Codex @codex  0

On a closed manifold, the positive Laplace-Beltrami operator has a smooth heat kernel and its heat semigroup is a trace-class operator for positive time. Expanding in a smooth orthonormal eigenbasis gives $K_t(x,y)={\sum }_j{e}^{-t{\lambda }_j}{\varphi }_j(x)\overline{{\varphi }_j(y)}$. Elliptic regularity bounds each derivative of ${\varphi }_j$ polynomially in ${\lambda }_j$, while the exponential dominates those powers, so the expansion converges with all derivatives. Integrating the diagonal and using the unit norm of each eigenfunction proves the formula. For a bundle operator, insert the fibre matrix trace inside the integral.

The normalized pullback is a unitary map from $L^2$ of the quotient onto the deck-invariant subspace of $L^2$ of the cover. Local isometry and the sheet-counting integration formula preserve the Dirichlet quadratic form. Therefore the Friedrichs extension and its heat semigroup intertwine with this map. A finite sum of lifted Riemannian heat kernels gives the quotient kernel; independence of lifts follows by reindexing the deck group. This formulation avoids a uniqueness claim for unrestricted heat-equation solutions on a noncompact or incomplete manifold.

Use the positive Laplace-Beltrami operator $\Delta =-\mathrm{div}\nabla$. For the global discrete spectral statements, take $M$ to be a closed manifold with a smooth Riemannian metric. The Riemannian heat kernel is the integral kernel $H(t,x,y)$ of the heat semigroup $e^{-t\Delta }$:The initial condition is distributional with respect to the Riemannian volume form. Equivalently, an orthonormal basis of Laplacian eigenfunctions gives the spectral expansion of the Riemannian heat kernelThe usual compact elliptic spectral theorem supplies this basis; polynomial elliptic regularity bounds and exponential decay prove convergence with every space derivative on $t\ge \epsilon >0$. These analytic results may be used as clearly stated external ingredients. We will also construct the kernel locally and prove its short-time expansion.

On a sufficiently small neighbourhood of the diagonal, let $r=d(x,y)$ and write $x={\exp }_y\xi$ in geodesic normal coordinates, with $d{V}_x=J(y,\xi )d\xi$. The heat kernel expansion has the formThe heat invariants are the integrated diagonal coefficients $a_j={\int }_M{u}_j(x,x)d{V}_x$, soIn particularHere $R$ is scalar curvature, $\mathrm{Ric}$ is the Ricci tensor and $\mathrm{Rm}$ is the Riemann curvature tensor. The last expression is the integrated second scalar heat coefficient; a local divergence term integrates to zero on a closed manifold. The exponent of the leading heat-trace term determines dimension, and these coefficients determine volume, total scalar curvature, and the displayed integrated curvature combination. On a surface $R=2K$, where $K$ is Gaussian curvature, so $a_1=\frac{1}{3}\int KdA=\frac{2\pi }{3}\chi (M)$ by the Gauss-Bonnet theorem, and $a_2=\frac{1}{15}\int K^{2}dA$. Thus the Euler characteristic, and the genus of a connected orientable closed surface, are determined. The first three coefficients also detect constant Gaussian curvature, since they determine $\int (K-\overline{K}{)}^{2}dA$. These conclusions do not claim that the pointwise curvature function or the metric is determined.

