Laplace resolvents and full generator domains

The generator of a uniformly bounded real-parameter action can be unbounded and only weak-star closed. Its resolvents still exist throughout both open half-planes, given by forward and backward Laplace integrals. This lesson proves Takesaki II, Lemma XI.1.19 with both inverse identities on the correct domains. The backward formula needs a minus sign under the convention of the earlier generator proof.

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The earlier proofs are BS0–1 for specified preadjoints and complete integrated maps, L24 Proposition4.1 for Banach-valued integration, CF1 for oriented fundamental calculus, and the earlier generator's full domain, closed graph and invariance. Here $dt$ is Lebesgue measure with $\int_0^s1\,dt=s$ as an oriented integral.

Scalar twists shift the generator

Retain $X=X_*^*$, the uniformly bounded weak-star continuous action $\alpha:\mathbb R\to\operatorname{GL}(X)$, and its weak-star generator $\delta$ from lesson 94. For real $b$, define $\alpha_t^{(b)}=e^{ibt}\alpha_t$. This is another action with the same uniform bound and predual continuity. The scalar product rule in (G2) gives

$D(\delta^{(b)})=D(\delta),\qquad \delta^{(b)}x=\delta x+ibx.\tag{L1}$

Thus an imaginary shift of the spectral parameter can be absorbed into an action. We will prove the stronger direct claim that every complex $\lambda$ with nonzero real part is resolvent, without first restricting $\lambda$ to the real axis.

The forward Laplace operator

For $\operatorname{Re}\lambda>0$, define, through the specified predual,

$R_\lambda^+x=\int_0^\infty e^{-\lambda t}\alpha_t x\,dt, \qquad\|R_\lambda^+\|\le\frac{C_\alpha}{\operatorname{Re}\lambda}.\tag{L2}$

The integral is a weak-star integral: on $X_*$, the norm-convergent Bochner integral $\int_0^\infty e^{-\lambda t}\beta_t\phi\,dt$ is its preadjoint. Indeed $1_{[0,\infty)}(t)e^{-\lambda t}$ is in $L^1(\mathbb R)$, and the predual orbit is norm continuous and bounded by $C_\alpha\|\phi\|$; the complete integral proof of BS1/L24 applies. Its norm bound is the integral of $C_\alpha e^{-\operatorname{Re}\lambda t}$. BS0 proves that its adjoint is normal, with the same norm. For real $s$, with oriented integration when $s<0$, the group law yields

$\alpha_sR_\lambda^+x =e^{\lambda s}R_\lambda^+x -e^{\lambda s}\int_0^s e^{-\lambda u}\alpha_u x\,du.\tag{L3}$

Take the weak-star derivative at $s=0$. The last integral divided by $s$ tends weak-star to $x$, while the first term has derivative $\lambda R_\lambda^+x$. Hence $R_\lambda^+x\in D(\delta)$ and

$(\lambda I-\delta)R_\lambda^+x=x\qquad(x\in X).\tag{L4}$

For the other inverse identity, let $x\in D(\delta)$. Domain invariance (G10) and scalar differentiation give

$\frac{d}{dt}\big(e^{-\lambda t}\alpha_t x\big) =-e^{-\lambda t}\alpha_t(\lambda I-\delta)x \quad\text{in the weak-star sense}.\tag{L5}$

Pair with an arbitrary predual functional and integrate from zero to $T$. The boundary term $e^{-\lambda T}\alpha_Tx$ tends to zero in norm by uniform boundedness and $\operatorname{Re}\lambda>0$. Letting $T\to\infty$ gives

$R_\lambda^+(\lambda I-\delta)x=x\qquad(x\in D(\delta)).\tag{L6}$

Equations (L4) and (L6) show that $\lambda I-\delta:D(\delta)\to X$ is bijective with bounded, normal, everywhere-defined inverse $R_\lambda^+$.

The backward Laplace operator and its sign

For $\operatorname{Re}\lambda<0$, apply the preceding argument to the reversed action $t\mapsto\alpha_{-t}$, whose generator is $-\delta$, with positive parameter $-\lambda$. The inverse of $\delta-\lambda I$ is $\int_0^\infty e^{\lambda t}\alpha_{-t}\,dt$. Therefore the inverse of $\lambda I-\delta$ is its negative:

$R_\lambda^-x :=-\int_0^\infty e^{\lambda t}\alpha_{-t}x\,dt, \qquad\|R_\lambda^-\|\le\frac{C_\alpha}{|\operatorname{Re}\lambda|}.\tag{L7}$

It is a normal weak-star integral and satisfies the two domain-sensitive identities

$(\lambda I-\delta)R_\lambda^-x=x\quad(x\in X), \qquad R_\lambda^-(\lambda I-\delta)x=x\quad(x\in D(\delta)).\tag{L8}$

The initial minus sign in (L7) is necessary. For the trivial action $\alpha_t=I$, the generator is $\delta=0$; then the integral without the minus sign equals $-1/\lambda$, whereas the inverse of $\lambda I$ is $1/\lambda$. This small model detects the sign before any unbounded-domain argument is used.

The spectrum lies on the imaginary axis

For a closed operator, $\lambda$ belongs to the resolvent set when $\lambda I-\delta:D(\delta)\to X$ has a bounded everywhere-defined inverse. Lessons 94 and (L2)–(L8) establish precisely that for every $\lambda$ off the imaginary axis. Thus

$\operatorname{Sp}(\delta)\subset i\mathbb R.\tag{L9}$

The estimates also give $\|(\lambda I-\delta)^{-1}\|\le C_\alpha/|\operatorname{Re}\lambda|$ on both half-planes. They do not assert that every point of the imaginary axis is spectral, nor that the generator is bounded or norm densely defined on $X$.

