Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 7 (Semigroups of Operators)

Consider the operators on all of $C_0(X)$ or $L^2(\mu )$, extending the displayed positive-function formula linearly. Here $C_0(X)$ is the space of continuous functions vanishing at infinity. The multiplier has modulus at most one, so $\Vert P_{t}f\Vert \le \Vert f\Vert$, and multiplication of exponentials gives $P_{s+t}=P_s{P}_t$ and $P_0=I$. For strong continuity on $C_0(X)$, choose a compact $K$ outside which $|f|$ is small. On $K$ the continuous $k$ is bounded, making $e^{-tk}\to 1$ uniform; outside $K$, $|(e^{-tk}-1)f|\le |f|$. For $L^2(\mu )$, use pointwise convergence and the dominated convergence theorem with bound $|(e^{-tk}-1)f{|}^2\le |f{|}^2$. Thus both are contraction semigroups, even when $k$ is unbounded.

The generator of a multiplication semigroup isIndeed pointwise difference quotients converge to $-kf$. A norm limit therefore forces the displayed domain condition, using an almost-everywhere convergent subsequence for the $L^2$ case. Conversely $|(e^{-tk}-1)/t|\le k$, so the dominated convergence theorem proves the $L^2$ limit when $kf\in L^2$. In $C_0(X)$ write the error as $kf$ times $1-(1-e^{-tk})/(tk)$, with the value at $k=0$ supplied by continuity. That second factor tends uniformly to zero where $k$ is bounded and has modulus at most one; the compact-set and small-tail argument applied to $kf$ proves uniform convergence. To prove directly that $L$ is a closed operator, take $f_n\to f$ and $L{f}_n\to g$. Pointwise limits in $C_0$, or a common almost-everywhere convergent subsequence in $L^2$, give $g=-kf$. Thus $f$ is in the appropriate generator domain and $Lf=g$. In $L^2$ the domain is dense because $f{1}_{\{k\le n\}}\to f$; in $C_0$ compactly supported continuous functions are a dense subspace contained in the domain.

A real multiplication operator on its maximal $L^2$ domain is an unbounded self-adjoint operator. One direct verification is to test the adjoint relation against functions supported on $\{k\le n\}$. If $h$ is in the adjoint operator domain with representing vector $g$, those tests force $g=-kh$ on each such set, hence $kh\in L^2$. The converse follows by integration. On the sigma-finite measure spaces used below, the spectrum of a real multiplication operator is its essential range; spectral values are detected by unit vectors supported where the multiplier is arbitrarily close to that value, while outside the essential range its reciprocal is a bounded resolvent multiplier.

For the heat semigroup on $L^2(\mathbb{R},dx)$, take the normalization $L_H=f^{\prime \prime }$. The unitary Fourier transform converts it to multiplication by $-{\xi }^2$, and converts the semigroup to multiplication by $e^{-t{\xi }^2}$. ConsequentlyThe Sobolev space domain is exactly $\{f:{\xi }^2\hat{f}\in L^2\}$. The continuous spectrum comes from the full essential range of $-{\xi }^2$, not from square-integrable Fourier eigenvectors.

For the Ornstein-Uhlenbeck semigroup on standard Gaussian measure, the normalized Probabilists' Hermite polynomials $h_n={\mathrm{He}}_n/\sqrt{n!}$ form an orthonormal basis. The construction below gives $L_{OU}=D^2-xD$ and $L_{OU}{h}_n=-n{h}_n$, so the Hermite coefficient map turns it into a real multiplication operator on ${\ell }^2({\mathbb{N}}_0)$. ThereforeFinally, the heat semigroup on the whole real line has no positive spectral gap, whereas the standard Ornstein-Uhlenbeck semigroup has gap one. For the first claim, take a nonzero smooth compactly supported $\varphi$ and set $f_R(x)=R^{-1/2}\varphi (x/R)$. Then $\Vert f_R{\Vert }_2^2=\Vert \varphi {\Vert }_2^2$, but $\Vert f_R^{\prime }{\Vert }_2^2=R^{-2}\Vert {\varphi }^{\prime }{\Vert }_2^2$. This disproves a uniform whole-line Poincaré inequality $\Vert f{\Vert }_2^2\le C\Vert f^{\prime }{\Vert }_2^2$; one can also choose $\int \varphi =0$. Lebesgue measure here is infinite, so it has no normalized probability mean to subtract. For Gaussian measure, the Hermite expansion instead gives the sharp Gaussian Poincaré inequalityEquality holds for affine functions. The identity on the right is interpreted on the energy-form domain, which is larger than the full operator domain.

