Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 8 (Semigroups of Operators)

A strongly continuous semigroup on the complex Banach space $E$ is a family $T_t\in \mathcal{L}(E)$ with $T_0=I$, $T_{s+t}=T_s{T}_t$ and $\Vert T_{t}f-f\Vert \to 0$ as $t\downarrow 0$ for every $f\in E$. Its semigroup generator is the linear operatorFor $f\in D(Z)$, boundedness of $T_t$ and the semigroup identity giveThus semigroup commutation with its unbounded generator meansThere is no assertion that $Z$ is defined on every vector. The orbit is differentiable on this domain, with $\frac{d}{dt}{T}_{t}f=T_{t}Zf$.

A contraction semigroup additionally satisfies $\Vert T_t\Vert \le 1$ for every $t\ge 0$. For $\mathrm{Re}\lambda >0$, define the Bochner integralStrong continuity gives a continuous integrand, and its norm is bounded by $e^{-t\mathrm{Re}\lambda }\Vert f\Vert$, so the integral exists andChanging variables in $T_h{L}_{\lambda }f$ yieldsAfter subtracting $L_{\lambda }f$ and dividing by $h$, the limit is $\lambda L_{\lambda }f-f$. ConsequentlyFor $f\in D(Z)$, integrate the derivative of $e^{-\lambda t}{T}_{t}f$ over $[0,\infty )$. Its limit at infinity is zero by the contraction bound, and its value at zero is $f$. ThereforeThus this is also the Laplace-transform formula for a semigroup resolvent, with $L_{\lambda }=(\lambda I-Z{)}^{-1}$.

To prove that $Z$ is a closed operator, suppose $f_n\in D(Z)$, $f_n\to f$ and $Z{f}_n\to g$. Fix one $\lambda$ with positive real part. The identity just proved and boundedness of $L_{\lambda }$ giveIts range lies in $D(Z)$, and applying $\lambda I-Z$ shows $Zf=g$. Thus the graph is closed. Linearity follows directly from linearity of the defining difference quotient, so $Z$ is a closed linear operator.

The closedness conclusion also holds without the contraction assumption. For a general strongly continuous semigroup, the Uniform boundedness principle gives a uniform operator bound on each compact time interval. Integrating the generator-orbit identity gives $T_t{f}_n-f_n={\int }_0^t{T}_{s}Z{f}_{n}ds$. Under $f_n\to f$ and $Z{f}_n\to g$, pass to the limit to obtain $T_{t}f-f={\int }_0^t{T}_{s}gds$. Divide by $t$ and let $t\downarrow 0$; strong continuity makes the right side converge to $g$, so $f\in D(Z)$ and $Zf=g$.

A linear operator $Z$ on a Banach space is called a dissipative operator whenOn a Hilbert space, using the inner product linear in its first argument,If $\mathrm{Re}⟨Zf,f⟩\le 0$, this proves the defining inequality. Conversely that inequality implies $2\lambda \mathrm{Re}⟨Zf,f⟩\le \Vert Zf{\Vert }^2$ for every $\lambda >0$; letting $\lambda \to \infty$ gives

For the next spectral assertion, dissipativity must be retained as a hypothesis. Let $Z$ be closed, densely defined and dissipative. For $\alpha >0$, the dissipative bound makes $\alpha I-Z$ injective. Its range is closed: if $(\alpha I-Z)f_n\to h$, the bound makes $f_n$ Cauchy, and then $Z{f}_n=\alpha f_n-(\alpha I-Z)f_n$ converges. Closedness produces a preimage of $h$. If the range is all of $H$, the same bound gives a bounded inverse, so $\alpha$ is in the resolvent set. If the closed range is proper, there is a nonzero vector in its orthogonal complement, and the definition of the adjoint of a densely defined operator givesThis proves that the positive spectrum of a dissipative operator is adjoint point spectrum:Without dissipativity the PDF's assertion is false. For example, multiplication by $x$ on $L^2(0,2)$ is bounded, closed and densely defined. The value $1$ is spectral: normalized functions supported closer and closer to $x=1$ make $\Vert (Z-I)f\Vert$ tend to zero. But $Z^{\ast }=Z$ has no eigenvector for eigenvalue $1$, since a square-integrable function supported on the singleton $\{1\}$ is zero.

