Strongly Continuous Semigroups and Hille Yosida — Examples

$T(0)=E(0)=I$ and $T(t+s)=E(t+s)=E(t)E(s)=T(t)T(s)$ for $s,t\ge 0$; moreover $\mathrm{\parallel }T(t)\mathrm{\parallel }=\mathrm{\parallel }E(t)\mathrm{\parallel }\le e^{t\mathrm{\parallel }A\mathrm{\parallel }}$ since $t\ge 0$, which is claim (i) and the group law restricted to $[0,\mathrm{\infty })$.

$t\mapsto T(t)$ is norm continuous on $[0,\mathrm{\infty })$, indeed $C^{\mathrm{\infty }}$ there with derivative $AE(t)$; since $\mathrm{\parallel }(T(t)-T(t_0))x\mathrm{\parallel }\le \mathrm{\parallel }T(t)-T(t_0)\mathrm{\parallel }\mathrm{\parallel }x\mathrm{\parallel }$, the family is strongly continuous, so it is a $C_0$-semigroup, and it is uniformly continuous as a norm-continuous family: claims (ii) and (iv) for real times follow from the same identities.

Generator: for $x\in X$ and $h>0$, $\parallel \frac{T(h)x-x}{h}-Ax\parallel \le \parallel \frac{E(h)-I}{h}-A\parallel \mathrm{\parallel }x\mathrm{\parallel }\le \frac{h}{2}\mathrm{\parallel }A{\mathrm{\parallel }}^2{e}^{h\mathrm{\parallel }A\mathrm{\parallel }}\mathrm{\parallel }x\mathrm{\parallel }\to 0$, so every $x\in X$ lies in the generator domain $D(A)$ of the semigroup and the generator acts by $x\mapsto Ax$; hence the generator is the bounded operator $A$ with $D(A)=X$, which is claim (iii).

Classical orbits: for $u(t):=E(t)x$ one has $u^{\mathrm{\prime }}(t)=E^{\mathrm{\prime }}(t)x=AE(t)x=Au(t)$ for every real $t$, and $u(0)=x$, so $u\in C^1(\mathbb{R};X)$ solves $u^{\mathrm{\prime }}=Au$ classically; $u$ is unique among such solutions by the same argument applied to the difference of two solutions, alternatively by the group law $E(-t)u(t)=x$.

All of (i)-(iv) and the classical-solution statement are established, with $\omega =\mathrm{\parallel }A\mathrm{\parallel }$ and $M=1$ in the exponential bound.

$T$ is a strongly continuous semigroup of isometries: $T(t)$ is linear, $T(t+s)f=T(t)T(s)f$ and $T(0)=I$ hold pointwise, $\mathrm{\parallel }T(t)f{\mathrm{\parallel }}_p=\mathrm{\parallel }f{\mathrm{\parallel }}_p$ because translation preserves the integral of $\mathrm{\mid }f{\mathrm{\mid }}^p$ [F4], and $T(t)f\to f$ in $L^p$ as $t\downarrow 0$ by [F1].

Let $f\in W^{1,p}(\mathbb{R})$ and $h>0$. The curve $u\mapsto {\tau }_{-u}{f}^{\mathrm{\prime }}$ is continuous from $[0,h]$ to $L^p$ by [F1], so $M_h:=\frac{1}{h}{\int }_0^h{\tau }_{-u}{f}^{\mathrm{\prime }}du\in L^p$ is defined by [F5], and $\mathrm{\parallel }{M}_h-f^{\mathrm{\prime }}{\mathrm{\parallel }}_p\le {\sup }_{0\le u\le h}\mathrm{\parallel }{\tau }_{-u}{f}^{\mathrm{\prime }}-f^{\mathrm{\prime }}{\mathrm{\parallel }}_p\to 0$ as $h\downarrow 0$.

