side is understood as the derivative of the curve $f$ in $E$! In case $L$ is a bounded linear operator this is just an ordinary differential equation in $E$, which is known to have the unique solution $f(t)=e^{tL}f(0)$. We are going to examine the case where $L$ is the generator $L$ of a continuous contraction semigroup $P_t$.
Let $F:\R_0^+\rar\R$ be continuous such that for all $t\in\R_0^+$ the right derivative $$ D_+F(t)\colon=\lim_{h\dar0}\frac{F(t+h)-F(t)}h $$ vanishes. Then $F$ is constant. Hint: for all $\a > 0$ the set $\{t:\forall s\leq t:|F(s)-F(0)|\leq\a s\}$ is closed, open and not empty. Solution by T. Speckhofer.
Let $P_t$ be a continuous contraction semigroup on $E$ with generator $L$ and $[a,b]\sbe\R^+$ a bounded interval. Then the following holds:
  1. For all $C^1$-functions $f:\R^+\rar\R$ and all $x\in E$ we have: $$ \int_a^bf(t)P_tx\,dt\in\dom L \quad\mbox{and}\quad L\int_a^bf(t)P_tx\,dt =f(b)P_bx-f(a)P_ax-\int_a^bf^\prime(t)P_tx\,dt~. $$
  2. For all $x\in E$ and all $t\geq0$ the function $u(t)\colon=\int_0^tP_sx\,ds$ is a $\dom L$-valued function and: $Lu(t)=P_tx-x$.
  3. For all $x\in\dom L$: $P_tLx=LP_tx$ and thus $P_t(\dom L)\sbe\dom L$. Moreover $\dom L$ is dense in $E$.
  4. For all $x\in\dom L$ the curve $t\mapsto P_tx$ is $C^1$ and we have $$ P_tx-x=\int_0^t P_sLx\,ds~. $$
$\proof$ 1. Put $y=\int_a^bf(t)P_tx\,dt$, then by proposition: $$ P_sy =\int_a^bf(t)P_{s+t}x\,dt =\int_{a+s}^{b+s}f(t-s)P_tx\,dt~. $$ By e.g. dominated convergence we now conclude that: \begin{eqnarray*} \lim_{s\dar0}\frac{P_sy-y}s &=&\lim_{s\dar0}\frac1s\left(\int_{a+s}^{b+s}f(t-s)P_tx\,dt- \int_a^b f(t)P_tx\,dt\right)\\ &=&\lim_{s\dar0}\frac1s\left(\int_a^b(f(t-s)-f(t))P_tx\,dt -\int_a^{a+s}f(t-s)P_tx\,dt+\int_b^{b+s}f(t-s)P_tx\,dt\right)\\ &=&-\int_a^bf^\prime(t)P_tx\,dt +\lim_{s\dar0}\left(\int_b^{b+s}\frac{f(t-s)-f(t)}s\,P_tx\,ds -\int_a^{a+s}\frac{f(t-s)-f(t)}s\,P_tx\,ds\right)\\ &&+\lim_{s\dar0}\frac1s\left(\int_b^{b+s}f(t)P_tx\,ds -\int_a^{a+s}f(t)P_tx\,ds\right)\\ &=&-\int_a^bf^\prime(t)P_tx\,dt+0+f(b)P_bx-f(a)P_ax~. \end{eqnarray*} This is the well known formula for the derivative of the parametric integral $s\mapsto\int_{a+s}^{b+s}f(t-s)P_tx\,dt$ from analysis (cf. e.g. exam).
2. is just the case $f(t)=1$.
3. First of all we have for $x\in\dom L$ by boundedness of $P_t$ and the semigroup property: $$ P_tLx =P_t\Big(\lim_{s\dar0}\frac{P_{t+s}x-P_tx}{s}\Big) =\lim_{s\dar0}\frac{P_{t+s}x-P_tx}{s} =\lim_{s\dar0}\frac{P_sP_tx-P_tx}{s} =LP_tx, $$ i.e. $P_tx\in\dom L$ and on $\dom L$: $LP_t=P_tL$. Finally by 1. $u(t)/t\in\dom L$ and since $t\mapsto P_tx$ is continuous $u(t)/t$ converges to $x$ as $t\dar0$; by 2. this implies that $\dom L$ is dense.
4. For $x\in\dom L$ and $x^*\in E^*$ we compute the right derivative of $$ F(t)\colon=x^*\Big(P_tx-x-\int_0^tP_sLx\,ds\Big) $$ By definition this is $x^*(LP_tx)-x^*(P_tLx)$ and by 3.: $D_+F(t)=0$. Finally by exam $F(t)=0$ and by Hahn-Banach: $$ P_tx-x=\int_0^tP_sLx\,ds~. $$ $\eofproof$
The following exams are about so called parametric integration!
Find conditions on a smooth function $f\in L_1(\R^+)$ such that $\int_0^\infty f(t)P_tx\,dt\in\dom L$ and $$ L\int_0^\infty f(t)P_tx\,dt =-\int_0^\infty f^\prime(t)P_tx\,dt-f(0)x~. $$
Suppose $f:I\times J\rar E$ is measurable such that for all $t\in J$ the function $f_t(s)\colon=f(s,t)$ is continuous and for all $s\in I$ there is some $\d > 0$ such that $t\mapsto\sup\{\norm{f(u,t)}:|u-s| < \d\}$ is integrable. Then $$ F(s)\colon=\int_Jf(s,t)\,dt \quad\mbox{is continuous.} $$ 2. If moreover $f_t$ is differentiable for all $t\in J$ and for all $s\in I$ there is some $\d > 0$ such that $t\mapsto\sup\{\norm{\pa_u f(u,t)}:|u-s| < \d\}$ is integrable, then $F$ is differentiable and $$ F^\prime(s)=\int_J\pa_sf(s,t)\,dt~. $$
Suppose $I,J$ are compact intervals, $f:J\times I\rar E$ continuous, $u,v:J\rar I$ $C^1$ and $\pa_sf$ continuous. Then the function $$ F(s)=\int_{v(s)}^{u(s)}f(s,t)\,dt $$ is differentiable and (suggested solution) $$ F^\prime(s)=f(s,u(s))u^\prime(s)-f(s,v(s))v^\prime(s) +\int_{v(s)}^{u(s)}\pa_sf(s,t)\,dt $$
Let $P_t$ be a contraction semigroup on $E$, such that for all $x^*\in E^*$: $\lim_{t\to0}x^*(P_tx)=x^*(x)$. Then $P_t$ is a continuous semigroup.
$\proof$ Let $F$ be the subspace $\{x\in E:\lim_{t\to0}P_tx=x\}$, the $F$ is closed, indeed if $x_n\to x$ and $\lim_{t\to0}P_tx_n=x_n$, then for any $\e > 0$ choose $n\in\N$ such that $\norm{x-x_n} < \e$, then $$ \limsup_{t\to0}\norm{P_tx-x} \leq\limsup_{t\to0}\norm{P_tx-P_tx_n}+\limsup_{t\to0}\norm{P_tx_n-x_n}+\norm{x-x_n} \leq2\e~. $$ By Hahn-Banach $F$ is also weakly closed. For separable spaces $E$ the mapping $t\mapsto P_tx$ is measurable and bounded and thus integrable over bounded intervals. For all $h,t > 0$ and all $x^*\in E^*$ we have \begin{eqnarray*} |x^*(P_tu(h)-u(h))| &=&\Big|\frac1h\int_0^hx^*(P_{t+s}x)-x^*(P_sx)\,ds\Big|\\ &=&\Big|\frac1h\int_{t}^{h+t}x^*(P_sx)\,ds-\frac1h\int_{0}^{h}x^*(P_sx)\,ds\Big|\\ &=&\Big|-\frac1h\int_{0}^{t}x^*(P_sx)\,ds+\frac1h\int_{h}^{h+t}x^*(P_sx)\,ds\Big| \leq\frac{2t}h\Vert x^*\Vert\Vert x\Vert, \end{eqnarray*} i.e. for all $h > 0$: $u(h)\in F$. As $\lim_{h\to0}x^*(u(h))=x^*(x)$, $F$ is also weakly dense. $\eofproof$
Suppose $S$ is locally compact. A contraction semigroup $P_t$ on $C_0(S)$ is a continuous semigroup if and only if for all $f\in C_0(S)$ and all $x\in S$: $\lim_{t\to0}P_tf(x)=f(x)$.
Suppose $E$ is a Hilbert space. A contraction semigroup $P_t$ on $E$ is a continuous semigroup if and only if for all $x\in E$: $\lim_{t\to0}\la P_tx,x\ra=\Vert x\Vert^2$.
Define for $f\in L_1(\R^+)$ and $x\in E$: $T(f)x\colon=\int_0^\infty f(t)P_tx\,dt$. Then $T:L_1(\R^+)\rar L(X)$ and $\ker T$ is an ideal in the convolution algebra $L_1(\R^+)$. Suggested solution.