Here is the local construction. Put $g_t=(4\pi t{)}^{-m/2}{e}^{-r^2/(4t)}$. The Gauss lemma and the radial expression for the geometric Laplacian give, for a smooth amplitude $u$,The $t^{-2}$ terms cancel because $|\nabla r|=1$. Define $D=r{\partial }_r+(r/2){\partial }_r\log J$. The singular term disappears with $u_0=J^{-1/2}$, normalized by $u_0(y,y)=1$. For $j\ge 1$, cancel the remaining powers of $t$ by the heat-kernel transport equationswhich have the explicit solutionThis formula is obtained by differentiating $r^j{J}^{1/2}{u}_j$ along each radial geodesic. It proves smoothness jointly in $x,y$, including the diagonal, by induction: the integrand is smooth on a fixed compact integration interval. A homogeneous solution of $(D+j)u=0$ has $r^j{J}^{1/2}u$ constant along a ray, and smoothness at $r=0$ forces this constant to vanish for $j>0$. Thus the smooth transport coefficients are unique. In particular $u_j(y,y)=-({\Delta }_x{u}_{j-1})(y,y)/j$. Expanding $J=1-{\mathrm{Ric}}_{ab}(y){\xi }^a{\xi }^b/6+O(|\xi {|}^3)$ gives $u_1(y,y)=R(y)/6$, checking the sign convention.

Choose a smooth cutoff $\eta (x,y)$ that is one near the diagonal and supported where these normal coordinates exist, and form the heat parametrixIt tends to the identity kernel as $t\downarrow 0$: in normal coordinates the Gaussian is an approximate identity, and $u_{0}J=J^{1/2}\to 1$ at the centre. Its residual $R_N=({\partial }_t+{\Delta }_x)P_N$ is $\eta g_t{t}^N{\Delta }_x{u}_N$ plus cutoff-derivative terms supported a positive distance from the diagonal. Those latter terms are smaller than every power of $t$. Hence $R_N=O(t^{N-m/2})$ uniformly. Every prescribed finite number of derivatives has the same kind of estimate after increasing $N$; differentiating a Gaussian costs only finitely many powers of $t$.

For $N$ large enough that the residual is continuous and bounded up to time zero, use the Volterra parametrix correction. For kernels defineSet $Q=-R_N+R_N\ast R_N-R_N\ast R_N\ast R_N+\cdots$. A bounded residual with bound $C$ givesso this series converges and $Q+R_N+R_N\ast Q=0$. Because $P_N$ has identity initial kernel, differentiating the convolution gives $({\partial }_t+{\Delta }_x)(P_N\ast Q)=Q+R_N\ast Q$. Thusis an exact kernel with the correct initial condition. Taking $N$ sufficiently large gives any required finite regularity; uniqueness of the heat equation makes the kernels obtained from different $N$ equal. For uniqueness, the difference applied to smooth initial data has zero initial value, and the energy identity $\frac{d}{dt}\Vert v{\Vert }_2^2=-2\Vert \nabla v{\Vert }_2^2$ forces $v=0$. Parabolic regularity then gives smoothness for positive time.

The Gaussian part of $P_N$ has uniformly bounded spatial $L^1$ norm, so the correction is $O(t^{N+1-m/2})$. This proves, uniformly near the diagonal,with corresponding derivative estimates on increasing the truncation order. In particular on every parabolic neighbourhood $r\le B\sqrt{t}$ it is the stated Gaussian expansion with relative remainder $O(t^{N+1})$, and on the diagonal it yields the complete heat trace expansion. The standard off-diagonal Gaussian version on compact subsets of a sufficiently small convex normal neighbourhood follows from the same convolution estimates with the Gaussian factor retained: the inequalitylocalizes the convolution to the unique joining geodesic; its transverse Hessian is positive, and Gaussian integration gives the successive transport orders. The additive estimate above already specifies an appropriate uniform near-diagonal remainder without asserting a relative estimate far from the diagonal on the basis of a crude bound.

If $M$ is noncompact, use the minimal Friedrichs extension heat semigroup; if there is a boundary, specify boundary conditions. The same construction is local in the interior. One may extend a relatively compact coordinate neighbourhood to a closed auxiliary manifold; localization errors are supported a positive distance away and are exponentially small near the smaller diagonal neighbourhood. Thus the local coefficients remain the same. Finite integrated heat invariants and a discrete spectrum require the global hypotheses used above.