For the given maximal-domain operator $Q$, finite-support sequences belong to $D(Q)$, so the domain is dense. Coordinatewise limits also show it is closed: if $f^{(j)}\to f$ and $Q{f}^{(j)}\to h$ in ${\ell }^2$, each defining coordinate equation passes to the limit, giving $Qf=h\in {\ell }^2$ and hence $f\in D(Q)$.

Put $g_n=2^n{f}_n$ and $h=Qf$. For $n\ge 1$,Since ${\sum }_{n\ge 1}{2}^{-n}|h_n|<\infty$ by Cauchy-Schwarz inequality, $g_n$ has a finite limit ${\ell }_f$. Its increments are square summable as well. Direct summation givesThe original inner-product series converges because $Qf,f\in {\ell }^2$. Taking $N\to \infty$ therefore givesThus $Q$ is dissipative, including its full maximal domain; the limiting boundary term has not been discarded.

To decide generation, solve $(\alpha I-Q)f=h$ for arbitrary $h\in {\ell }^2$ and fixed $\alpha >0$. The first coordinate gives $g_0=f_0=h_0/(\alpha +1)$, and the rest giveBecause all denominators are at least one,Hence $g_n$ is bounded, so $f_n=2^{-n}{g}_n$ belongs to ${\ell }^2$. The coordinate equations give $Qf=\alpha f-h\in {\ell }^2$, proving $f\in D(Q)$ and surjectivity of $\alpha I-Q$. Together with density, closedness and dissipativity, the Lumer-Phillips theorem proves the contraction generator from an exponentially weighted difference operator conclusion:

The domain is crucial. The formal transposed matrix suggests a positive-eigenvalue solutionwhich is square summable because the products converge. It is not a true adjoint eigenvector: the admissible test vector $f_n=2^{-n}$ has $Qf=(-1,0,0,\dots )$, so $⟨Qf,u⟩=-\overline{u_0}$, while $\alpha ⟨f,u⟩=\alpha \overline{u_0}{\sum }_n{4}^{-n}{\prod }_{k<n}(1+\alpha 4^{-k})$. These cannot agree for $\alpha >0$ unless $u_0=0$. This exemplifies how boundary terms exclude formal adjoint eigenvectors, and explains why a calculation using only finite-support test vectors would give the wrong generation answer.

For the Volterra integration operator, Cauchy-Schwarz inequality givesIntegrating and reversing the order yieldsThus $J$ is bounded, with $\Vert J\Vert \le 1/\sqrt{2}$. Since $f\in L^2(0,1)\subseteq L^1(0,1)$, $Jf$ is absolutely continuous and $(Jf{)}^{\prime }=f$ almost everywhere. If $Jf=0$ as an $L^2$ element, its continuous representative is zero everywhere, so its derivative is zero almost everywhere. Hence $J$ is one-to-one.

Its range is exactly $\{u\in H^1(0,1):u(0)=0\}$, and $Au=i{u}^{\prime }$. This contains the smooth compactly supported functions in $(0,1)$, so $A$ is a densely defined operator. Closedness can also be seen directly: if $u_n\to u$ and $A{u}_n\to v$ in $L^2$, then $u_n=J(-iA{u}_n)\to J(-iv)$, giving $u=J(-iv)\in D(A)$ and $Au=v$.

For every $\lambda \in \mathbb{C}$,If $\lambda \ne 0$, the allowed fact $\sigma (J)=\{0\}$ makes $J-(i/\lambda )I$ invertible. At $\lambda =0$, $\lambda J-iI=-iI$ is invertible too. Thus the bounded operatorhas range in $D(A)$ and is a two-sided inverse of $\lambda I-A$. Equivalently, solving the initial-value differential equation givesThis is bounded for every fixed $\lambda$ on the finite interval. Therefore Volterra inverse differentiation has empty spectrum:The familiar nonempty-spectrum theorem for bounded operators does not apply to this unbounded operator.

