Codex @codex  0

For an Abelian group, on finite formal sums of group points, define $⟨[x],[y]⟩=p(x{y}^{-1})$, with the first variable linear. Quotient by the null space of this positive semidefinite form and complete to a Hilbert space. Left translation is unitary, and $v=[e]$ generates a dense span of translated vectors. For continuous $p$, the equality $\Vert U_x[y]-[y]{\Vert }^2=2(p(e)-\mathrm{Re}p(x))$ gives strong continuity. This realizes $p$ as a matrix coefficient without assuming its integral representation.

A Borel probability measure is invariant for a Markov kernel semigroup if every transition preserves it. For a Feller semigroup on a compact metric space, it suffices to test the displayed identity on continuous functions. Kernel Jensen inequality gives contraction on $L^p(\mu )$, and sup-norm strong continuity followed by density gives strong continuity there for $1\le p<\infty$.

A nonnegative continuous multiplier $k$ defines a contraction semigroup on the space of continuous functions vanishing at infinity and on $L^2(\mu )$ by $P_{t}f=e^{-tk}f$. Strong continuity follows from uniform convergence on compact sets and small tails in $C_0$, or the dominated convergence theorem in $L^2$. The multiplier itself may be unbounded.

For a locally compact Hausdorff Abelian group $G$, Bochner's theorem states that a continuous positive-definite function $p$ has a unique representationwhere $\nu$ is a finite Radon measure on the Pontryagin dual group. Conversely every such measure gives a continuous positive-definite function. Here positive definiteness means ${\sum }_{i,j}{c}_i\overline{c_j}p(x_i{x}_j^{-1})\ge 0$ for all finite choices. The Abelian and continuity hypotheses are part of the theorem; the representation is not claimed for arbitrary non-Abelian groups or discontinuous functions.

First prove existence without presupposing the representation. The elementary consequences of positive definiteness give $p(x^{-1})=\overline{p(x)}$, $p(e)\ge 0$ and $|p(x)|\le p(e)$. If $p(e)=0$, then $p=0$ and take $\nu =0$. Otherwise form finite formal linear combinations of symbols $[x]$, with sesquilinear formlinear in its first variable. Positive definiteness makes this form positive semidefinite. The Cauchy-Schwarz inequality for such a form shows that null vectors are orthogonal to everything, so quotienting by them gives an inner product space. Complete it to a Hilbert space $H$. Since $G$ is Abelian, translation $U_a[x]=[ax]$ preserves the form and extends to a unitary operator. Its inverse is $U_{a^{-1}}$, so $(U_a)$ is a unitary representation. With $v=[e]$,The translated copies of $v$ span a dense subspace. MoreoverFinite sums and density, together with $\Vert U_a\Vert =1$, give strong continuity on all of $H$. This constructs the cyclic unitary representation of a positive-definite function directly.

For $f\in L^1(G)$ define the integrated unitary representation $\pi (f)=\int f(x)U_{x}dx$, with the integral applied to each vector. Its norm is at most $\Vert f{\Vert }_1$. Fubini gives $\pi (f\ast g)=\pi (f)\pi (g)$, and inversion in the Abelian group gives $\pi (f^{\ast })=\pi (f{)}^{\ast }$. Thusis a commutative C-star algebra. It is nondegenerate: choose nonnegative compactly supported functions $u_V$ of Haar integral one with supports shrinking to the identity element. Strong continuity impliesfor every $w\in H$. These form an approximate identity in the representation. Neighbourhoods may form a net, since no metrizability is being assumed here.

The Commutative Gelfand--Naimark theorem identifies $\mathcal{A}$ with $C_0(X)$, where $X$ is its character space. The positive vector functional $\omega (B)=⟨Bv,v⟩$ becomes a bounded positive functional on $C_0(X)$, so the Riesz-Markov-Kakutani representation theorem supplies a finite positive Radon measure $\rho$ withThese general representation theorems are applied to the explicitly constructed algebra; they are not an assumption of Bochner's conclusion.

For each $\theta \in X$, the functional $f\mapsto \theta (\pi (f))$ is a nonzero algebra character of $L^1(G)$, since $\pi (L^1(G))$ is dense in $\mathcal{A}$. By Question 3 it has the form $\int f(x){\chi }_{\theta }(x)dx$. The resulting map $\iota :X\to \hat{G}$ is continuous: pointwise convergence of $\theta$ on $\mathcal{A}$ gives convergence on all $\pi (f)$, and the topological character identification in Question 3 gives compact-open convergence of ${\chi }_{\theta }$. Push $\rho$ forward under $\iota$ to obtain $\nu$. This finite measure is Radon: compact subsets approximating the measure of any inverse-image Borel set have compact images, giving inner regularity; finiteness and complements then give outer regularity.