Closed linear operators

The generator $L:\dom L\rar E$ of a continuous contraction semigroup $P_t$ is closed.
$\proof$ If $x_n\in\dom L$, $x_n\rar x$ and $y_n=Lx_n\rar y$, then by proposition: $$ P_tx_n-x_n =\int_0^t P_sy_n\,ds \quad\mbox{and thus by e.g. dominated convergence}\quad P_tx-x =\int_0^t P_sy\,ds~. $$ Therefore we get by continuity of $s\mapsto P_sy$: $$ \lim_{t\dar0}\frac{P_tx-x}t =\lim_{t\dar0}\frac1t\int_0^tP_sy\,ds =y $$ i.e. $x\in\dom L$ and $Lx=y$. $\eofproof$
For any $x\in\dom L$ we have by proposition for all integrable functions $f:\R^+\rar\R$ and all $0\leq a < b < \infty$: $$ L\int_a^b f(t)P_tx\,dt =\int_a^b f(t)LP_tx\,dt =\int_a^b f(t)P_tLx\,dt =\int_a^b f(t)\ttd t P_tx\,dt~. $$ If moreover $f$ is $C^1$, then integration by parts gives you the formula in proposition.1. Thus the point of proposition.1 is that $x$ may be arbitrary.
Suppose $L$ is a closed operator. A subspace $D$ of $\dom L$ is called a core for $L$ if $D$ is a dense subspace of $\dom L$ in the graph norm $\Vert x\Vert_L\colon=\Vert x\Vert+\norm{Lx}$.
Suppose $L$ is the generator of a continuous contraction semigroup and $D\sbe\dom L$ is a dense $P_t$-invariant subspace of $E$, i.e. for all $t > 0$: $P_t(D)\sbe D$. Then $D$ is a core for $L$.
$\proof$ We have to prove that the closure $\cl D$ of $D$ in $(\dom L,\Vert.\Vert_L)$ is $\dom L$. So take any $x\in\dom L$ and choose a sequence $x_n$ in $D$ such that $\norm{x_n-x}=0$. We claim that $t\mapsto P_tx$ is a continuous curve in $(\dom L,\Vert.\Vert_L)$: indeed, by proposition: $$ \norm{P_tx-x}_L =\norm{P_tx-x}+\norm{LP_tx-Lx} =\norm{P_tx-x}+\norm{P_tLx-Lx} $$ This implies that $\int_0^t P_sx_n\,ds\in\cl D$. By proposition we get: \begin{eqnarray*} \lim_n\bigg\Vert\int_0^tP_sx_n\,ds-\int_0^tP_sx\,ds\bigg\Vert_L &=&\lim_n\bigg\Vert\int_0^tP_s(x_n-x)\,ds\bigg\Vert +\lim_n\bigg\Vert\int_0^tLP_s(x_n-x)\,ds\bigg\Vert\\ &=&\lim_n\bigg\Vert\int_0^tP_s(x_n-x)\,ds\bigg\Vert +\lim_n\Vert P_t(x_n-x)+(x_n-x)\Vert=0~. \end{eqnarray*} This shows that $\int_0^t P_sx\,ds\in\cl D$. Finally again by proposition: $$ \lim_{t\to0}\bigg\Vert\frac1t\int_0^tP_sx\,ds-x\bigg\Vert_L =\lim_{t\to0}\bigg\Vert\frac1t\int_0^tP_sx\,ds-x\bigg\Vert +\lim_{t\to0}\Vert t^{-1}(P_tx-x)-Lx\Vert=0~. $$ $\eofproof$
If $X$ is a complete vector field, then $C_c^\infty(M)$ is a core for $X$.
For a diffusion operator $H$ the space $C_c^\infty(M)$ is usually not invariant under $P_t=e^{-tH}$. However in case of the Laplacian on $\R^d$ the so called Schwartz space is invariant under $e^{-t\D}$.
The Schwartz space ${\cal S}(\R^d)$ is a core for $-\D$.