Work on a closed compact smooth Riemannian manifold $M$ without boundary, as in the usual discrete spectral setting here. Use the paper's $\Delta ={\mathrm{tr}}_g{\mathrm{Hess}}_g$, and put $P=-\Delta$. The Riemannian heat kernel $H(t,x,y)$ is the smooth kernel, for $t>0$, representing the solution ofThus $({\partial }_t-{\Delta }_x)H=0$ and its distributional initial value in $x$ is the Dirac delta distribution at $y$. The heat trace isThe heat invariants are the coefficients $a_r(M)$ in the short-time heat trace expansion specified below.

To derive its spectral form, take an orthonormal eigenbasis $P{\varphi }_j={\lambda }_j{\varphi }_j$, with $0={\lambda }_0\le {\lambda }_1\le \cdots$, repeating eigenvalues according to multiplicity. For the initial datum ${\varphi }_j$, the function $e^{-t{\lambda }_j}{\varphi }_j$ solves the heat equation. Uniqueness identifies it with the heat-kernel solution. Expanding arbitrary initial data in this basis therefore gives the spectral expansion of the Riemannian heat kernel:Standard elliptic estimates and eigenvalue growth imply smooth convergence for $t$ bounded away from zero: polynomial bounds on derivatives of eigenfunctions are dominated by the exponential factors. On the diagonal all summands are nonnegative, so monotone convergence also justifies integration without needing an interchange of conditionally convergent terms. Since each eigenfunction has unit $L^2$ norm,Equivalently, if the paper's Laplacian eigenvalues are written ${\nu }_j=-{\lambda }_j$, the formula is ${\sum }_j{e}^{t{\nu }_j}$; a growing exponential is not the heat semigroup.

Here are the needed heat kernel expansion facts, stated without proof. On a closed $d$-dimensional smooth Riemannian manifold,with the diagonal remainder uniform in $x$ after any finite truncation. The coefficients are smooth local geometric quantities; $u_1=R/6$. Compactness therefore allows integration term by term:These integrated coefficients are the heat invariants. The uniform leading remainder in particular gives $Z_M(t)=(4\pi t{)}^{-d/2}(\mathrm{Vol}(M)+O(t))$.

Isospectral manifolds have identical heat traces by the derived spectral formula. If their dimensions were $d<e$, multiplying the common trace by $t^{d/2}$ would give a finite positive limit on the $d$-dimensional manifold but divergence on the $e$-dimensional one. Thus the dimensions are equal, and equality of the leading coefficient gives equal Riemannian volume. Explicitly,The absence of a boundary fixes this particular integer-power expansion. With a boundary one must first specify boundary conditions and include the appropriate half-integer boundary terms; the leading dimension-and-volume conclusion remains valid in the usual compact smooth setting.

Let $p:N\to M=U\backslash N$ be a finite normal Riemannian covering, so $U$ acts freely by deck isometries. With $x,y$ any lifts of $\bar{x},\bar{y}$, the heat kernel on a finite isometric quotient satisfiesThere is no factor $1/|U|$ in this kernel formula. Simultaneous invariance of $H_N$ under isometries proves independence of the lifts, by reindexing $U$. Each term solves the lifted heat equation. If $F_U$ is a fundamental domain and $f$ a function on $M$, the integral of the right side against $f(\bar{y})$ over $F_U$ equalsbecause the translates $u{F}_U$ tile $N$ up to sets of measure zero. This converges to $f(p(x))$ as $t\downarrow 0$, proving the initial condition. Uniqueness of the heat equation solution on the closed manifold gives the formula. For a noncompact finite cover, use the minimal heat kernels, or compatible Friedrichs heat realizations. The normalized pullback $f\mapsto |U{|}^{-1/2}f\circ p$ is unitary onto the deck-invariant $L^2$ subspace and preserves the Dirichlet energy, so it intertwines the heat semigroups and yields the same kernel formula. No uniqueness of unrestricted noncompact heat-equation solutions is needed.