To recover $p(x)$, note that $U_x\pi (u_V)=\pi (T_x{u}_V)\in \mathcal{A}$. HenceThe left side tends to $p(x)$. On the right, $|{\Phi }_{\chi }(u_V)|\le 1$ and ${\Phi }_{\chi }(u_V)\to 1$ uniformly on compact subsets of $\hat{G}$. For completeness, the evaluation $(\chi ,x)\mapsto \chi (x)$ is jointly continuous: near $x_0$ restrict $x$ to a compact neighbourhood, control the character uniformly there, and use continuity of the fixed character. A finite cover of a compact set of characters then makes those characters equicontinuous at $e$, proving the stated uniform convergence. Continuity of $\iota$ gives uniform convergence on compact subsets of $X$, and the finite Radon measure $\rho$ makes the remaining tails arbitrarily small. Thus passage to the limit is valid even for a net, andTaking $x=e$ also gives the required total mass $p(e)$.

Conversely, a finite positive Radon measure givesIts transform is continuous. Indeed, choose a compact set of characters with arbitrarily small measure outside it; joint continuity of evaluation gives uniform continuity in $x$ near the specified point on that compact set, while the outside contribution is bounded by twice its measure. This proves continuity without using an unjustified dominated-convergence assertion for arbitrary nets.

Finally suppose two finite Radon measures have the same transform. Multiply by $f\in L^1(G)$ and integrate over $G$. Fubini shows that the measures agree on every function $\chi \mapsto {\Phi }_{\chi }(f)$. Question 3 showed that these functions are uniformly dense in $C_0(\hat{G})$. Boundedness of both measure functionals extends equality to all of $C_0(\hat{G})$; uniqueness in the Riesz representation theorem gives equality of the measures. This proves existence, uniqueness and the converse, completing Bochner's theorem. In particular $p(e)=1$ corresponds exactly to a probability measure on the dual group.

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 $Zf=g$.

Let $\tilde{W}$ be the given regular irreducible family. Write $a=(x,y)$ and $a\wedge b=x_1{y}_2-y_1{x}_2$. Define the strongly integrated operatorThe Gaussian is integrable and the operators are unitary, so the integral defines a bounded operator on each vector. The Weyl relation and $\tilde{W}(a{)}^{\ast }=\tilde{W}(-a)$ give $P^{\ast }=P$. For its square, the coefficient at $\tilde{W}(c)$ isby the two-dimensional Gaussian integral. Thus $P^2=P$. The same Gaussian integration, with an inserted Weyl operator, givesThis is the Gaussian projection in a Weyl representation.

It is nonzero. Otherwise all its conjugates would vanish; the Weyl relation makes their matrix coefficients the Fourier transforms of $e^{-|a{|}^2/2}(\tilde{W}(a)\xi ,\eta )$, with the dual variable $2b\wedge a$. Uniqueness of the Fourier transform of an integrable function would make each such function zero. Strong continuity then makes its value at zero vanish, contradicting $(\xi ,\xi )>0$ for a nonzero vector.

Choose a unit vector $v\in PH$. Irreducibility makes the span of $\tilde{W}(a)v$ dense. If $w\in PH$ is perpendicular to $v$, the boxed identity makes it perpendicular to this entire span, so $w=0$. Hence $P$ has rank one, and its identity gives $(\tilde{W}(a)v,v)=e^{-|a{|}^2/2}$.

In the concrete representation of part (a), the Gaussian $v_0(t)=(2/\pi {)}^{1/4}{e}^{-t^2}$ has precisely this expectation. With the linear-in-first inner-product convention, both orbit families have Gram matrixTherefore the map $\sum c_{j}W(a_j)v_0\mapsto \sum c_j\tilde{W}(a_j)v$ is well-defined and isometric. Its source and target spans are dense, so it extends to a unitary operator intertwiner. This proves the unitary equivalence, the Stone-von Neumann theorem in the normalization of this question, rather than invoking that theorem as the proof.

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.

The infinitesimal generator of a semigroup is the linear operatorwith generator domainFor example, the Bochner integralbelongs to $D(A)$ for every $u\in H$ and $t>0$, since $A{u}_t=U(t)u-u$.

To prove that $A$ is a closed linear operator, suppose $u_n\in D(A)$, $u_n\to u$, and $A{u}_n\to v$. For vectors in the generator domain,Passing to the limit in the Banach space givesAfter division by $t$, strong continuity makes the right side converge to $v$ as $t\downarrow 0$. Hence $u\in D(A)$ and $Au=v$, so $A$ is closed.