Generators determine the semigroups

Let $L$ be the generator of a continuous contraction semigroup $P_t$. Suppose $f:[0,t_0]\rar\dom L$ solves the equation $f^\prime(t)=Lf(t)$ for all $t\in[0,t_0]$. Then $f(t_0)=P_{t_0}f(0)$. In particular the generator determines the semigroup.
$\proof$ 1. For any $x^*\in E^*$ put $F(t)\colon=x^*(P_tf(t_0-t))$. Then $F(0)=x^*(f(t_0))$ and for the right derivative we get: \begin{eqnarray*} D_+F(t) &=&\lim_{h\dar0}x^*\Big( \frac1h\Big(P_{t+h}f(t_0-t-h)-P_tf(t_0-t)\Big)\Big)\\ &=&\lim_{h\dar0} x^*\Big(P_{t+h}\Big(\frac{f(t_0-t-h)-f(t_0-t)}h\Big)\Big) +\lim_{h\dar0} x^*\Big(\frac1h\Big(P_{t+h}-P_t)f(t_0-t)\Big)\Big)\\ &=&-x^*(P_tLf(t_0-t))+x^*(LP_tf(t_0-t))~. \end{eqnarray*} By proposition this vanishes. Since $F$ is continuous it follows by exam that: $F(t_0)=F(0)$, i.e. $x^*(P_{t_0}f(0))=x^*(f(t_0))$ and hence: $f(t_0)=P_{t_0}f(0)$.
Now let $P_t$ and $Q_t$, respectively, be continuous contraction semigroups with generator $L$. Then for all $x\in\dom L$ the functions $f(t)\colon=P_tx$ and $g(t)\colon=Q_tx$ are differentiable and satisfy: $$ f(0)=g(0),\quad f^\prime(t)=Lf(t) \quad\mbox{and}\quad g^\prime(t)=Lg(t)~. $$ By 1. we must have $g=f$. $\eofproof$
If $f(0)\in\dom L$ the function $f(t)\colon=P_tf(0)$ is the unique solution of the Cauchy problem $f^\prime(t)=Lf(t)$; thus we will write $$ P_t=e^{tL}~. $$ Beware, proposition doesn't give you any condition on a closed operator $L$ to generate a continuous contraction semigroup, it simply presupposes that $L$ generates a continuous contraction semigroup $P_t$. Necessary and sufficient conditions for a closed linear operator to generate a continuous contraction semigroup are given in the celebrated Hille-Yosida Theorem.
Suppose $T:E\rar E$ is bounded and $P_t$ a continuous contraction semigroup with generator $L$. $T$ and $P_t$ commute if and only if $T$ and $L$ commute and $T(\dom L)\sbe\dom L$. Suggested solution.
$\eofproof$
The heat equation $\ttd tu(t)=-\D u(t)$ on $\R^d$ and $\TT^d$, respectively, has a unique solution given $u(0)\in E$ for all of the spaces $E=L_p,C_0,H^s$.
Duhamel's formula generalizes the well known formula from ODE for solutions of inhomogeneous equations $\ttd tu(t)=f(t)+Lu(t)$ for bounded linear operators $L$ to generators of continuous contraction semigroups.
Suppose $L$ is the generator of a continuous contraction semigroup $P_t$ on $E$ and $f:\R_0^+\rar E$ is a $C^1$ curve in $E$. Then the unique solution of the inhomogeneous equation $$ \ttd tu(t)=f(t)+Lu(t) \quad u(0)=x\in\dom L $$ is given by Duhamel's formula: $$ u(t)=P_tx+\int_0^tP_{t-s}f(s)\,ds~. $$
$\proof$ $\eofproof$
Let $L$ be the generator of a continuous contraction semigroup $P_t$ on $E$. Show by integration by parts that for $x\in\dom L^{n+1}$: $$ \frac{t^n}{n!}L^nx =\int_0^t\frac{(t-s)^{n-1}}{(n-1)!}P_sL^nx\,ds -\int_0^t\frac{(t-s)^n}{n!}P_sL^{n+1}x\,ds~. $$ 2. Deduce Taylor's formula: $$ P_tx=\sum_{k=0}^n\frac{t^k}{k!}L^kx+\int_0^t\frac{(t-s)^n}{n!}P_sL^{n+1}x\,ds~. $$ 3. Prove for $n=1$ the Kallman-Rota inequality: for all $x\in\dom L^2$: $$ \norm{Lx} \leq\inf_{t > 0}\Big(\frac2t\Vert x\Vert+\frac t2\Vert L^2x\Vert\Big) =2\sqrt{\Vert x\Vert\Vert L^2x\Vert}~. $$ 4. Generalize this inequality.
Let $A,B$ and $L$ be generators of continuous contraction semigroups $e^{tA},e^{tB}$ and $e^{tL}$ on $E$ such that for all $x$ in a core of $L$: $Lx=Ax+Bx$. Then we have for all $x\in E$ and all $t\geq0$: $$ \lim_{n\to\infty}(e^{\frac tnA}e^{\frac tnB})^nx=e^{tL}x~. $$
Prove Trotter's product formula for bounded operators $A$ and $B$.
Suppose $P_t^\l$ and $P_t$ are continuous contraction semigroups on $E$ with generators $L_\l$ and $L$, respectively. If for all $x$ in a core $F$ of $L$ there is some $\d > 0$, such that for all $\l < \d$: $x\in\dom L_\l$ and $\lim_{\l\to0}L_\l x=Lx$, then for all $t_0>0$ and all $x\in E$: $$ \lim_{\l\to0}\sup_{t\leq t_0}\tnorm{P_t^\l x-P_tx}=0~. $$

The Resolvent
Let $L$ ba a closed linear operator on a complex Banach space $E$. The resolvent set is defined by $$ \{z\in\C:\,z-L:\dom L\rar E\mbox{ is a bijection}\}~. $$ If $z$ is in this set the operator $U_z\colon=(z-L)^{-1}:E\rar\dom L$ is called the resolvent operator and $\Spec(L)\sbe\C$ is defined to be the complement of the resolvent set. The family of operators $U_z$, $z\notin\Spec(L)$ is called the resolvent of $L$ and the complementary set $\Spec(L)$ is called the spectrum of $L$.
For $z\notin\Spec(L)$ the operator $(z-L)^{-1}$ is defined on all of $E$ and its graph is the graph of $z-L$; thus by the Closed Graph Theorem the resolvent is a family of bounded linear operators.
If $L$ is closed, then a subspace $F$ of $\dom L$ is a core iff $(z-L)F$ is dense for all $z\notin\Spec(L)$. Suggested solution.
The resolvent set $\Spec(L)^c$ of a closed operator is open. Moreover the mapping $z\mapsto U_z$ is analytic on $\Spec(L)^c$ and for all $z,w\in\Spec(L)^c$ the so called resolvent equation holds: $$ U_z-U_w=(w-z)U_zU_w~. $$
$\proof$ Suppose $z\notin\Spec(L)$, then $\Vert U_z\Vert\leq C$ and for all $|w-z| < 1/C$: $$ w-L =w-z+U_z^{-1} =U_z^{-1}(1+(w-z)U_z) $$ and since $|w-z|\Vert U_z\Vert < 1$ we conclude that $w\in\Spec(L)^c$ and $$ (w-L)^{-1}=\sum_{k=0}^\infty(-1)^k(w-z)^kU_z^{k-1}~. $$ This proves that $\Spec(L)^c$ is open and $z\mapsto U_z$ is analytic on $\Spec(L)^c$. Finally since both $U_z$ and $U_w$ are bijections from $\dom L$ onto $E$ the resolvent equation is equivalent to $(w-L)-(z-L)=(w-z)$, which trivally holds. $\eofproof$
Now suppose $R$ is an open subset of $\C$ and $U_z$, $z\in R$, is a family of linear operators satisfying the resolvent equation - such a family is called a pseudo resolvent.
Let $U_z$ be a pseudo resolvent such that $(a,\infty)\sbe R$.
  1. $\im U_z$ and $\ker U_z$ do not depend on $z$.
  2. If $\norm{U_z}\rar0$ for $z\rar\infty$, then for all $x\in\im U_z$: $\lim_{z\to\infty}zU_zx=x$.
  3. If $\norm{U_z}\rar0$ for $z\rar\infty$ and $\im U_z$ is dense for some $z$, then $U_z$ is the resolvent of a closed linear operator $L$ with $\dom L=\im U_z$.
$\proof$ 1. The resolvent equation $U_w=U_z((z-w)U_w+1)$ immediately implies that for all $z,w\in R$: $\im U_z\spe\im U_w$. Moreover, if $U_wx=0$ for some $w\in R$, then $U_zx=0$ for all $z\in R$. Hence: $\ker U_z\spe\ker U_w$.
2. Since $zU_zU_wx=\frac{z}{z-w}(U_w-U_z)x$ we infer that: $$ \lim_{z\to\infty}zU_zU_w x =U_w x~. $$ 3. By 2. we conclude that $U_z$ is injective. Put $D=\im U_z$ and define $L:D\rar E$ by $L\colon=z-U_z^{-1}$. $L$ is obviously closed. We check that it doesn't depend on $z\in R$: For $x\in D$ we have $$ wx-U_w^{-1}x=zx-U_z^{-1}x \quad\mbox{iff}\quad (w-z)U_wU_zx=U_zx-U_wx $$ which is exactly the resolvent equation. Hence $L$ doesn't depend on $z\in R$, i.e. for all $z\in R$: $U_z=(z-L)^{-1}$. $\eofproof$
Let $L$ be a closed linear operator such that for some $\l\notin\Spec(L)$ the operator $U_\l$ is compact. Then $\im L$ is of finite codimension and $\ker L$ is finite dimensional. Suggested solution.
Suppose $P_t$ is a continuous contraction semigroup, then $\Spec(L)\sbe\{z:\Re z\leq0\}$ and for all $\Re z > 0$ and all $x\in E$: $$ U_zx=\int_0^\infty e^{-zs}P_sx\,ds~. $$
$\proof$ For $x\in E$ put $R_zx\colon=\int_0^\infty e^{-zs}P_sx\,ds$ and $y=R_zx$. Then by proposition for all $n > 0$: $$ L\int_0^ne^{-zt}P_tx\,dt =z\int_0^ne^{-zt}P_tx\,dt+e^{-zn}P_nx-x $$ As $n$ tends to $\infty$ we conclude by e.g. dominated convergence and closedness of $L$: $LR_zx=zR_z-x$. Therefore $\im R_z\sbe\dom L$ and for all $x\in E$: $(z-L)R_zx=x$. It suffices to prove that $(z-L):\dom L\rar E$ is injective, because then $R_zx=(z-L)^{-1}x$. So take any $x\in\dom L$ such that $Lx=zx$. Then $x(t)\colon=e^{zt}x\in\dom L$ is a smooth curve in $\dom L$ und $x^\prime(t)=Lx(t)$. By proposition: $x(t)=P_tx$ and thus: $\norm{P_tx}=e^{t\Re z}\Vert x\Vert$; since $\Re z > 0$ this is only possible if $x=0$. $\eofproof$
Prove by induction on $n\in\N$ that for all $\Re z > 0$ and all $x\in E$: $$ U_z^n=\frac1{\G(n)}\int_0^\infty s^{n-1}e^{-zt}P_sx\,ds~. $$
Thus the Bessel potential operators $$ U_z^\a x\colon=\frac1{\G(n)}\int_0^\infty s^{\a-1}e^{-zt}P_sx\,ds~. $$ of an arbitrary continuous contraction semigroup $P_t$ just interpolate the powers of the resolvent!
If $P:E\rar E$ is a linear contraction, then $(\l+1-P)^{-1}$ is the resolvent of the Poissonization.