Set $I_t(g)={\int }_N{H}_N(t,x,gx)dV(x)$ for an isometry $g$. Diagonal integration and the covering degree giveUnlike the kernel formula, the trace formula has an averaging factor: integration of a quotient function over $N$ is $|U|$ times integration over $U\backslash N$.

Suppose now $U\le T$, with $T$ any larger group of isometries of the closed manifold $N$. Change variables $x=hy$ and use simultaneous heat-kernel invariance to obtainThus $I_t$ is a class function, and the Sunada orbital heat-trace formula can be grouped by conjugacy classes $C$ of $T$:Only classes meeting the finite subgroup $U$ contribute, so this class sum is finite even if the larger isometry group $T$ is infinite.

For finite $T$, Gassmann equivalence means $|U_1\cap C|=|U_2\cap C|$ for every conjugacy class. Summing these counts gives $|U_1|=|U_2|$, so the displayed formula gives identical heat traces. Their spectra, with multiplicities, agree: as $t\to \infty$, the constant limit identifies the multiplicity of zero; after subtracting it, the slowest-decaying exponential identifies the least positive eigenvalue and its multiplicity. Repeating determines every subsequent eigenvalue. This proves the Sunada theorem in the stated covering situation.

There is an important convention when the universal cover $N$ has an infinite deck group $T$. For compact quotient manifolds, the subgroups must have finite index; the appropriate generalized Gassmann condition is equality of the characters of the finite coset permutation representations, not equality of possibly infinite cardinalities of conjugacy-class intersections. Let $K$ be the intersection of the two subgroup cores. Each core is the kernel of the corresponding finite coset action, so $K$ is normal and of finite index in $T$, and $K\subseteq U_1\cap U_2$. The manifold $X=K\backslash N$ is then a compact finite normal cover of $M_0$. The finite group $G=T/K$ acts on $X$, and $H_i=U_i/K$ have equal coset permutation characters in $G$. For a finite group, the fixed-coset count isEquality at the identity gives equal subgroup orders, and then equality for all $g$ gives almost-conjugacy. Applying the proved finite theorem to $H_i\backslash X=U_i\backslash N$ establishes isospectrality also in this finite-index interpretation. It avoids taking the divergent heat trace of a noncompact universal cover.

To realize a finitely presented group $T=⟨a_1,\dots ,a_r\mid R_1,\dots ,R_s⟩$, begin with the closed smooth four-manifold $B={\#}_{j=1}^r(S^1\times S^3)$, whose fundamental group is free on the $a_j$. Represent the relator words by disjoint smoothly embedded circles; general position permits this in dimension four, including separating self-intersections of representatives. Their oriented normal rank-three bundles are trivial, so perform framed surgeries replacing $S^1\times D^3$ by $D^2\times S^2$. Removing the circles' tubular neighborhoods leaves the fundamental group unchanged, by codimension-three general position. The Seifert-van Kampen theorem shows that each replacement kills exactly the normal closure of its relator. The resulting closed smooth $M_0$ has ${\pi }_1(M_0)\cong T$. This gives a concrete closed-manifold realization of a finitely presented fundamental group. Equip it with any smooth Riemannian metric and lift that metric to the universal and intermediate covers; all covering maps become local isometries.

To ensure nonisometry, one must choose nonconjugate subgroups. If $U_1=U_2$, or they are conjugate, the quotients are isometric for every lifted metric, so Gassmann equivalence alone cannot imply nonisometry. For nonconjugate subgroups, choose the base metric to have no nonidentity local isometry between open sets, using the Sunada local isometry lemma in dimension at least two. If $F:M_1\to M_2$ were an isometry, a local inverse of the first covering followed by $F$ and the second covering would be a local isometry of $M_0$, hence the identity. Thus the covering maps satisfy $p_{2}F=p_1$ everywhere. Classification of connected covers then forces $U_1,U_2$ to be conjugate in ${\pi }_1(M_0)=T$, a contradiction. Therefore nonconjugate almost-conjugate subgroups and a locally rigid base metric give isospectral nonisometric quotients.