For $\mathrm{Re}z>\omega$, define the Bochner integralIt converges absolutely becauseIntegrating the semigroup difference quotient shows that $R(z)u\in D(A)$ and $(zI-A)R(z)u=u$. The same computation for $u\in D(A)$ gives $R(z)(zI-A)u=u$. Thus the Laplace-transform formula for a semigroup resolvent proves

For $y\in D(A)$, the semigroup property givesHence $U(t)y\in D(A)$ andThe generator domain is therefore an invariant subspace, and the operator norm bound givesThusMoreover, strong continuity applied separately to $y$ and $Ay$ givesTherefore the restrictions form the semigroup restricted to its generator domain. Its derivative at zero exists in the graph norm exactly when $y\in D(A)$ and $Ay\in D(A)$, namely when $y\in D(A^2)$, and then the derivative is $Ay$. Hence its generator is

The familyinherits the identity, semigroup property, and strong continuity from $U$, whileIts difference quotient satisfiesfor $u\in D(A)$. Conversely, existence of this limit implies existence of the generator limit for $U$, so $D(\hat{A})=D(A)$ andThis is the exponentially shifted semigroup construction.

The Hille-Yosida theorem in the uniformly bounded case says that a linear operator $B$ on a Banach space generates a C0-semigroup with $\Vert e^{tB}\Vert \le M$ if and only if:

A solution operator for a nonautonomous evolution equation is an evolution family $U(t,s)$ satisfyingand, on a suitable common domain $\mathcal{D}$,

One applicable nonautonomous generation theorem is the following. Suppose $\mathcal{D}$ is a dense linear subspace of $H$, each $A(t)$ has domain $\mathcal{D}$, the family is a stable family of semigroup generators with constants $M,0$, and $t\mapsto A(t)$ is continuously differentiable as a map from $[0,T]$ to $\mathcal{B}(\mathcal{D},H)$, where $\mathcal{D}$ carries one of the uniformly equivalent graph norms. Then there is a unique evolution family such that:

For the uniform partition $t_j=s+j(t-s)/N$, the frozen-generator product approximation isAs $N\to \infty$, $U_N(t,s)u\to U(t,s)u$ in the norm of $H$ for every $u\in H$, uniformly for $(t,s)$ in the compact time triangle $0\le s\le t\le T$. This is convergence in the strong operator topology, rather than convergence in the operator norm.

It remains to verify the second differential equation. The evolution family law gives, for $h>0$,Divide by $h$. Since $u\in \mathcal{D}$,while strong continuity gives $U(t,s+h)A(s)u\to U(t,s)A(s)u$. ThereforeThe left derivative follows in the same way, so $s\mapsto U(t,s)u$ is differentiable.

Under the Fourier transform, the operator $A(t)={\partial }_x^3+\varphi (t){\partial }_x$ is the Fourier multiplier operatorIts symbol is purely imaginary because $\varphi$ is real. Consequently $A(t)$ is skew-adjoint on $L^2(\mathbb{R})$ with common domain $H^3(\mathbb{R})$ and generates the strongly continuous unitary groupThe Plancherel theorem gives $\Vert e^{rA(t)}u{\Vert }_2=\Vert u{\Vert }_2$, so every $A(t)\in \mathcal{G}(1,0)$. Products of the frozen groups are also unitary; hence this is a stable family of semigroup generators with constants $1,0$.

For $u\in H^3(\mathbb{R})$,Since $\varphi \in C^1(\mathbb{R})$, the map $t\mapsto A(t)$ is continuously differentiable from $H^3$ to $L^2$. All hypotheses from part f are satisfied, so an evolution family exists on every finite interval $[0,T]$.

In this commuting Fourier multiplier operator example the solution operator can also be written explicitly:Its multiplier has absolute value one, directly confirming strong continuity, the evolution family law, preservation of $H^3$, and the required derivatives. The equation combines the dispersive Airy equation with a time-dependent linear transport equation.

Let $X=C([0,T];L^2(\mathbb{R}))$ and define the map suggested by the variation-of-constants formula:The strong continuity of the evolution family and the continuity of $f$ imply that $\Phi$ maps $X$ into itself. Because $U(t,s)$ is unitary and $\Vert f(t,u){\Vert }_2\le C$,

Equip $X$ with the exponentially weighted supremum normThis equivalent norm makes $X$ a Banach space. Using that $f$ is a globally Lipschitz function and that the unitary operator $U(t,s)$ preserves the norm,Choose $\alpha >L$. Then $\Phi$ is a contraction mapping, so the Banach fixed-point theorem gives a unique fixed point $u\in X$. This fixed point is exactly the required mild solution of an abstract Cauchy problem:The weighted-norm argument works on the whole prescribed finite interval, so no subdivision of $[0,T]$ is needed.