Invariant functionals

Suppose $P_t$ is a continuous contraction semigroup on $E$. $x^*\in E^*$ is $P_t$-invariant, i.e. $P_t^*x^*=x^*$ for all $t > 0$, if and only if for some $\l_0 > 0$: $\l_0 U_{\l_0}^*x^*=x^*$.
$\proof$ Let $x^*$ be invariant under $\l_0U_{\l_0}^*$, then for all $x\in E$: $\l x^*(U_\l x)=\l\l_0x^*(U_{\l_0}U_\l)$. By the resolvent equation we have $$ \l U_{\l_0}U_\l =\l_0 U_{\l_0}U_\l-U_\l+U_{\l_0} $$ and thus: \begin{eqnarray*} \l x^*(U_\l x) &=&\l_0x^*(\l_0U_{\l_0}U_\l x-U_\l x+U_{\l_0}x)\\ &=&\l_0x^*(U_\l x)-x^*(U_\l x)+x^*(U_{\l_0}x) =x^*(\l_0 U_{\l_0}x) =x^*(x)~. \end{eqnarray*} Hence we have for all $\l > 0$ and all $x\in E$: $$ \int_0^\infty e^{-\l t} x^*(P_tx)\,dt =\frac1\l x^*(x) =\int_0^\infty e^{-\l t}x^*(x)\,dt $$ i.e. the Laplace transform of the functions $t\mapsto e^{-t}x^*(x)$ and $t\mapsto e^{-t}x^*(P_t(x))$ coincide. Since the Laplace transform determines integrable functions, we conclude that $x^*(P_tx)=x^*(x)$. $\eofproof$
This proposition usually applies to continuous contraction semigroup on $C_b(S)$ or $C_0(S)$: If $P_t=e^{tL}$ is a continuous contraction semigroup on $C_b(S)$ or $C_0(S)$, then $\mu\in M_1(S)$ is an invariant probability measure if and only if there exists some $\l > 0$ such that $\l U_\l^*\mu=\mu$, i.e. for all $f\in C_b(S)$ or all $f\in C_0(S)$: $$ \int f\,d\mu =\l U_\l^*\mu(f) =\int\l U_\l xf\,d\mu =\int_0^\infty\l e^{-\l t}\int P_tf\,d\mu\,dt~. $$ Assuming in addition that $P_t^*:M_1(S)\rar M_1(S)$ this is equivalent to: $\l U_\l^*:M_1(S)\rar M_1(S)$ has a fixed point.
If $X$ is a complete vector field with flow $\theta$ on a manifold $S$ and $P_tf\colon=f\circ\theta_t$, $t\in\R$, the corresponding continuous contraction group on $C_0(S)$, then the adjoint $P_t^*:M(S)\rar M(S)$ is given by $P_t^*\mu(A)=\mu_{\theta_t}(A)=\mu(\theta_{-t}(A))$ and $$ \l U_\l^*\mu(A)=\int_0^\infty\l e^{-\l t}\mu(\theta_{-t}(A))\,dt~. $$ If there exists a non negative smooth function $\r:S\rar\R$ such that $\divergence(\r X)=0$, then $P_t^*\mu=\mu$ and $P_t$ is a continuous contraction group on $L_2(\mu)$, where $\mu(dx)=\r(x)\,v(dx)$, cf. exam. In this case the dual group is given by $P_t^*f=f\circ\theta_{-t}$ with generator $-X$. Thus $X$ is skew symmetric.
Suppose $A$ is a skew symmetric operator on the real Hilber space $L_2(\mu)$, then $iA$ is a symmetric operator on the complex Hilbert space $L_2(\mu)$ of complex valued functions.
Every real Hilbert $E$ space admits a complexification $E^\C\colon=\C\otimes E$: $\la\a\otimes x,\b\otimes y\ra\colon=\a\bar\b\la x,y\ra$. Every element of $E^\C$ is of the form $\a(1\otimes x)+\b(i\otimes y)$ for some $\a,\b\in\R$ and $x,y\in E$ - for this element we usually write $\a x+i\b y$, which could be confusing unless misleading! If $F$ is another real Hilbert space and $A:E\rar F$ a bounded linear operator, then $A^\C:E^C\rar F^C$, $A^\C(\a\otimes x)\colon=\a\otimes Ax$ is a bounded linear operator and $A^\C(x+iy)=Ax+iAy$; moreover: $\Vert A^\C\Vert=\Vert A\Vert$.
The space $S$ of all self-adjoint linear operators on a complex Hilbert space $E$ is a real vector-space. Determine its complexification.
If $E$ is a real Banach space, then there is usually no norm $\Vert.\Vert_\C$ on its complexification $E^\C$ such that for every real Banach space $F$ and every bounded linear operator $A:E\rar F$: $\norm{A^\C}=\norm A$. Verify that $$ \norm{x+iy}_\C \colon=\sup\{\norm{x\cos\theta+y\sin\theta}:\theta\in[0,2\pi)\} $$ is indeed a norm on $E^C$ and $$ \max\{\Vert x\Vert,\Vert y\Vert\} \leq\norm{x+iy}_\C \leq\sqrt{\Vert x\Vert^2+\Vert y\Vert^2}~. $$ Moreover: $\norm{A^\C}\leq\sqrt2\norm A$.