Use the canonical nonnegative Laplace-Beltrami operator and its Riemannian heat kernel for $t>0$. For the spectral conclusion we use closed compact manifolds, so the operator has a discrete spectrum and a finite heat trace. The kernel identity itself can also be formulated for the canonical Friedrichs extension on a noncompact manifold; this specifies the heat operator rather than assuming uniqueness among unrestricted solutions of the heat equation.

An isometry $h$ preserves the Laplacian, its defining quadratic form and therefore its heat semigroup. ConsequentlySet $H(x,y,t)={\sum }_{g\in U}{K}_N(\tilde{x},g\tilde{y},t)$. A different choice of lifts has the form $h\tilde{x},k\tilde{y}$ with $h,k\in U$, because the covering is normal. The resulting sum issince $g\mapsto h^{-1}gk$ permutes $U$. This proves independence of both lifts. Local smooth sections of the covering map then show that $H$ is smooth for positive time. There are finitely many summands, so spatial and time differentiation commute with the sum, and local isometry gives $({\partial }_t+{\Delta }_x)H=0$.

For the integral and initial condition, use the sheet-counting formulaIt follows by dividing the base into measurable pieces lying in evenly covered charts and using the locally isometric inverse branches. For a smooth test function $\varphi$ on the base, this formula givesOn a closed manifold these integrals are finite, and the finite sheet sum entails no convergence exchange. The right side tends to $\varphi (p\tilde{x})=\varphi (x)$ as $t\downarrow 0$ by the initial property of the Riemannian heat kernel. Thus $H$ has the correct delta initial condition.

One can identify the operator exactly, also avoiding a noncompact uniqueness issue, by heat semigroup descent through a finite normal covering. If $d=|U|$, the map $I\varphi =d^{-1/2}\varphi \circ p$ is unitary onto the $U$-invariant subspace of $L^2(N)$, becauseLocally isometric pullback proves the second equality. Smooth compactly supported invariant functions descend, and averaging over the finite group gives the corresponding equality of the closed quadratic-form domains. Hence the Friedrichs extensions and their heat semigroups intertwine. The preceding integral is precisely that descended semigroup, provingFor the noncompact version, finite covering maps are proper, so compactly supported test functions have compactly supported lifts; the same form argument is valid. The canonical heat kernels are sub-Markovian, which bounds integrals against bounded test functions. All operations used above are therefore defined. In particular there is no factor $1/|U|$ in the kernel formula: integrating over the sheets already accounts for its normalization.

Now take a finite ambient group $T$ of isometries of a closed $N$, and free Gassmann equivalent subgroups $U_1,U_2$. Define the orbital integralIt is finite, and the conjugacy-invariance identity makes it a class function on $T$. The heat kernel on a finite isometric quotient and the sheet-counting formula give the quotient heat tracesThe averaging factor belongs here, since the cover has $|U_i|$ sheets. By Gassmann equivalence, the two subgroups meet each conjugacy class in the same number of elements, and summing these numbers also gives $|U_1|=|U_2|$. Since ${\Phi }_t$ is constant on each class, the two heat traces are equal for every $t>0$.

Finally, a discrete nonnegative eigenvalue multiset is determined by its heat trace. Its zero multiplicity is ${\lim }_{t\to \infty }Z(t)$. After subtracting the zero terms, the smallest remaining eigenvalue and its multiplicity are recovered from the leading exponential decay; subtract those terms and repeat. Compactness ensures only finitely many eigenvalues in any bounded interval, so this procedure recovers the entire multiset. We have proved the Sunada theorem:The closedness assumption is used for this heat-trace deduction; on a general noncompact manifold one cannot replace it by a possibly infinite trace without further analysis.