For $f\in W^{1,p}$ and $h>0$ the difference quotient $q_h:=\frac{T(h)f-f}{h}\in L^p$ equals $M_h$ almost everywhere. Indeed, for every $\phi \in C_c^{\mathrm{\infty }}(\mathbb{R})$, translation invariance [F4] gives ${\int }_{\mathbb{R}}{q}_h\phi =\frac{1}{h}({\int }_{\mathbb{R}}f(s)\phi (s-h)ds-{\int }_{\mathbb{R}}f(s)\phi (s)ds)=\frac{1}{h}{\int }_{\mathbb{R}}f(s)(\phi (s-h)-\phi (s))ds$; writing $\phi (s-h)-\phi (s)=-{\int }_0^h{\phi }^{\mathrm{\prime }}(s-u)du$ and applying Fubini [F5] and the weak-derivative identity of [F2] with the test function $\phi (\cdot -u)$, ${\int }_{\mathbb{R}}{q}_h\phi =-\frac{1}{h}{\int }_0^h{\int }_{\mathbb{R}}f(s){\phi }^{\mathrm{\prime }}(s-u)dsdu=\frac{1}{h}{\int }_0^h{\int }_{\mathbb{R}}{f}^{\mathrm{\prime }}(s)\phi (s-u)dsdu=\frac{1}{h}{\int }_0^h{\mathrm{\Lambda }}_{\phi }({\tau }_{-u}{f}^{\mathrm{\prime }})du$; by [F5] this equals ${\mathrm{\Lambda }}_{\phi }(M_h)={\int }_{\mathbb{R}}{M}_h\phi$. Two $L^p$ functions with the same pairing with every test function coincide almost everywhere by [F6].

Therefore $\parallel \frac{T(h)f-f}{h}-f^{\mathrm{\prime }}{\parallel }_p=\mathrm{\parallel }{M}_h-f^{\mathrm{\prime }}{\mathrm{\parallel }}_p\to 0$ as $h\downarrow 0$ for every $f\in W^{1,p}(\mathbb{R})$; by the definition of the generator [F7], $W^{1,p}(\mathbb{R})\subseteq D(A)$ and $Af=f^{\mathrm{\prime }}$ for $f\in W^{1,p}(\mathbb{R})$.

Conversely, suppose $f\in D(A)$, so that $q_h\to g$ in $L^p$ for some $g$. For every $\phi \in C_c^{\mathrm{\infty }}(\mathbb{R})$, $\mathrm{\mid }\int (q_h-g)\phi \mathrm{\mid }\le \mathrm{\parallel }{q}_h-g{\mathrm{\parallel }}_p\mathrm{\parallel }\phi {\mathrm{\parallel }}_{p^{\mathrm{\prime }}}\to 0$ by [F3], so $\int g\phi ={\lim }_h\int q_h\phi$. On the other hand the identity of [step 1.3] (which used only $f\in L^p$) gives $\int q_h\phi =\frac{1}{h}\int f(s)(\phi (s-h)-\phi (s))ds$, and for $0<h\le 1$ the integrand is supported in a fixed compact interval $K$ and bounded there by $\mathrm{\mid }f\mathrm{\mid }{\sup }_K\mathrm{\mid }{\phi }^{\mathrm{\prime }}\mathrm{\mid }$, whose integral over $K$ is finite because $f\in L^p(K)\subseteq L^1(K)$; since $\frac{\phi (s-h)-\phi (s)}{h}\to -{\phi }^{\mathrm{\prime }}(s)$ pointwise, dominated convergence [F4] gives $\int g\phi =-\int f{\phi }^{\mathrm{\prime }}$ for every test function $\phi$. By the definition of the weak derivative [F2], $g$ is the weak derivative of $f$, so $f\in W^{1,p}(\mathbb{R})$ and $g=f^{\mathrm{\prime }}$ almost everywhere.

Combining [step 2.1] and [step 2.2], the generator of the right-translation semigroup is $Af=f^{\mathrm{\prime }}$ with $D(A)=W^{1,p}(\mathbb{R})$, and the difference quotients converge to $f^{\mathrm{\prime }}$ in $L^p$ for every $f\in W^{1,p}(\mathbb{R})$; the semigroup is strongly continuous by [step 1.1]. The verification assumes Countable Choice, inherited from the translation-continuity and distribution-embedding inputs.

For every $t\ge 0$ the map $T(t)$ is linear and $\mathrm{\parallel }T(t)f{\mathrm{\parallel }}_p^p={\int }_0^{\mathrm{\infty }}{e}^{-tps}\mathrm{\mid }f(s){\mathrm{\mid }}^{p}ds\le \mathrm{\parallel }f{\mathrm{\parallel }}_p^p$, so $T(t)$ is a contraction; the pointwise identities $e^{-(t+r)s}=e^{-ts}{e}^{-rs}$ and $e^0=1$ give $T(t+r)=T(t)T(r)$ and $T(0)=I$.