The Hille-Yosida theorem

Cf. e.g. wikipedia.
Suppose $L:\dom L(\sbe E)\rar E$ is any linear operator. $L$ is the generator of a continuous contraction semigroup if and only if the following conditions hold:
  1. $L$ is a closed linear operator (in particular $\dom L$ is dense).
  2. For all $\l > 0$ and all $y\in E$ there is exactly one $x\in\dom L$ such that $(\l-L)x=y$ and this solution satisfies: $\Vert x\Vert\leq\l^{-1}\norm y$.
The second condition is equivalent to $U_\l:\dom L\rar E$ is a bijection and $\Vert U_\l\Vert\leq\l^{-1}$. The proof goes essentially as follows: if $L$ satisfies all of the above conditions, the operators $L_\l\colon=\l LU_\l$ are bounded, because $$ \l LU_\l =\l(-(\l-L)+\l)U_\l =\l(\l U_\l-1) \quad\mbox{i.e.}\quad \norm{L_\l}\leq2\l $$ Thus $\l LU_\l$ generate a continuous semigroup and we are left to prove that this semigroup 'converges' to a continuous contraction semigroup with generator $L$.
A linear operator $L$ is said to be dissipative, if for all $\l > 0$: $$ \norm{\l x-Lx}\geq\l\Vert x\Vert~. $$
If for all pairs $(x,x^*)\in E\times E^*$ satisfying $\Vert x\Vert=\norm{x^*}=x^*(x)=1$: $\Re x^*(Lx)\leq0$, then $L$ is dissipative. Also, if $L$ admits a closure $\bar L$, then $\bar L$ is dissipative if $L$ is: For $(x_n,Lx_n)\to(x,\bar Lx)$ we have $\norm{\l x_n-Lx_n}\geq\l\Vert x_n\Vert$ and therefore: $\norm{\l x-\bar Lx}\geq\l\Vert x\Vert$.
Let $L$ be a dissipative operator on a Banach space $E$, which admits a closure $\bar L$. If for some $\l_0 > 0$ the subspace $\im(\l_0-L)$ is dense, then there is exactly one continuous contraction semigroup $P_t$ on $E$ with generator $\bar L$.
$\proof$ Since $\im(\l_0-\bar L)\spe\im(\l_0-L)$ we may assume w.l.o.g. that $L=\bar L$. For all $\l > 0$ we have: $\norm{\l x-Lx}\geq\l\Vert x\Vert$ an therefore $\norm{(\l_0-L)^{-1}:\im(\l_0-L)\rar E}\leq1/\l_0$. We claim that $\im(\l_0-L)$ is closed, i.e. $\im(\l_0-L)=E$. Indeed, if $y_n\in\im(\l_0-L)$ converges to $y$, then $y_n=(\l_0-L)x_n$ and $x_n$ is Cauchy in $\dom L$, because $$ \norm{y_n-y_m}=\norm{(\l_0-L)(x_n-x_m)}\geq\l_0\norm{x_n-x_m}~. $$ Now since $\l_0-L$ is closed, we conclude that $(x,y)=\lim(x_n,y_n)\in\G(\l_0-L)$.
It remains to prove that for all $\l\colon=\l_0+\e > 0$: $\im(\l-L)=E$. As in the proof of lemma we have $$ \l-L =(\l_0-L)(1+\e(\l_0-L)^{-1}) $$ If $|\e| < \l_0$, then $1+\e(\l_0-L)^{-1}$ is an isomorphism, hence $$ (\l-L)(1+\e(\l_0-L)^{-1})^{-1}=\l_0-L~. $$ It follows that for $|\e| < \l_0$: $$ \im(\l-L) \spe\im((\l-L)(1+\e(\l_0-L)^{-1})^{-1}) =\im(\l_0-L) =E~. $$ Iterating the argument for $\l_0\to(3/2)\l_0,(3/2)^2\l_0,\ldots$ proves the assertion. $\eofproof$
Suppose $h:\R\rar\R$ is measurable and locally bounded. Put $L:C_c^\infty(\R)(\sbe L_1(\R))\rar L_1(\R)$, $Lf(x)=h(x)f(x)$ and $\dom L=\{f\in L_1(\R): hf\in L_1(\R)\}$. Verify that $L$ is closed and $\dom L$ dense. If for all $x$: $h(x)\leq 0$ then $L$ is dissipative and the generator of a continuous contraction semigroup $P_t$ on $L_1(\R)$.
1. Given $f\in L_1(\R)$ the sequence $f_n\colon=fI_{[h\leq n]}$ is in $\dom L$ and $f_n$ converges in $L_1(\R)$ to $f$, i.e. $\dom L$ is dense.
2. Suppose $(f_n,Lf_n)$ converges in $\G(L)$ to $(0,g)$, then $hf_n$ converges in $L_1(\R)$ to $g$ and a subsequence $f_{n(k)}$ of $f_n$ converges on the set $[h\neq0]$ a.e. to $g(x)/h(x)$; another subseqence of $f_{n(k)}$ converges a.e. to zero. Hence $g$ equals zero on $[h\neq0]$. As $g$ must be zero on $[h=0]$, we conclude that $g=0$ and thus $L$ is closed.
3. We have for all $\l > 0$: $$ \norm{\l f-Lf} =\int|\l-h(x)||f(x)|\,dx \geq\int\l|f(x)|\,dx =\l\norm f $$ 4. By 1. $\im(1-L)$ is dense. The semigroup generated by $L$ is $P_tf=e^{th}f$.
Let ${\cal S}(\R^d)$ denote the Schwartz space on $\R^d$ and put $H_0f\colon=\sum\pa_j^4f$. Then $-H_0$ is a dissipative operator and $\im(1+H_0)$ is dense in $L_2(\R^d)$. Thus its closure $-H$ is the generator of a continuous contraction semigroup on $L_2(\R^d)$. Suggested solution.
Let $X_1,\ldots,X_n$ be complete vector fileds on $M$ and $\r$ a smooth density of a probability measure $\mu$ on $M$ such that for all $j=1,\ldots,n$: $\divergence(\r X_j)=0$. Then $X_j$ is skew symmetric on $L_2(\mu)$ by exam and $H_0\colon=-\sum X_j^2$ is a positive, symmetric linear operator defined on a dense subspace $\dom H_0$ of $L_2(\mu)$. By exam the operator $H$ defined by $\G(H)=\cl{\G(H_0)}$ is positive symmetric on $\dom H\spe\dom H_0$. If $\im(1+H_0)$ is dense, then $-H$ is the generator of a continuous contraction semigroup.
Let $H_0$ be the operator $H_0u\colon=-u^\dprime$ on the interval $I=(0,1)$. Verify that $-H_0$ is dissipative. However, putting $\dom H_0=C_c^\infty(I)$ the image $\im(1+H_0)$ is orthogonal to the functions $w_1(x)=e^x$ and $w_2(x)=e^{-x}$ in $L_2(I)$. Hence $\im(1+H_0)$ is not dense in $L_2(I)$.
On the other hand we know from ODE that for any $f\in C_c^\infty(I)$ there is some bounded smooth $u$, actually $$ u(x)=\frac12\int_0^1 e^{-|x-y|}f(y)\,dy+C_1w_1(x)+C_2w_2(x) $$ such that $(1+H_0)u=f$ and we may choose the constants $C_1$ and $C_1$ such that $u(0)=u(1)=0$. Hence, putting $\dom H_0=\{u\in C^\infty(\cl I): u(0)=u(1)=0\}$ we conclude that $\im(1+H_0)$ is dense. In this case the closure $-\bar H_0$, which exists by exam is the generator of a continuous contraction semigroup.