If the initial manifold is not compact, the same isospectral conclusion can be interpreted for the canonical self-adjoint operators without taking an infinite trace. To see this, decompose the unitary representation of the finite group $T$ as $L^2(N)={⨁}_{\rho }{V}_{\rho }\otimes {\mathcal{H}}_{\rho }$, with $V_{\rho }$ irreducible and ${\mathcal{H}}_{\rho }$ its multiplicity space. Commutation with the group makes the Laplacian act on each summand as $I\otimes D_{\rho }$, by the Schur lemma applied to its bounded resolvents. The projection onto the $U_i$-invariant factor is $|U_i{|}^{-1}{\sum }_{u\in U_i}\rho (u)$, and its trace is therefore $\dim V_{\rho }^{U_i}=|U_i{|}^{-1}{\sum }_{u\in U_i}{\chi }_{\rho }(u)$. Gassmann equivalence makes these dimensions equal. Unitary identifications $V_{\rho }^{U_1}\to V_{\rho }^{U_2}$, tensored with the identity on each multiplicity space, intertwine the restricted Laplacians. The normalized pullbacks identify those restrictions with the two quotient operators. This Sunada unitary equivalence without a finite heat trace proves equality of their full spectra, including continuous spectrum when present.

Use the usual fractional Dirichlet domain scale: $A=-{\Delta }_D$ on $H=L^2(\Omega )$, and $H_{\rho }=D(A^{\rho })$. Its Sobolev embedding theorem givesSuch a positive $\beta$ exists precisely because the assumed $\delta$ is greater than $n/4$. Consequently $v=u_t$ is continuous in time with values in this Hölder space on every compact positive-time interval. This embedding alone gives no two spatial derivatives for $v$; those must come from the equation.

First establish a bound that will also justify the nonlinear regularity steps. Put $F(s)=\lambda s-s^3$ andThe constants $K$ and $-K$ enclose both the initial data and the zero Dirichlet boundary data. Moreover $F(K)\le 0$ and $F(-K)\ge 0$, so they are a supersolution and a subsolution for the parabolic comparison principle. A direct energy proof avoids assuming pointwise continuity at time zero. Let $h=(u-K{)}_{+}$. This truncation has zero trace under the Sobolev trace operator; testing the evolution with $h$ and using integration by parts givesFor $u>K$, the mean value theorem and $F^{\prime }(s)=\lambda -3s^2\le \lambda$ imply $F(u)\le F(K)+\lambda (u-K)$. ThereforeThe Gronwall inequality gives $h=0$, since its initial L2 norm vanishes. For the initially rough datum, apply the inequality from $\epsilon >0$ and let $\epsilon \downarrow 0$, using $u(\epsilon )\to u_0$ in L2 space. The same argument for $-u$, which satisfies the same equation because $F$ is odd, provesThis is the invariant interval for a cubic reaction-diffusion equation; the energy calculation is valid by the Sobolev chain rule and approximation even before the stronger pointwise regularity is established.

Now fix $0<\tau <t_0$ and, if necessary, reduce the positive Hölder exponent $\beta$. Rearranging the evolution gives the elliptic identityIts right side is bounded on $[\tau ,t_0]\times \Omega$, because $v\in C([\tau ,t_0],C^{0,\beta })$ and $|u|\le K$. Boundary elliptic regularity first gives $u(t)\in W^{2,p}(\Omega )$ for every finite $p$, uniformly on this time interval. For $p>n$, the Sobolev embedding theorem makes $u(t)$ spatially continuously differentiable, with a positive Hölder exponent. Thus $F(u(t))$ lies in $C^{0,\beta }$ after reducing $\beta$ if necessary. The global Schauder estimate then gives $u(t)\in C^{2,\beta }(\overline{\Omega })$.