The subspace $H_0$ is the kernel of the bounded functional $f\mapsto Jf(1)={\int }_0^{1}f$, so it is closed. Its image under $J$ isthe zero-trace Sobolev space. If $u_n\in D(A_0)$, $u_n\to u$ and $A_0{u}_n\to v$, then $-i{A}_0{u}_n\in H_0$ converges to $-iv\in H_0$, and $u=J(-iv)\in D(A_0)$. Thus $A_0$ is closed. It is densely defined because it also contains the smooth compactly supported functions.

For $u,v\in D(A_0)$, integration by parts and their zero endpoint traces giveHence $A_0$ is closed and symmetric.

For any $v\in C^1([0,1])$, the same formula gives $⟨A_{0}u,v⟩=⟨u,i{v}^{\prime }⟩$, with no endpoint term because $u$ vanishes there. Consequently $v\in D(A_0^{\ast })$ and $A_0^{\ast }v=i{v}^{\prime }$, proving the requested inclusion. In fact, testing against compactly supported smooth $u$ shows that every element of $D(A_0^{\ast })$ has weak derivative in $L^2$; conversely the integration-by-parts formula applies to every $v\in H^1(0,1)$. Thuswith no boundary restrictions.

For any $\lambda \in \mathbb{C}$, the nonzero function $v_{\lambda }(x)=e^{-i\lambda x}$ belongs to this domain and satisfies $i{v}_{\lambda }^{\prime }=\lambda v_{\lambda }$. Conversely this first-order equation has only scalar multiples of that exponential. ThereforeFor any $\lambda$, the adjoint eigenfunction with eigenvalue $\overline{\lambda }$ annihilates $\mathrm{Ran}(\lambda I-A_0)$, so this range is not all of $H$. Explicitly the resolvent equation with zero initial value has the solution $R_{\lambda }h$ above, and its endpoint value is zero exactly whenThus the range is a proper closed hyperplane for every $\lambda$. This proves that the two-endpoint symmetric derivative has whole complex spectrum:There are no eigenvalues of $A_0$ itself: the eigenfunction equation together with $u(0)=0$ forces the exponential coefficient to vanish. The operator is symmetric but not self-adjoint, since its adjoint has a strictly larger domain; symmetry alone does not force a real spectrum.

Use the joint energy conventionOn the generator domain it is $-⟨Lf,g⟩$. The symmetric Feller semigroup has symmetric two-point measure ${\nu }_s(dx,dy)=\mu (dx)P_s(x,dy)$, with both marginals equal to $\mu$. Expanding the product of increments gives the finite-time identityfor real functions. This supplies a proof that works for a general symmetric Markov kernel, without assuming a diffusion chain rule.

For $a\ge b\ge 0$, Cauchy-Schwarz inequality givesInterchanging $a,b$ gives the same inequality in the other order. Apply it with $a=f(x),b=f(y)$, integrate against ${\nu }_s$, and take $s\downarrow 0$. This proves the power inequality for symmetric Markov energiesInitially take bounded nonnegative functions in the form domain. On their bounded range, the power maps are Lipschitz; their finite-time squared-increment energies are therefore bounded by a constant times that of $f$. This puts the powers in the form domain and makes both joint energy limits well defined. Truncation gives the corresponding extended-energy interpretation when needed. In particular the proof applies to the bounded positive functions used below.

Normalize the logarithmic Sobolev inequality asStart with a bounded $f$ bounded away from zero, and put $u(t)=P_{t}f$, $q=q(t)$ and $F(t)=\int u^{q}d\mu$. Positivity and preservation of constants keep $u$ within the same positive bounds. Symmetry realizes $P_t$ as a self-adjoint contraction semigroup on $L^2(\mu )$, so for $t>0$ its orbit lies in the generator domain and ${\partial }_{t}u=Lu$; this justifies the following differentiation away from zero.

Differentiate both the evolving function and exponent:Since $\log \Vert u{\Vert }_q=q^{-1}\log F$, it follows thatApply the logarithmic Sobolev inequality to $g=u^{q/2}$, and then the joint energy inequality:ThereforeThe specified exponent satisfies $q(0)=2$ and $q^{\prime }=4(q-1)/c_{\mathrm{LS}}$. Thus the right side is zero and the evolving norm is nonincreasing. Integrating from a positive time $\delta$ and letting $\delta \downarrow 0$ givesHere the initial limit follows from strong $L^2$ continuity, boundedness of $f$ and $P_{\delta }f$, and $q(\delta )\to 2$.