Strong continuity: for fixed $f\in X$, $\mathrm{\parallel }T(t)f-f{\mathrm{\parallel }}_p^p={\int }_0^{\mathrm{\infty }}\mathrm{\mid }{e}^{-ts}-1{\mathrm{\mid }}^p\mathrm{\mid }f(s){\mathrm{\mid }}^{p}ds\to 0$ as $t\downarrow 0$ by [F2], since $e^{-ts}\to 1$ pointwise and $\mathrm{\mid }{e}^{-ts}-1{\mathrm{\mid }}^p\le 2^p$ for $t\ge 0$, so the integrand is dominated by the $L^1$ function $2^p\mathrm{\mid }f{\mathrm{\mid }}^p$.

Inclusion $\{qf\in X\}\subseteq D(A)$: if $qf\in X$, then $\parallel \frac{T(h)f-f}{h}-qf{\parallel }_p^p={\int }_0^{\mathrm{\infty }}\mid \frac{e^{-hs}-1}{h}+s{\mid }^p\mathrm{\mid }f(s){\mathrm{\mid }}^{p}ds\to 0$ by [F2], because $\frac{e^{-hs}-1}{h}\to -s$ pointwise and, by the inequality $1-e^{-x}\le x$ for $x\ge 0$, the bracket is at most $2s$, so the integrand is dominated by $(2s\mathrm{\mid }f\mathrm{\mid }{)}^p\in L^1$.

Converse: suppose the difference quotients converge in $X$ to some $g$, and let $\phi \in C_c^{\mathrm{\infty }}(0,\mathrm{\infty })$. By [F2] and the boundedness of $s$ on $\mathrm{supp}\phi$, ${\int }_0^{\mathrm{\infty }}\frac{e^{-hs}-1}{h}f\phi ds\to -{\int }_0^{\mathrm{\infty }}sf\phi ds$, while ${\int }_0^{\mathrm{\infty }}\frac{T(h)f-f}{h}\phi ds\to {\int }_0^{\mathrm{\infty }}g\phi ds$ because $\mathrm{\parallel }\frac{T(h)f-f}{h}-g{\mathrm{\parallel }}_p\to 0$ and $\phi \in L^{p^{\mathrm{\prime }}}$; hence ${\int }_0^{\mathrm{\infty }}(g-qf)\phi ds=0$ for every test function. The locally integrable function $g-qf$ has $sf\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1(0,\mathrm{\infty })$, so [F3] gives $g=qf$ almost everywhere; in particular $qf=g\in X$ and $f\in D(A)$ with $Af=qf$.

Unboundedness and proper domain: for $n\ge 1$ let $f_n$ be the normalised nonnegative bump supported in $[n,n+1]$ with $\mathrm{\parallel }{f}_n{\mathrm{\parallel }}_p=1$. Then $\mathrm{\parallel }A{f}_n{\mathrm{\parallel }}_p^p={\int }_n^{n+1}{s}^p\mathrm{\mid }{f}_n(s){\mathrm{\mid }}^{p}ds\ge n^p$, so $\mathrm{\parallel }A{f}_n{\mathrm{\parallel }}_p\ge n\to \mathrm{\infty }$ while $\mathrm{\parallel }{f}_n{\mathrm{\parallel }}_p=1$, and no constant bounds $A$ on its domain. Moreover $D(A)\mathrm{\ne }X$: the function $f(s):=s^{-1-1\mathrm{/}p}{\mathbf{1}}_{(1,\mathrm{\infty })}$ lies in $L^p(0,\mathrm{\infty })$ because ${\int }_1^{\mathrm{\infty }}{s}^{-p-1}ds=1\mathrm{/}p$, while $sf(s)=s^{-1\mathrm{/}p}$ is not in $L^p$ because ${\int }_1^{\mathrm{\infty }}{s}^{-1}ds=\mathrm{\infty }$, so $f\in X\setminus D(A)$.

Together with [step 1.1] and [step 2.1], the displayed claims follow: $T$ is a strongly continuous contraction semigroup on $L^p(0,\mathrm{\infty })$ whose generator has domain $\{f:qf\in X\}=\{f:sf\in L^p\}$ and acts by $Af=qf=-sf$, and this operator is unbounded.