Adjoint operators

The adjoint of a densely defined linear operator $A:\dom A(\sbe E)\rar E$ is defined by $$ \dom A^*=\{x^*\in E^*:x\mapsto x^*(Ax)\mbox{ is continuous}\} $$ Since $\dom A$ is dense there is for all $x^*\in\dom A^*$ exactly one $A^*x^*\in E^*$ such that for all $x\in\dom A$: $x^*(Ax)=A^*x^*(x)$. It follows that $$ \G(A^*) \colon=\{(A^*x^*,x^*): x^*\in\dom A^*\} =\{(y^*,x^*):\,\forall x\in\dom A: y^*(x)+x^*(-Ax)=0\} =\colon\G(-A)^\perp~. $$ Notice the swaped components of $\G(A^*)$! Usually $\dom A^*$ is not dense but in case $E$ is reflexive it can be proved that $\dom A^*$ is dense and thus $A^*$ is closed, moreover if $A$ is closed then $A=A^{**}$.
If $P_t$ is a continuous contraction semigroup on a reflexive Banach space $E$ with generator $L$, then $P_t^*$ is a continuous contraction semigroup on $E^*$ with generator $L^*$.
Compute the adjoint of $A:C_c^\infty(\R)(\sbe L_1(\R))\rar L_1(\R)$, $Af(x)=xf(x)$ and show that $\dom A^*$ is not dense in $L_\infty(\R)$.
Find an example of a continuous contraction semigroup on $\ell_1$ such that its dual is not a continuous semigroup on $\ell_\infty$.
Take $P_tf(n)=e^{-nt}f(n)$, then $P_t^*g(n)=e^{-tn}g(n)$ but $$ \sup_n|P_tg(n)-g(n)|=\sup_n|e^{-tn}-1||g(n)| $$ doesn't converge to $0$ for all $g\in\ell_\infty$, i.e. $P_t^*$ is not a continuous semigroup on $\ell_\infty$.

Self adjoint operators

For a Hilbert space $E$ the adjoint $A^*$ is again defined on a dense subspace $\dom A^*$ of $E$ and thus we call $A$ self-adjoint (cf. e.g. wikipedia) if $A^*=A$, i.e. $$ \dom A^*\colon=\{y\in E:x\mapsto\la Ax,y\ra\mbox{ is continuous}\}=\dom A $$ and for all $x\in\dom A$: $A^*x=Ax$, i.e. for all $x,y\in\dom A$: $\la Ax,y\ra=\la x,Ay\ra$, i.e. $A$ is symmetric. In general the converse is not true: a closed, symmetric linear operator need not be self-adjoint: e.g. the closure of the operator in exam is not self-adjoint. Moreover a closed symmetric operator need not have a self-adjoint extension! Anyhow, if the closure of a symmetric operator $A$ is self-adjoint, then $A$ is called essentially self-adoint.
Compute the adjoint of $A:C_c^\infty(\R)(\sbe L_2(\R))\rar L_2(\R)$, $Af(x)=xf(x)$.
Compute the adjoint of $A:C_c^\infty(\R)(\sbe L_2(\R))\rar L_2(\R)$, $Af=f^\prime$. Verify that $f\in\dom A^*$ iff $f$ has a weak derivative and in this case $A^*f$ equals the neagtive of the weak derivative of $f$.
By means of distributions this generalizes as follows:
Let ${\cal D}(M)\colon=C_c^\infty(M)$ be the space of test functions (the topology being the strict inductive limit of Fréchet spaces, thus ${\cal D}(M)$ is an LF-space) and its dual ${\cal D}^\prime(M)$ the space of distributions. Suppose $H:{\cal D}(M)(\sbe L_2(\mu))\rar L_2(\mu)$ is a diffusion operator on $M$ with speed probability measure $\mu$. Then $H:{\cal D}(M)\rar{\cal D}(M)$ is a continuous linear operator and thus its dual $H^\prime:{\cal D}^\prime(M)\rar{\cal D}^\prime(M)$ is a continuous linear operator. Imbed $L_2(\mu)$ into ${\cal D}^\prime(M)$ via $f\mapsto T_f$: $T_f(\vp)\colon=\int\vp f\,d\mu$. Then $f\in\dom H^*$ if and only if $H^\prime(T_f)=T_g$ for some $g\in L_2(\mu)$, in which case: $g=H^*f$.
Suppose $A:\dom A(\sbe E)\rar E$ is a closed linear operator on a Hilbert space $E$. Then we have an orthogonal decomposition: $$ E\oplus E=\{(x,-Ax)+(A^*y,y):x\in\dom A,\,y\in\dom A^*\}~. $$ The inner product on $E\oplus E$ is given by $\la(x_1,x_2),(y_1,y_2)\ra=\la x_1,y_1\ra+\la x_2,y_\ra$. Suggested solution.
If $H$ is a self-adjoint operator on a Hilbert space $E$, then $iH$ is the generator of a continuous contraction groups $U_t$, $t\in\R$; moreover all operators $U_t$ are isometries. If $H$ is a positive, self-adjoint operator on a Hilbert space $E$, then $-H$ is the generator of a self-adjoint continuous contraction semigroup.
The converse statements are also true, i.e. if $U_t$ is a continuous unitary group on a Hilbert space $E$, then the generator of $U_t$ is $iH$ for some self-adjoint operator $H$. If $P_t$ is a self-adjoint continuous contraction semigroup on a Hilbert space $E$, then the generator of $P_t$ is $-H$ for some positive, self-adjoint operator $H$. This can be used to verify essential self-adjointness of an operator!
Let $H_0$ be the operator $H_0u\colon=-u^\dprime$ on the interval $I=(0,1)$ and $\dom H_0=\{u\in C^\infty(\cl I): u(0)=u(1)=0\}$. Then $H_0$ is essentially self-adjoint.
Let $A$ be a symmetric linear operator defined on a dense subspace $\dom A$ of a Hilbert space $E$. Suppose $U_t$ is a continuous group of isometries on $E$ such that $$ U_t(\dom A)\sbe\dom A \quad\mbox{and}\quad \forall x\in\dom A:\quad\lim_{t\to0}\frac{U_tx-x}t=iAx~. $$ Then $A:\dom A\rar E$ is essentially self-adjoint, $$ \dom(A^*)=\Big\{x\in E:\lim_{t\to0}\frac{U_tx-x}t\quad\mbox{exists}\Big\} \quad\mbox{and}\quad \lim_{t\to0}\frac{U_tx-x}t=iA^*x~. $$ Moreover, if $A(\dom A)\sbe\dom A$, then for all $n\in\N$ the operator $A^n:\dom A\rar E$ is essentially self-adjoint.
Let $H_0$ be the operator $H_0u\colon=-u^\dprime$ on $\R$ and $\dom H_0=C_c^\infty(\R)$. Then $H_0$ is essentially self-adjoint on $L_2(\R)$. Hint: Choose $Au=-iu^\prime$, then $A^2=H_0$ and $U_tf(x)=f(x+t)$.
Verify by means of Hahn-Banach that $\im A$ is dense iff $\ker A^*=\{0\}$. Solution by T. Speckhofer.
Suppose $H\geq1$ is a positive self-adjoint operator on the Hilbert space $E$. Let $Q(H)$ be the Hilbert space $\dom(H^{1/2})$ furnished with the inner product: $$ \la x,y\ra_1\colon=\la H^{1/2}x,H^{1/2}y\ra~. $$ Put $F\colon=Q(H)\oplus_2 E$, $D$ the subspace $D\colon=\dom(H)\oplus Q(H)$ of $F$ and $A:D(\sbe F)\rar F$ the operator $$ (u,v)\mapsto(v,-Hu) $$
  1. $iA:D\rar F$ is a self-adjoint operator.
  2. Define $Q_t:F\rar F$ by $$ Q_t=\left( \begin{array}{cc} \cos(tH^{1/2})&H^{-1/2}\sin(tH^{1/2})\\ -H^{1/2}\sin(tH^{1/2})&\cos(tH^{1/2}) \end{array}\right)~. $$ Then $Q_t$ is a group of isometries with generator $A$.
  3. $x(t)\colon=\cos(tH^{1/2})x+H^{-1/2}\sin(tH^{1/2})y$ is the only solution to the wave equation $x^\dprime(t)=-Hx(t)$ satisfying $x(0)=x$, $x^\prime(0)=y$ and there is some $a > 0$ such that $\sup_t\tnorm{x(t)e^{-at}}$ is bounded.
1. For all $(x,y)\in D$ we have: $$ \la iA(x,y),(u,v)\ra =i\la (y,-Hx),(u,v)\ra =i\la y,u\ra_1-i\la x,v\ra_1 =\la(x,y),iA(u,v)\ra $$ For $(u,v)\in F$ the mapping $(x,y)\mapsto\la A(x,y),(u,v)\ra$ is continuous iff $x\mapsto\la Hx,v\ra$ is continuous on $Q(H)$ and $y\mapsto\la y,u\ra_1$ is continuous on $E$. This holds if and only if $v\in\dom H^{1/2*}=\dom H^{1/2}$ and $u\in\dom H^*=\dom H$.
2. Put $R=H^{1/2}$, $C_t=\cos(tR)$, $S_t=\sin(tR)$ and $R^{-1}S_t$ are defined by the spectral theorem for unbounded self-adjoint operators. We have $$ Q_t(x,y)=(C_tx+R^{-1}S_ty,-RS_tx+C_ty) $$ and therefore \begin{eqnarray*} \norm{Q_t(x,y)}_F^2&=& \tnorm{RC_tx}^2 +\tnorm{RS_tx}^2 +\tnorm{C_ty}^2 +\tnorm{S_ty}^2\\ \norm{Q_t(x,y)-(x,y)}_F^2 &=&\tnorm{(C_t-1)Rx+S_ty}^2 +\tnorm{(C_t-1)y-RS_tx}^2\\ \norm{Q_t(x,y)-(x,y)-tA(x,y)}_F^2 &=&\tnorm{(C_t-1)Rx+(S_t-tR)y}^2 +\tnorm{(C_t-1)y-(S_t-tR)Rx}^2 \end{eqnarray*} The assertion follows from the spectral theorem and dominated convergence.
3. Let $x(t)$ be a solution to the equation $x^\dprime(t)=-Hx(t)$ satisfying $x(0)=x^\prime(0)=0$. Then we get for $\l > a$ and $y(\l)\colon=\int_0^\infty x(t)e^{-\l t}\,dt$: $$ \l^2 y(\l) =\l\int_0^\infty x^\prime(t)e^{-\l t}\,dt =\int_0^\infty x^\dprime(t)e^{-\l t}\,dt =-\int_0^\infty Hx(t)e^{-\l t}\,dt =-Hy(\l)~. $$ As $H\geq0$ it follows that for all $\l > a$: $y(\l)=0$, i.e. the Laplace transform of $t\mapsto x(t)e^{-at}$ vanishes.
Suppose $H$ is a positive self-adjoint operator on the Hilber space $E$ and $f:\R\rar E$ is $C^1$. Then (cf. Duhamel's principle): $$ u(t)\colon=c\int_0^t H^{-1/2}\sin\Big(c(t-s)H^{1/2}\Big)f(s)\,ds $$ solves the inhomogeneous wave equation $(c^{-2}\pa_t^2+H)u(t)=f(t)$. Suggested solution.
Suppose $a_1,\ldots,a_n\in\UU(N)$ satisfy the following conditions: $$ \forall j,k: a_j^2=-1, a_ja_k+a_ka_j=0 $$ then the operator $D\colon=\sum a_j\pa_j$ on $L_2(\R^n,\C^N)$ is called a Dirac-operator. Prove that $D$ is symmetric, $D^2=\D$ and (formally) $$ e^{itD} =\cos(t\D^{1/2})+iD(\D^{-1/2}\sin(t\D^{1/2}) $$ Thus if $x(t)=\cos(t\D^{1/2})x$ and $y(t)=\D^{-1/2}\sin(t\D^{1/2})y$ are the solution to the wave equation $x^\dprime(t)=-\D x(t)$ satisfying $x(0)=x$, $x^\prime(0)=0$ and $y(0)=0$, $y^\prime(0)=y$, then $$ z(t)\colon=x(t)+iDy(t) $$ is a solution of the equation $z^\prime(t)=iDz(t)$.