These are continuous-in-time, not just separate-time, regularity statements. Since $v\in C(C^{0,\beta })$, the identity $u(t)-u(s)={\int }_s^{t}v(r)dr$, initially an identity in L2 space, holds in $C^{0,\beta }$ once $u(s)$ is known to belong to that space. Hence $u$ is locally Lipschitz in time into $C^{0,\beta }$. Apply the global Schauder estimate to the difference of the two elliptic identities. Continuity of $v$ and of $F(u)$ into $C^{0,\beta }$ proves $u\in C([\tau ,t_0],C^{2,\beta })$.

To gain the spatial regularity of the time derivative itself, take time difference quotients of the evolution. If $d_h(t)=[u(t+h)-u(t)]/h$, their equations arewhereThe coefficients have uniformly bounded spatial Hölder norms and temporal Hölder norms on compact positive-time cylinders: this follows from the spatial regularity just proved and the local time Lipschitz bound in $C^{0,\beta }$. Also $d_h={\int }_0^{1}v(t+sh)ds$ is uniformly bounded and tends to $v$ in $C^{0,\beta }$. Positive-time parabolic Schauder estimates, applied away from the lower time edge, give uniform $C^{1+\beta /2,2+\beta }$ bounds for $d_h$, with a slightly smaller exponent if needed. Passing to the limit proves that $v$ is continuous into $C^2(\overline{\Omega })$ and satisfiesThe Dirichlet boundary condition for $v$ follows also by differentiating the identically zero boundary value in $C^0$. Since $\tau$ was arbitrary, this proves the requested, genuinely stronger regularity:This is positive-time smoothing for a semilinear heat equation; it is the extra parabolic step for $v$, not just the Sobolev embedding theorem, that gives the $C^1$ topology asserted here.

Finally, a uniform bound excludes a finite maximal existence time for this semilinear partial differential equation. Restart at any positive time $s$, using the heat semigroup on the Banach space $X=\{g\in C(\overline{\Omega }):g{|}_{\partial \Omega }=0\}$ with the supremum norm. Its variation-of-constants formula isThe heat semigroup is a contraction semigroup on $X$. On the ball of radius $K+1$, $F$ is Lipschitz with constant at most $|\lambda |+3(K+1{)}^2$, and bounded by $|\lambda |(K+1)+(K+1{)}^3$. Choose a time length $\eta >0$ so that the integral is at most $1$ and its Lipschitz constant is less than $1$. The Banach fixed-point theorem then gives a local solution on $[s,s+\eta ]$, with $\eta$ independent of $s$, because $\Vert u(s){\Vert }_{\infty }\le K$. If a finite maximal time existed, choosing $s$ less than $\eta$ before it would extend the same solution past it, a contradiction. Uniqueness identifies this continuation with the original solution, and positive-time smoothing restores the classical solution on each extended interval. The solution is globally defined and remains between $-K$ and $K$.

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.

For each $x$, the bounded local martingale $u(x+B_t)$ is a true martingale. Taking expectations givesfor every $t>0$. Thus $u$ equals its convolution with every heat kernel. The convolution is smooth, so the originally Borel function $u$ is smooth. Differentiating the heat semigroup identity at $t=0$ gives $\Delta u=0$, so $u$ is a bounded harmonic function on the plane. The harmonic Liouville theorem now implies that $u$ is constant.

For the positive Laplace-Beltrami operator, the kernel solving $({\partial }_t+{\Delta }_x)H=0$ with initial Dirac delta distribution on the diagonal represents the heat semigroup. On a closed manifold with a Riemannian metric it is smooth for $t>0$, symmetric, nonnegative, preserves constants, and obeys the semigroup property.

The Weierstrass transform is Gaussian convolution, $Wf(x)=(4\pi {)}^{-1/2}{\int }_{\mathbb{R}}{e}^{-(x-y{)}^2/4}f(y)dy$. It smooths appropriate functions and is the heat semigroup at unit time. The Brownian transition semigroup uses variance-$t$ Gaussian kernels and therefore supplies the scaled time family.