On a compact metric space $X$, a conservative Feller semigroup is a strongly continuous semigroup of positive linear maps on $C(X)$ satisfying $P_{t}1=1$. Positivity means $f\ge 0⟹P_{t}f\ge 0$. These conditions imply sup-norm contraction. Conversely, for a contraction semigroup with $1\in D(L)$ and $L1=0$, differentiation on the generator domain gives $\frac{d}{dt}{P}_{t}1=P_{t}L1=0$, so $P_{t}1=1$. On the real space $C(X)$, if $0\le f\le 1$, then $\Vert 1-P_{t}f{\Vert }_{\infty }\le \Vert 1-f{\Vert }_{\infty }\le 1$, implying $P_{t}f\ge 0$. Scaling proves positivity for every nonnegative $f$. This is the unital contraction positivity criterion. For complex $C(X)$ the same conclusion follows because each functional $f\mapsto P_{t}f(x)$ has norm one and value one on $1$, hence is a positive linear functional.

For each $t,x$, the Riesz-Markov-Kakutani representation theorem represents that functional by a unique Borel probability measure $p_t(x,dy)$:Continuity of $P_{t}f$ gives weak continuity of $x\mapsto p_t(x,\cdot )$. Approximating indicators of open sets increasingly by continuous functions proves Borel measurability in $x$; a monotone-class argument then gives a Markov kernel. The semigroup property and uniqueness of representing measures yield $p_0(x,\cdot )={\delta }_x$ and the Chapman-Kolmogorov equationThese are the required transition probabilities.

An invariant probability measure for a semigroup satisfies ${\int }_X{P}_{t}fd\mu ={\int }_{X}fd\mu$ for every $t\ge 0$ and $f\in C(X)$. Equivalently its distribution is preserved by the Markov kernel. For $1\le p<\infty$, Jensen inequality gives $|P_{t}f{|}^p\le P_t(|f{|}^p)$ pointwise. Integrating and using invariance provesAlso $\Vert P_{t}f-f{\Vert }_{L^p(\mu )}\le \Vert P_{t}f-f{\Vert }_{\infty }\to 0$. Continuous functions are dense in these $L^p$ spaces for a finite Borel measure on a compact metric space, so the contraction extends to a strongly continuous semigroup on $L^p(\mu )$.

For real $f$ with $f,f^2\in D(L)$, the kernel form of Jensen inequality gives $P_t(f^2)-(P_{t}f{)}^2\ge 0$, with equality at $t=0$. Take its right derivative to obtain the generator square inequalityThis is nonnegativity of the associated carré du champ.

Each time-average functional ${\varphi }_n$ is positive with ${\varphi }_n(1)=1$, hence represents a Borel probability measure ${\mu }_n$. The compactness of probability measures on a compact metric space gives a subsequence converging weakly to a probability measure $\mu$. Explicitly, $C(X)$ is a separable Banach space, so Banach-Alaoglu theorem and a countable dense family give a subsequence converging on every continuous function, rather than merely a net. For fixed $t\ge 0$, the semigroup property gives the Krylov-Bogolyubov time-average argument:Its absolute value is at most $2t\Vert f{\Vert }_{\infty }/n$. Pass to the subsequential limit, observing that $P_{t}f\in C(X)$, to get ${\int }_X{P}_{t}fd\mu ={\int }_{X}fd\mu$. Thus the limiting measure is invariant.

For a densely defined operator $A$ on a complex Hilbert space, unbounded self-adjointness means equality $A=A^{\ast }$ including equality of the operator domains. Symmetry alone asserts only $A\subseteq A^{\ast }$ and is insufficient. Use an inner product linear in its first argument. If $\mathrm{Im}z\ne 0$, symmetry givesThus $A-z$ is bounded below. Since $A$ is closed, its range is closed; its orthogonal complement is $\ker (A^{\ast }-\overline{z})=\ker (A-\overline{z})=0$, so the range is also dense and therefore all of $H$. The inverse is bounded by $|\mathrm{Im}z{|}^{-1}$. The nonreal resolvent estimate for a self-adjoint operator proves $\sigma (A)\subseteq \mathbb{R}$.

For bounded self-adjoint $T$, its numerical range of an operator and numerical radius areEvery number in $W(T)$ is real. If $z\notin \overline{W(T)}$, its distance $\delta$ from that closed set is positive, and $\Vert (T-z)u\Vert \ge \delta \Vert u\Vert$. The same bound applies to $T-\overline{z}$, so the range is dense as well as closed and $T-z$ has a bounded inverse. Hence the numerical-range spectral enclosure isThe closure is necessary in infinite dimension: the endpoints of the numerical range need not themselves be attained.