Positivity, Irreducibility and Ergodicity

 Throughout this section we assume that $\mu$ is a probability measure on $(S,\F)$, $1\leq p < \infty$ and $E$ always denotes one of the spaces $L_p(\mu)$.

Positivity

If a semigroup $P_t$ on $E$ is of the form $P_tf=f\circ\theta_t$, then we know (cf. section) that if $P_tf=f$ for some function $f\in E$, then for all Borel sets $B$ of $\R$: the set $[f\in B]$ is invariant. For an arbitrary semigroup this is no more true. In this section we study a condition, which still guarantees the existence of invariant sets provided the semigroup admits an invariant function:
A subspace $F$ of $E$ is said to be a sublattice, if $f\in F$ implies $|f|\in F$. An operator $A:E\rar E$ is said to be positive, if $f\geq0$ implies $Af\geq0$ - here and in the following we mean by $f\geq0$: $\mu(f < 0)=0$.
Suppose $F$ is a sublattice, then $f\in F$ implies $f^+,f^-\in F$ and $f,g\in F$ implies: $f\wedge g,f\vee g\in F$. Hint: $f\vee g=(f+g+|f-g|)/2$.
Suppose $A$ is positive, then $\norm A=\sup\{\norm{Af}:f\geq0,\norm f=1\}$. Solution by T. Speckhofer.
For the remainder of this section we will assume that $P_t$ is a positive continuous contraction semigroup on $E$ such that $P_t1=1$ - we say $P_t$ is conservative - and we will denote its generator by $L$.
The subspace $\ker L\colon=\{f\in\dom L:Lf=0\}$ is a closed sublattice of $E$.
$\proof$ Put $F_t\colon=\{f:P_tf=f\}$, then $\ker L=\bigcap_tF_t$. Since $F_t$ is closed, it suffices to check that $F_t$ is a sublattice: So take any $f\in F_t$; by positivity we conclude that $0\leq|f|=|P_tf|\leq P_t|f|$. If $|f|\neq P_t|f|$, then $\norm{|f|} < \norm{P_t|f|}$; hence $\norm{P_t} > 1$. $\eofproof$
Suppose $T$ is a linear operator such that $$ \norm{T:L_1(\P)\rar L_1(\mu)}, \norm{T:L_\infty(\P)\rar L_\infty(\mu)}\leq1~. $$ Then there exists a positive linear operator $P$, such that for all $X\in L_1(\P)$: $|T(X)|\leq P|X|$ and $$ \norm{P:L_1(\P)\rar L_1(\mu)}, \norm{P:L_\infty(\P)\rar L_\infty(\mu)}\leq1~. $$
This is most easily demonstrated on $\ell_1^n$: Let $e_1,\ldots,e_n$ be the canonical basis of $\ell_1^n$, i.e. $e_j(k)=\d_{jk}$, then $Pe_k\geq|T(e_k)|=|\sum_j t_{jk}e_j|$ and thus $Pe_k(j)\geq|t_{jk}|$. Hence the matrix $p_{jk}=Pe_k(j)$ of $P$ with respect to the basis $e_1,\ldots,e_n$ must satisfy: $p_{jk}\geq|t_{jk}|$. On the other hand putting $p_{jk}=|t_{jk}|$ it is readily verified that for all $x\in\ell_1^n$: $|Tx|\leq P|x|$. Finally: \begin{eqnarray*} \norm{T:\ell_1^n\rar\ell_1^n} &=&\sup_k\sum_j|t_{jk}| =\norm{P:\ell_1^n\rar\ell_1^n} \quad\mbox{and analogously}\\ \norm{T:\ell_\infty^n\rar\ell_\infty^n} &=&\sup_k\sum_j|t_{kj}| =\norm{P:\ell_\infty^n\rar\ell_\infty^n} \end{eqnarray*}