For a bounded nonnegative $f$, apply the estimate to $f+ϵ$ and let $ϵ\downarrow 0$. For arbitrary $f\in L^2(\mu )$, truncate $|f|$ to bounded nonnegative functions. Their images converge in $L^2$ to $P_t|f|$; an almost-everywhere convergent subsequence and Fatou lemma give the same $L^{q(t)}$ estimate for that limit. Positivity of the Markov kernel gives $|P_{t}f|\le P_t|f|$, also for complex $f$, soThis proves that the logarithmic Sobolev inequality implies hypercontractivity, with the numerical exponent matched to the stated joint-energy normalization.

Write $d{\gamma }_1(x)=(2\pi {)}^{-1/2}{e}^{-x^2/2}dx$. On the dense polynomial algebra in $L^2({\gamma }_1)$, the Gaussian creation and annihilation operators areGaussian integration by parts gives $⟨a^{-}f,g⟩=⟨f,a^{+}g⟩$. Their closed realizations are adjoints. More explicitly, for the orthonormal basis $h_n={\mathrm{He}}_n/\sqrt{n!}$ of normalized Probabilists' Hermite polynomials,Thus the closed annihilation operator has domain $\{\sum c_n{h}_n:\sum n|c_n{|}^2<\infty \}$, and the closed creation operator has domain $\{\sum c_n{h}_n:\sum (n+1)|c_n{|}^2<\infty \}$. These describe the same set of $L^2$ functions, the Gaussian Sobolev space, though the two operators differ. Directly on polynomials, $[a^{-},a^{+}]=I$, and

Use the standard carré du champ operator convention $\Gamma (f,g)=\frac{1}{2}(L(fg)-fLg-gLf)$. The product rule for this semigroup generator givesso the requested squared-gradient and joint energy formulas, for real functions in the standard algebra, areFor complex functions the joint energy is sesquilinear, with $\overline{g^{\prime }}$ in the second factor. If “squared gradient” is defined without the factor $1/2$ in the product-rule expression, its value is instead $2f^{\prime }{g}^{\prime }$ and the joint energy is one half of its integral; the joint energy and logarithmic Sobolev constant used here remain as displayed.

For the Gaussian logarithmic Sobolev inequality, first take smooth bounded $f$ with bounded first and second derivatives, bounded away from zero. The Mehler formula for the Ornstein-Uhlenbeck semigroup isIt preserves ${\gamma }_1$ and differentiating in $x$ gives the Ornstein-Uhlenbeck gradient commutation identityPut $u_t=P_{t}f$. Differentiate its entropy functional. Since $\int u_{t}d{\gamma }_1=\int fd{\gamma }_1$ is constant, and Gaussian integration by parts applies,For such data, Mehler's formula and dominated convergence give $P_{t}f(x)\to \int fd{\gamma }_1$ as $t\to \infty$. The uniform positive upper and lower bounds justify integrating this limit inside the entropy, whose limit is zero. Hence

Apply the allowed weighted Cauchy-Schwarz inequality with $g=f^{\prime }$:After integration, invariance of ${\gamma }_1$ givesIntegrating in $t$ yieldsFor a real smooth compactly supported $g$, take $f=g^2+ϵ$. ThenLetting $ϵ\downarrow 0$ provesSmooth approximation and cutoff in the Gaussian Sobolev norm extend this to the full form domain. The entropy is lower semicontinuous under the resulting $L^2$ convergence: the square densities converge in $L^1$, their masses converge, and, along an almost-everywhere convergent subsequence, Fatou's lemma applies to $h\log h$ after adding the bound $1/e$. The same inequality for complex $g$ follows by applying it to $|g|$ and using the weak-derivative bound $|(|g|{)}^{\prime }|\le |g^{\prime }|$. Thus the constant is at most $2$.

For sharpness of the Gaussian logarithmic Sobolev constant, take $g_s(x)=e^{sx/2}$ with real $s\ne 0$. These functions belong to the Gaussian Sobolev space. The Gaussian moment-generating function givesThereforeThe entropy-to-energy ratio is exactly $2$, so no smaller constant can work. Combining this lower bound with the proved inequality gives