The operator $L$. The $L^2$ operator associated with the symmetric Dirichlet form is the symmetric-case operator of The $L^2$ operator associated with a symmetric elliptic form for coefficients $a^{ij}={\delta }^{ij}$, $b\equiv 0$, $c=0$ and ellipticity constant $\theta =1$ (Uniformly elliptic divergence-form operators and their sesquilinear forms); it is densely defined, symmetric and lower bounded (The associated elliptic operator is densely defined, symmetric and lower bounded), while $a(u,u)={\int }_{\mathrm{\Omega }}\mathrm{\mid }\mathrm{\nabla }u{\mathrm{\mid }}^2\ge 0$. The explicit Gårding constant of this form is $\beta =1\mathrm{/}2$ (Garding's inequality for a divergence-form elliptic operator); fix ${\mu }_0:=1\ge \beta$. Then $L$ is self-adjoint and $L+{\mu }_0:D(L)\to H$ is bijective with compact inverse (The symmetric elliptic form operator is self-adjoint with compact resolvent). In particular $L$ is closed and $I+L=L+1$ is bijective.

Spectral basis and positive eigenvalues. With ${\mu }_0=1$, [F5] gives eigenvalues ${\lambda }_1\le {\lambda }_2\le \cdots \to +\mathrm{\infty }$ and an orthonormal basis $(e_j)$ with $L{e}_j={\lambda }_j{e}_j$. For an eigenvector $u\mathrm{\ne }0$ of ${\lambda }_j$, [F1] gives ${\lambda }_j\mathrm{\parallel }u{\mathrm{\parallel }}^2=(Lu,u)=a(u,u)\ge 0$. If ${\lambda }_j=0$, then $\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_{L^2}=0$, and Poincaré on $H_0^1(\mathrm{\Omega })$ gives $\mathrm{\parallel }u{\mathrm{\parallel }}_{L^2}\le C_P\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_{L^2}=0$ (The Poincare inequality for zero-boundary Sobolev closures on domains bounded in one direction); this contradicts $u\mathrm{\ne }0$. Thus $0<{\lambda }_1\le {\lambda }_2\le \cdots \to +\mathrm{\infty }$, and $A{e}_j=-{\lambda }_j{e}_j$.

Generation. The operator $A=-L$ is densely defined, closed and dissipative: closedness and density follow from [step 1.1], while for $u\in D(A)$, [F1] and [F3] give $\mathrm{Re}⟨Au,u⟩=-a(u,u)=-{\int }_{\mathrm{\Omega }}\mathrm{\mid }\mathrm{\nabla }u{\mathrm{\mid }}^2\le 0$. Also $\mathrm{Ran}(I-A)=\mathrm{Ran}(I+L)=H$ by [step 1.1].

By [F2] with ${\lambda }_0=1$, $A$ generates a strongly continuous semigroup of contractions $T$ on $H$.

Orbit of each eigenvector. For fixed $j$, $v_j(t):=e^{-{\lambda }_{j}t}{e}_j$ belongs to $D(A)$, is $C^1$, has $v_j(0)=e_j$, and satisfies $v_j^{\mathrm{\prime }}(t)=-{\lambda }_j{v}_j(t)=A{v}_j(t)$. By uniqueness for the classical homogeneous problem in [F4], $T(t)e_j=e^{-{\lambda }_{j}t}{e}_j$. By linearity, if $f_N:={\sum }_{j\le N}(f,e_j)e_j$, then $T(t)f_N={\sum }_{j\le N}{e}^{-{\lambda }_{j}t}(f,e_j)e_j$.

Arbitrary $L^2$ data and initial trace. The finite sums $f_N$ converge to $f$ in $H$, so $c_j:=(f,e_j{)}_{L^2}$ is square-summable. Since $\mathrm{\mid }{e}^{-{\lambda }_{j}t}\mathrm{\mid }\le 1$, the spectral series converges in $H$ for each $t\ge 0$. For every $N$, contraction and orthonormality bound the distance between $T(t)f$ and this series by $\mathrm{\parallel }f-f_N{\mathrm{\parallel }}_H+({\sum }_{j>N}\mathrm{\mid }{c}_j{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}$, uniformly in $t\ge 0$. This tends to $0$, so the expansion holds in $H$; strong continuity also gives $T(t)f\to f$ in $H$ at $0$.