Put $\beta =\sup W(T)$ and $B=\beta I-T$, a bounded positive semidefinite operator. Choose unit vectors $u_n$ with $⟨B{u}_n,u_n⟩\to 0$. The Cauchy-Schwarz inequality for the positive form $⟨B\cdot ,\cdot ⟩$ impliesTherefore $\beta$ is an approximate eigenvalue. Apply the same argument to $T-\alpha I$, where $\alpha =\inf W(T)$, to get the other endpoint. This proves the numerical-range endpoint approximate-eigenvalue theorem. Each endpoint is in the spectrum: a bounded inverse would forbid unit vectors with vanishing residual.

If an unbounded self-adjoint operator $A$ is positive semidefinite, then for every real $z<0$, $\Vert (A-z)u\Vert \ge (-z)\Vert u\Vert$. The range argument above again proves invertibility, excluding negative values from the spectrum. For the converse, suppose $\sigma (A)\subseteq [0,\infty )$. Its resolvent operator $R=(I+A{)}^{-1}$ is bounded and self-adjoint. The resolvent spectral mapping identity puts $\sigma (R)\subseteq [0,1]$: for $w\ne 0$, the relevant spectral parameter of $A$ is $1/w-1$, and nonreal values are excluded by self-adjointness. By the endpoint result just proved, $W(R)\subseteq [0,1]$. Positive-form Cauchy-Schwarz inequality then givesFor $u=Rf\in D(A)$ this yields $⟨Au,u⟩=⟨f-Rf,Rf⟩\ge 0$. Thus the spectral positivity criterion for a self-adjoint operator isThis resolvent proof uses the bounded numerical-range result to handle the full unbounded domain.

Identify $G$ with the Boolean hypercube $\{-1,1{\}}^d$, with coordinatewise multiplication and normalized Haar measure assigning mass $2^{-d}$ to each point. Its Bernoulli function ${ϵ}_i(\omega )={\omega }_i$ is a Rademacher random variable. The Walsh functions on a hypercube are the Walsh characters $w_A={\prod }_{i\in A}{ϵ}_i$, indexed by all subsets $A\subseteq \{1,\dots ,d\}$, including $w_{\varnothing }=1$. Independence of the sign coordinates gives $⟨w_A,w_B⟩=\mathbb{E}{w}_{A△B}=1_{A=B}$. There are $2^d$ characters, equal to the dimension of $L^2(G)$, so they form an orthonormal basis.

Flipping coordinate $i$ negates $w_A$ exactly when $i\in A$. Thus the coordinate-flip generator on a hypercube satisfiesFor $f={\sum }_A\hat{f}(A)w_A$, Parseval identity gives $⟨f,Lf⟩=-{\sum }_A|A||\hat{f}(A){|}^2\le 0$. Equivalently its Dirichlet form of a Markov chain iswhere ${\omega }^{(i)}$ flips coordinate $i$.

Set $S(\omega )={\sum }_i{a}_i{ϵ}_i(\omega )$ and $F(\omega )=\Vert S(\omega )\Vert$. Since $F(-\omega )=F(\omega )$, its Walsh expansion contains only sets $A$ of even size. Every nonconstant such set has $|A|\ge 2$, giving the even-function spectral gap on a hypercubeAt each $S(\omega )$ the Hahn-Banach theorem supplies a real supporting linear functional $\ell$ of norm at most one with $\ell (S)=\Vert S\Vert$; for a complex normed space use the real part of a complex norming functional. At $S=0$ take $\ell =0$. Convexity of the norm gives $F({\omega }^{(i)})-F(\omega )\ge \ell (S({\omega }^{(i)})-S(\omega ))$. Since $LS=-S$, sum these inequalities to get $LF\ge -F$. Therefore $\mathcal{E}(F)=-\mathbb{E}[FLF]\le \mathbb{E}{F}^2$. Combining the two bounds yields the sharp Rademacher second-moment inequalityNo smoothness of the norm is required. The constant $2$ is sharp: two equal nonzero real coefficients give modulus $0$ or $2|a|$ with equal probabilities. This is the second-versus-first moment case of the Kahane-Khintchine inequality for arbitrary normed vector spaces.