Irreducibility and ergodicity

Let $P_t$ be a continuous contraction semigroup on $E=L_p(\mu)$, $1\leq p <\infty$, and assume that it's conservative, positive and a contraction on $L_\infty(\mu)$. Now let's assume in addition that there is $A\in\F$ such that for all $f\in E$ the condition $$ \supp(f)\colon=[f\neq0]\sbe A\in\F \quad\mbox{implies for all $t$:}\quad \supp(P_tf)\sbe A~. $$ Then $P_tI_A\leq I_A$, for $\supp(P_tI_A)\sbe A$ and $P_t$ is a contraction on $L_\infty(\mu)$. Hence $$ 1 =P_t(I_A+I_{A^c}) \leq I_A+P_t(I_{A^c}) \quad\mbox{i.e.}\quad P_t(I_{A^c})\geq I_{A^c}~. $$ Now if $P_t(I_{A^c})\neq I_{A^c}$, then: $\norm{P_t(I_{A^c})} > \norm{I_{A^c}}$, which is impossible. It follows $P_tI_{A^c}=I_{A^c}$ and thus $P_tI_A=I_A$. In case $P_t$ is ergodic, i.e. any $P_t$-invariant function is constant, this can only happen if $\mu(A)=0$ or $\mu(A^c)=0$.
Suppose $A\in\F$ has the property that for all $f\in E$ satisfying $\supp(f)\sbe A$ implies for all $t$: $\supp(P_tf)\sbe A$. If this property of $A$ implies $\mu(A)=0$ or $\mu(A^c)=0$, then we call the semigroup $P_t$ irreducible.
So we've just verified that if $P_t$ is ergodic, then $P_t$ is irreducible. Next we are up to proving the converse in case $\mu$ is finite: Assume $P_t$ is not ergodic, then there is some non-constant function $f\in L_p(\mu)$ such that for all $t$: $P_tf=f$. Since $P_t$ is conservative, it follows that $\dim\ker L\geq2$.
If $\mu$ is finite and $P_t$ irreducible on $L_p(\mu)$, then $\dim\ker L=1$ and thus $P_t$ is ergodic.
$\proof$ As $P_t$ is conservative and $\mu$ finite $\ker L$ contains all constant functions and thus $\dim\ker L\geq1$. For all $f\in\ker L$ satisfying $f\geq0$ and $f\neq0$ we put: $$ C(f)\colon=\bigcup_{a > 0}\{g:\,|g|\leq af\}~. $$ Then for $g\in C$: $|P_tg|\leq P_t|g|\leq aP_tf=af$, i.e. $P_t(C)\sbe C$. But the closure of $C$ in $E=L_p(\mu)$, $1\leq p < \infty$, is $\cl{C}=\{g:\supp(g)\sbe\supp(f)\}$; as the right hand side is closed it contains the left hand side. Conversely if $\supp(g)\sbe\supp(f)$, then put $f_n(x)=g(x)$ if $|g(x)|\leq nf(x)$, $f_n(x)=-nf(x)$ if $g(x) < -nf(x)$ and $f_n(x)=nf(x)$ if $g(x) > nf(x)$. $f_n\in C(f)$ and by e.g. dominated convergence $f_n\to g$.
Consequently $\mu(\supp f)=\mu(\O)$ - we say $f$ is strictly positive. For an arbitrary $f\in\ker L$ the functions $f^+$ and $f^-$ are again in $\ker L$ and by what we've just proved: $f$ must either be strictly positive or strictly negative.
Finally, let $f$ be any non-constant and strictly positive function in $\ker L$. For all $\l\in\R$ the function $f-\l 1$ is in $\ker L$ and thus it must either be strictly positive or strictly negative. On the other hand as $\l$ increases the sign of $f-\l 1$ must change. But this can only happen if there is some $\l\in\R$ such that $f-\l 1=0$, i.e. $f=\l 1$. $\eofproof$
Suppose $\mu(S)=1$, then $P_t$ is not ergodic iff it's not irreducible and this holds if and only if there is some set $A\in\F$ satisfying $0 < \mu(A) < 1$ and for all $t$: $P_tI_A=I_A$.
Later on (cf. lemma) we will prove that the space of invariant functions is the space of functions measurable with respect to the $\s$-algebra of invariant sets. So there is actually a whole $\s$-algebra of invariant sets. For finite measure spaces the concepts of ergodicity and irreducibility coincide!

Conductance

Suppose we've got a Markovian transition function $P(x,A)$ with invariant probability measure $\mu$ and a set $A\in\F$ satisfying $0 < \mu(A) < 1$ and $PI_A=I_A$, then $$ 0 =\int_{A^c}PI_A\,d\mu =\int_{A^c}P(x,A)\,\mu(dx) =\P^\mu(X_n\in A^c,X_{n+1}\in A) $$ Thus starting in $A^c$ the Markov chain will never ever visit $A$; there is no 'flow' from $A^c$ into $A$. On the other hand if the function $x\mapsto P(x,A)$ is strictly positive (with respect to the invariant measure $\mu$), then $$ \int_{A^c}P(x,A)\,\mu(dx) > 0 $$ and therefore the corresponding Markov operator $P$ is ergodic on $L_p(\mu)$. This in particular applies to the Markov chain in exam
If for all $A$ satisfying $0 < \mu(A) < 1$ there is some $n\in\N$ such that $\P^\mu(X_0\in A,X_n\in A^c) > 0$, then the corresponding Markov operator $P$ is ergodic on $L_p(\mu)$. This applies to the Markov chain in exam
The probability $\P^\mu(X_n\in A,X_{n+1}\notin A)$ measures the flow from $A$ to $A^c$ and thus $$ K(A) \colon=\P^\mu(X_0\in A,X_1\notin A) =\int_{A}P(x,A^c)\,\mu(dx) $$ is called the conductance of the set $A$ and the infimum $$ K\colon=\inf\Big\{\frac{K(A)}{\mu(A)\mu(A^c)}:0 < \mu(A) < 1\Big\}~. $$ is called the conductance of the Markov chain. For a reversible chain we have $K(A)=K(A^c)$ and therefore: $$ K\geq\inf\Big\{\frac{K(A)}{\mu(A)}:0 < \mu(A)\leq1/2\Big\}~. $$
If $X_0,X_1$ are independent under $\P^\mu$, then the coductance is $1$.
For a stochastic matrix $P=(p(x,y))$ with invariant probability $\mu$ we have $$ K(A)=\sum_{x\in A}P(x,A^c)\mu(x) \quad\mbox{in particular}\quad K(\{x\})=(1-p(x,x))\mu(x) \quad\mbox{and thus}\quad K\leq\min_x\frac{1-p(x,x)}{1-\mu(x)}, $$
For reversible chains the conductance can be expresses in terms of the Dirichlet-form: $$ K(A) =\int_{A}P(x,A^c)\,\mu(dx) =\int I_{A}PI_{A^c}\,d\mu =\int I_{A^c}PI_{A}\,d\mu =\int(1-I_A)PI_A\,d\mu =Q(I_A)~. $$ Hence: $Q(I_A)\geq K\mu(A)(1-\mu(A))$.
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