Positive-time smoothing in graph norm. Write $c_j=(f,e_j{)}_{L^2}$ and $u_N(t):={\sum }_{j\le N}{e}^{-{\lambda }_{j}t}{c}_j{e}_j$. Fix $0<\delta <T_0<\mathrm{\infty }$. For $M<N$ and $t\in [\delta ,T_0]$, orthonormality gives $\mathrm{\parallel }{u}_N(t)-u_M(t){\mathrm{\parallel }}_H^2={\sum }_{M<j\le N}{e}^{-2{\lambda }_{j}t}\mathrm{\mid }{c}_j{\mathrm{\mid }}^2\le {\sum }_{j>M}\mathrm{\mid }{c}_j{\mathrm{\mid }}^2\mathrm{.}$ Also $A{u}_N(t)=-{\sum }_{j\le N}{\lambda }_j{e}^{-{\lambda }_{j}t}{c}_j{e}_j$, so [F9] and continuity on bounded intervals give $\mathrm{\parallel }A(u_N(t)-u_M(t)){\mathrm{\parallel }}_H^2={\sum }_{M<j\le N}{\lambda }_j^2{e}^{-2{\lambda }_{j}t}\mathrm{\mid }{c}_j{\mathrm{\mid }}^2\le C_{\delta }^2{\sum }_{j>M}\mathrm{\mid }{c}_j{\mathrm{\mid }}^2,C_{\delta }:={\sup }_{\lambda \ge 0}\lambda e^{-\delta \lambda }<\mathrm{\infty }\mathrm{.}$ Both tails tend to zero uniformly on $[\delta ,T_0]$. Thus $(u_N,A{u}_N)$ converges uniformly there in $H\oplus H$. By [F2] the operator $A$ is closed; its graph is closed, so the limit pair is $(u(t),Au(t))$ for $u(t)=T(t)f$. Consequently $u(t)\in D(A)$ for every $t>0$, and $t\mapsto u(t)$ is continuous on every compact positive-time interval in the graph norm [F6].

Classical evolution at positive times. Fix $0<\delta <T_0$ and put $x_{\delta }:=u(\delta )\in D(A)$ by [step 6.1]. For $t\in [\delta ,T_0]$, $u(t)=T(t-\delta )x_{\delta }$ by the semigroup law. By [F7], this orbit is differentiable for $t>\delta$ and satisfies $u^{\mathrm{\prime }}(t)=AT(t-\delta )x_{\delta }=Au(t)$; since $\delta$ can be chosen below any positive time, $u$ is a classical solution on $(0,T_0]$ (and on each closed interval bounded away from $0$). For general $f\in H$ no graph-norm trace at $0$ is asserted; if $f\in D(A)$, [F4] gives the classical solution on $[0,T_0]$.

The semigroup orbit is the unique mild solution of the homogeneous abstract Cauchy problem by [F4], and its initial value is attained in $H$ by [step 5.1]. The positive-time graph-norm and differentiability conclusions are those of [steps 6.1 and 7.1]; no spatial $H^2$ identification of $D(A)$ is made for an arbitrary bounded open $\mathrm{\Omega }$.

For every $t>0$ the vector $f_t$ has norm $\mathrm{\parallel }{f}_t{\mathrm{\parallel }}_p^p={\int }_0^t{t}^{-1}ds=1$, so $f_t$ is a unit vector of $L^p(\mathbb{R})$.

$T(t)f_t=t^{-1\mathrm{/}p}{\mathbf{1}}_{(-t,0)}$: indeed $(T(t)f_t)(s)=f_t(s+t)=t^{-1\mathrm{/}p}{\mathbf{1}}_{(0,t)}(s+t)$, and $s+t\in (0,t)$ exactly when $s\in (-t,0)$.

The two indicators ${\mathbf{1}}_{(-t,0)}$ and ${\mathbf{1}}_{(0,t)}$ are disjoint up to the null set $\{0\}$, so $\mathrm{\parallel }T(t)f_t-f_t{\mathrm{\parallel }}_p^p=t^{-1}{\int }_{\mathbb{R}}\mid {\mathbf{1}}_{(-t,0)}-{\mathbf{1}}_{(0,t)}{\mid }^{p}ds=t^{-1}(\lambda ((-t,0))+\lambda ((0,t)))=t^{-1}(t+t)=2$; hence $\mathrm{\parallel }T(t)f_t-f_t{\mathrm{\parallel }}_p=2^{1\mathrm{/}p}$.