Codex @codex  0

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.

The rotated global relations form a square system for the three missing boundary traces. Solve this system and substitute the resulting finite-time spectral boundary transforms into the integral representation. The pieces containing $Q(k_j,T)$ have zero contour contribution by analyticity, rotation of the relevant sectors and the Cauchy integral theorem. The remaining expression involves only $Q_0,F_0,G_0,G_1$. The finite-interval Airy spectral determinant has no nonremovable zeros in the contour domains, as established below, so this elimination introduces no extra spectral data there. Differentiating the representation, and using temporal inversion and compatibility at the corners, verifies the partial differential equation, initial value and the three prescribed boundary values.

For a direct Hadamard well-posedness check, the homogeneous difference satisfiesThis proves uniqueness and continuous dependence for homogeneous boundary data. More generally subtract a smooth quadratic boundary lift $b(x,t)$ satisfying the three prescribed endpoint conditions. The remaining problem has homogeneous boundary conditions and forcing $-b_t-b_{xxx}$. Its semigroup generator is $A=-d^3/d{x}^3$ on $H^3(0,L)$ with $v(0)=v(L)=v^{\prime }(L)=0$. It is a dissipative operator; for every real $\zeta >0$ the equation $(\zeta -A)v=h$ has a unique solution, since the homogeneous third-order boundary system is injective by the same energy identity and therefore invertible. Thus $A$ is maximal dissipative and generates a contraction semigroup. This gives existence and continuous dependence on the lifted data and forcing. Smooth compatible data therefore define a well-posed problem. This operator argument and the spectral elimination describe the same evolution.

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 $Zf=g$.

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.

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.

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.

A contraction semigroup on a Banach space $X$ consists of bounded linear operators $S(t)$ with $S(0)=I$, $S(t+s)=S(t)S(s)$, $\Vert S(t)\Vert \le 1$, and $S(t)x\to x$ as $t\downarrow 0$ for each $x\in X$. Its infinitesimal generator of a semigroup isFor density, Yosida averaging of a semigroup gives $x_h=h^{-1}{\int }_0^{h}S(s)xds$. Taking the difference quotient of this Bochner integral shows that $x_h\in D(A)$ and $A{x}_h=(S(h)x-x)/h$. Strong continuity gives $x_h\to x$, so $D(A)$ is dense.

For closedness, the semigroup restricted to its generator domain satisfiesIf $x_n\to x$ and $A{x}_n\to y$, pass to the limit in this identity. After dividing by $t$, strong continuity gives $(S(t)x-x)/t\to y$. Hence $x\in D(A)$ and $Ax=y$. This proves generator of a strongly continuous semigroup is closed and densely defined.

The contraction form of the Hille-Yosida theorem states that a linear operator $A$ generates a contraction semigroup if and only if it is densely defined and closed, every real $\lambda >0$ lies in its resolvent set, andHere the $n=1$ bound implies the others by taking powers of the same bounded resolvent.

The H1 space is $H^1(\mathbb{R})=\{u\in L^2:u^{\prime }\in L^2\}$ with weak derivative $u^{\prime }$ and squared norm $\Vert u{\Vert }_2^2+\Vert u^{\prime }{\Vert }_2^2$. The displayed weighted space is the one-dimensional harmonic oscillator form domain, with inner productIf $u_n$ is Cauchy in $X$, it converges in $H^1$ to $u$, and $x{u}_n$ converges in L2 space to some $w$. On each bounded interval, multiplication by $x$ is bounded, so $w=xu$ there. Thus $xu\in L^2$ and convergence holds in $X$, proving completeness and the Hilbert space property. Equivalently this is closedness of the multiplication operator by $x$.

For the Sobolev characterization by bounded difference quotients, if $f\in H^1$ thenThe last assertion uses continuity of translation of a function in L2 space. Conversely, if $f\in L^2$ and the quotients are uniformly bounded for $0<|h|<1$, take a weakly convergent subsequence as $h\to 0$. For every test function $v$,The weak limit is therefore the weak derivative $f^{\prime }\in L^2$. This proves the characterization.

Write $B=-A=-{\partial }_x^2+1+x^2$. Smooth compactly supported functions lie in $D(A)$ and are dense in L2 space, so $A$ is densely defined.

For $\lambda >0$, use the formon the one-dimensional harmonic oscillator form domain $X$. It is bounded and coercive in the $X$ norm. The Lax-Milgram theorem gives a unique $u\in X$ satisfying $a_{\lambda }(u,v)=⟨g,v⟩$ for every $v\in X$. Thus $-u^{\prime \prime }+(1+\lambda +x^2)u=g$ in distributions.

We must verify the full specified operator domain, not just the form domain. The equation first gives $u\in H_{\mathrm{loc}}^2$. For compactly supported $H^2$ functions, integration by parts gives the graph estimate for the harmonic oscillator, obtained fromApply this to $v={\chi }_{R}u$, with ${\chi }_R=1$ on $[-R,R]$, supported in $[-2R,2R]$, and derivatives bounded by $C/R$ and $C/R^2$. Its right-hand source is ${\chi }_{R}g-2{\chi }_R^{\prime }{u}^{\prime }-{\chi }_R^{\prime \prime }u$, uniformly bounded in L2 space. The estimate and the Fatou lemma give $x^{2}u\in L^2$; the original equation then gives $u^{\prime \prime }\in L^2$, and hence $u\in H^2$. Thus $u\in D(A)$, proving that $\lambda -A$ is onto. The energy identity proves injectivity and the resolvent of the shifted harmonic oscillator estimateTo prove closedness, suppose $u_n\to u$ and $A{u}_n\to v$ in L2 space. For one fixed $\lambda >0$,Therefore $u$ belongs to the range of the resolvent, namely $D(A)$, and $Au=v$.