For the complex-circle assertion, write each independent Steinhaus random variable as ${\eta }_i={ϵ}_i\cos {\theta }_i+i{\delta }_i\sin {\theta }_i$, where ${\theta }_i$ is uniform on $[0,\pi /2]$ and all quadrant signs ${ϵ}_i,{\delta }_i$ are independent. Conditional on the angles, ${\sum }_i{a}_i{\eta }_i$ is a Rademacher sum with $2d$ complex coefficients $a_i\cos {\theta }_i$ and $i{a}_i\sin {\theta }_i$. Its conditional second moment is ${\sum }_i|a_i{|}^2$, independent of the angles. The preceding inequality gives its conditional first moment at least $\sqrt{\frac{1}{2}{\sum }_i|a_i{|}^2}$. Average over the angles and square to obtain the Steinhaus first-moment lower bound

On the polynomial subspace of $L^2(\gamma )$, the Gaussian creation and annihilation operators are $a^{-}=D$ and $a^{+}=x-D$. Gaussian integration by parts makes them adjoints there and gives $[a^{-},a^{+}]=I$. The Gaussian number operator isThe Probabilists' Hermite polynomials are ${\mathrm{He}}_n(x)=(-1{)}^n{e}^{x^2/2}{D}^n{e}^{-x^2/2}=(a^{+}{)}^{n}1$. Their generating function is $e^{sx-s^2/2}={\sum }_n{\mathrm{He}}_n(x)s^n/n!$, giving $D{\mathrm{He}}_n=n{\mathrm{He}}_{n-1}$. The commutator also yieldsThe orthogonality follows by repeated integration by parts; the leading coefficient is one, so these polynomials span every polynomial. By the allowed density assumption $h_n={\mathrm{He}}_n/\sqrt{n!}$ is an orthonormal basis. Define the closed number operator by $N\sum c_n{h}_n=\sum n{c}_n{h}_n$ on $\sum n^2|c_n{|}^2<\infty$. This diagonal multiplication operator is self-adjoint and positive semidefinite; polynomial truncations show it is the closure of the polynomial operator.

The Ornstein-Uhlenbeck semigroup is $P_t=e^{-tN}$, or $P_{t}f={\sum }_n{e}^{-nt}{c}_n{h}_n$. It also has the Mehler formula for the Ornstein-Uhlenbeck semigroupTo verify the formula, apply its right side to $e^{sx-s^2/2}$: the Gaussian moment-generating function makes the result $e^{s{e}^{-t}x-s^2{e}^{-2t}/2}$, proving the eigenvalue identity on every Hermite polynomial. Positivity and invariance of Gaussian measure give $L^2$ contraction by Jensen inequality, so polynomial density extends the equality to all $L^2(\gamma )$. The coefficient expansion givesFor continuously differentiable $f$ with bounded derivative, differentiate the Mehler expectation by the dominated convergence theorem to get the Ornstein-Uhlenbeck gradient commutation identityThe same identity extends to the Gaussian Sobolev form domain. Mehler's formula also gives pointwise convergence to the Gaussian mean for such $f$, since bounded derivative permits at most linear growth.

The Gaussian Dirichlet energy is the energy form of $N$:For polynomials this follows from $N=(a^{-}{)}^{\ast }{a}^{-}$, and closure extends it to the form domain. For the given continuously differentiable $f$, Gaussian integration by parts yields $⟨f^{\prime },h_{n-1}⟩=\sqrt{n}{c}_n$, so Parseval identity proves the derivative-energy equality directly. Bounded derivative puts $f$ in the Gaussian Sobolev space, hence in that form domain; it need not be in the full operator domain, so $⟨f,Nf⟩$ is not always a legitimate initial definition.

Use the entropy functional ${\mathrm{Ent}}_{\gamma }(h)=\int h\log hd\gamma -(\int hd\gamma )\log \int hd\gamma$. Begin with bounded smooth $h>0$ bounded away from zero, and put $h_t=P_{t}h$. Invariance and Gaussian integration by parts give the Ornstein-Uhlenbeck entropy dissipation identityThe gradient identity and the permitted weighted Cauchy-Schwarz inequality, applied with weight $h$, implyThe entropy tends to zero as $t\to \infty$ by Mehler's formula and bounded convergence. Integrating in time and using invariance yields ${\mathrm{Ent}}_{\gamma }(h)\le \frac{1}{2}\int |h^{\prime }{|}^2/hd\gamma$. Set $h=f^2$ to obtain the sharp Gaussian logarithmic Sobolev inequalityFor the stated possibly unbounded $f$, apply the argument to smooth positive clipped approximations with a common positive lower bound and uniformly bounded derivatives. They can converge pointwise together with their derivatives and have a common linear-growth bound. Since Gaussian measure integrates every polynomial moment, dominated convergence theorem passes both the entropy and the derivative energy to the limit. This also justifies use of the supplied inequality when $2f{f}^{\prime }$ itself is unbounded. The normalization $\Vert f{\Vert }_1=1$ concerns $f$, and does not imply $\int f^{2}d\gamma =1$; the second term in the entropy definition must be retained.