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The Yosida averaging of a semigroup gives $x_h=h^{-1}{\int }_0^{h}S(t)xdt\in D(A)$ with $x_h\to x$, proving that the generator domain is dense. For $x\in D(A)$, the Bochner integral identity $S(t)x-x={\int }_0^{t}S(s)Axds$ holds. If $x_n\to x$ and $A{x}_n\to y$, pass to the limit in this identity and divide by $t$. The strongly continuous semigroup property makes the resulting limit $y$, so $x\in D(A)$ and $Ax=y$. This proves that the infinitesimal generator of a semigroup is a closed operator.

If real $f$ and $f^2$ belong to the generator domain of a Feller semigroup, Jensen inequality gives $P_t(f^2)-(P_{t}f{)}^2\ge 0$. This expression is zero at $t=0$, so its right derivative is nonnegative, proving the inequality. It is equivalent to positivity of the carré du champ.

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.

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.

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

A C0-semigroup on a Banach space $X$ is a family $U(t)\in \mathcal{B}(X)$ such thatfor every $x\in X$. Its infinitesimal generator of a semigroup iswith generator domain

For $M\ge 1$ and $\omega \in \mathbb{R}$, write $A\in \mathcal{G}(M,\omega )$ when $A$ generates a $C_0$-semigroup satisfying $\Vert U(t)\Vert \le M{e}^{\omega t}$. The Hille-Yosida theorem states that this holds exactly when $A$ is closed and densely defined,and, for every real $\lambda >\omega$ and every integer $n\ge 1$,The estimates for every resolvent power, rather than only $n=1$, are essential when $M>1$.