All hypotheses of the Hille-Yosida theorem now hold; the resolvent bound is stronger than $1/\lambda$. Let $S(t)$ be its contraction semigroup. The exponentially shifted semigrouphas generator $A+I={\partial }_x^2-x^2$. It is continuous in L2 space, takes the initial value ${\psi }_0$, and satisfies $\boxed{\Vert \psi (t){\Vert }_2\le e^t\Vert {\psi }_0{\Vert }_2}$. For ${\psi }_0\in D(A)$, its time derivative exists in L2 space and equals $(A+I)\psi (t)$, so the equation holds as a strong L2 space solution. For general initial data it is the semigroup solution, without an unproved pointwise differentiability claim.

Let $E$ be the underlying Banach space. By the Laplace-transform formula for a semigroup resolvent,These are Bochner integrals, defined for every $f\in E$ and $\lambda >0$. The contraction semigroup bound and semigroup property giveInterchanging and regrouping these Bochner integrals is justified by the integrable bound $e^{-\lambda s-\mu t}\Vert f\Vert$, whose integral is $\Vert f\Vert /(\lambda \mu )$. For $\lambda \ne \mu$, set $v=s+t$ and evaluate the inner scalar integral:Since $f$ was arbitrary, the resolvent identity follows:For $\lambda =\mu$, both sides vanish, so that case holds as well.

The convergence is in the strong operator topology: for each fixed $f\in E$, one must prove $\Vert \lambda R_{\lambda }f-f\Vert \to 0$. First, the contraction semigroup bound givesThus $\lambda R_{\lambda }$ is a contraction. This is the useful scaling of the contraction estimate in the source's hint. Since ${\int }_0^{\infty }\lambda e^{-\lambda t}dt=1$,Fix $\epsilon >0$. By strong continuity, choose $\delta >0$ so that $\Vert P_{t}f-f\Vert <\epsilon$ for $0\le t\le \delta$. Split the integral there and use $\Vert P_{t}f-f\Vert \le 2\Vert f\Vert$ for the tail:Let $\lambda \to \infty$ and then $\epsilon \downarrow 0$. ThereforeThe distinction from operator norm convergence is essential. For example, the multiplication semigroup on ${\ell }^2(\mathbb{N})$ defined by $(P_{t}f{)}_n=e^{-nt}{f}_n$ is a strongly continuous semigroup of contractions: coordinatewise convergence and dominated convergence applied to ${\sum }_n|f_n{|}^2$ prove strong continuity. Its resolvent operator is $(R_{\lambda }f{)}_n=f_n/(\lambda +n)$, andThus the claimed convergence cannot in general be read as convergence in operator norm.

For $Au=u^{\prime \prime }-(1+x^2)u$ on $D(A)=\{u\in H^2(\mathbb{R}):x^{2}u\in L^2\}$, the form $a_{\lambda }(u,v)=\int u^{\prime }\overline{v^{\prime }}+(1+\lambda +x^2)u\bar{v}$ is coercive on the one-dimensional harmonic oscillator form domain when $\lambda >0$. The Lax-Milgram theorem gives a unique weak solution of $(\lambda -A)u=g$. The graph estimate for the harmonic oscillator puts it in $D(A)$, and $a_{\lambda }(u,u)\ge (1+\lambda )\Vert u{\Vert }_2^2$ gives the displayed resolvent bound. The domain is dense and the bounded resolvent proves closedness, so the Hille-Yosida theorem gives a contraction semigroup. An exponentially shifted semigroup with generator $A+I$ solves the heat equation with oscillator potential.

A real-linear contraction on $C(X)$ that preserves $1$ is positive: for $0\le f\le 1$, $\Vert 1-Tf{\Vert }_{\infty }\le 1$ forces $Tf\ge 0$, then scale. In the complex case, a unital norm-one functional is real on real functions, since $|1+it\varphi (g)|\le \Vert 1+itg{\Vert }_{\infty }=1+O(t^2)$ for both signs of $t$. The same positivity argument applies pointwise to $\varphi (f)=Tf(x)$. This turns a unital contraction semigroup into a Feller semigroup.

For a contraction semigroup with semigroup generator $L$, the bounded operators $L_{\lambda }$ approximate $L$ on its generator domain. The Laplace-transform formula for a semigroup resolvent and strong continuity give $\lambda R_{\lambda }g\to g$ for every vector $g$: split ${\int }_0^{\infty }\lambda e^{-\lambda t}(P_{t}g-g)dt$ into a short interval and an exponentially small tail. For $f\in D(L)$, integration by parts gives $R_{\lambda }Lf=\lambda R_{\lambda }f-f$, so $L_{\lambda }f=\lambda R_{\lambda }Lf\to Lf$. No operator norm convergence of $\lambda R_{\lambda }$ is implied.