Problem. Let $\alpha_tz=e^{i\omega t}z$ on $\mathbb C$. Compute $R_\lambda^+$ and $R_\lambda^-$ and compare them with the algebraic inverse.

Solution. The generator is multiplication by $i\omega$. For $\operatorname{Re}\lambda>0$, $\int_0^\infty e^{-(\lambda-i\omega)t}dt=(\lambda-i\omega)^{-1}$. For $\operatorname{Re}\lambda<0$, $-\int_0^\infty e^{(\lambda-i\omega)t}dt=(\lambda-i\omega)^{-1}$ as well. Both agree with $(\lambda I-\delta)^{-1}$. $\square$

The mathematical source is M. Takesaki, Theory of Operator Algebras II, Lemma XI.1.19, printed pages 325–326 (edition record). The backward display on page326 needs the minus sign for the convention $(\lambda I-\delta)^{-1}$; the trivial-action test and the full reversed-action proof (L7)–(L8) establish it. The proof above retains both inverse identities at their exact domains and supplies all integral and earlier generator inputs.

Resolvents on an ordinary Banach space

Let $E,T,A$ be the norm-continuous-orbit group and full norm generator from the norm-domain theorem, with $C=\sup_t\|T_t\|<\infty$. For $\operatorname{Re}\lambda>0$, the vector integral $Q_\lambda x=\int_0^\infty e^{-\lambda t}T_tx\,dt,\qquad \|Q_\lambda\|\leq\frac{C}{\operatorname{Re}\lambda} \tag{NR1}$ exists in $E$ by the complete Banach integration theorem. Linearity and its norm bound make $Q_\lambda$ a bounded operator on all of $E$.

The translated integral satisfies, for real $h$, $T_hQ_\lambda x =e^{\lambda h}Q_\lambda x -e^{\lambda h}\int_0^h e^{-\lambda u}T_ux\,du . \tag{NR2}$ This follows by the scalar change of variable $u=t+h$, with oriented integration if $h<0$; bounded maps pass through the vector integral. The last integral divided by $h$ converges to $x$ in norm by orbit continuity. Taking the norm derivative at zero gives $Q_\lambda E\subset D(A),\qquad (\lambda I-A)Q_\lambda x=x\quad(x\in E). \tag{NR3}$ For $x\in D(A)$, (NG3) gives $\frac{d}{dt}\bigl(e^{-\lambda t}T_tx\bigr) =-e^{-\lambda t}T_t(\lambda I-A)x \quad\text{in norm}. \tag{NR4}$ Integrating to $b>0$, then letting $b\to\infty$, is justified by $\|e^{-\lambda b}T_bx\|\leq C e^{-b\operatorname{Re}\lambda}\|x\|\to0$. It proves $Q_\lambda(\lambda I-A)x=x$. Thus both inverse identities hold, on all of $E$ and on the entire domain $D(A)$, respectively.

For $\operatorname{Re}\lambda<0$, apply this proof to $t\mapsto T_{-t}$, whose norm generator is $-A$ on the same domain, and parameter $-\lambda$. The result is $(\lambda I-A)^{-1}x =-\int_0^\infty e^{\lambda t}T_{-t}x\,dt,\qquad \|(\lambda I-A)^{-1}\|\leq\frac{C}{|\operatorname{Re}\lambda|}. \tag{NR5}$ Together with (NR1)–(NR4), this proves the resolvent assertion on both open half-planes and $\sigma(A)\subset i\mathbb R$, now for an ordinary Banach space with its norm-closed, norm-dense generator domain.

Scalar twists have the same full norm domain. For $b\in\mathbb R$, put $S_t=e^{ibt}T_t$. For $x\in D(A)$, $\frac{S_hx-x}{h} =e^{ibh}\frac{T_hx-x}{h}+\frac{e^{ibh}-1}{h}x \longrightarrow Ax+ibx .$ Conversely, the same identity applied to $T_t=e^{-ibt}S_t$ shows that existence of the norm derivative for $S$ implies $x\in D(A)$. Its generator is therefore $A+ibI$ on exactly $D(A)$, and $(\lambda I-(A+ibI))^{-1} =((\lambda-ib)I-A)^{-1}\qquad(\operatorname{Re}\lambda\ne0).$ Both sides act on all of $E$ with range $D(A)$, by the inverse identities just proved.

In particular, if $T_t$ are isometries, then for every nonzero real $s$, $(I-sA)^{-1}x=\int_0^\infty e^{-u}T_{su}x\,du,\qquad \|(I-sA)^{-1}\|\leq1 . \tag{NR6}$ For $s>0$ this is (NR1) with $\lambda=1/s$ and a positive scaling; for $s<0$ it is (NR5) with the same parameter and the corresponding backward scaling. The inverse has range exactly $D(A)$, by the two inverse identities. No weak-star compactness is used.

For the $c_0$ model (NG7), these formulas give $((\lambda I-A)^{-1}x)_n=\frac{x_n}{\lambda-in}. \tag{NR7}$ This sequence is in $c_0$, since its absolute value is at most $|x_n|/|\operatorname{Re}\lambda|$. It lies in the full domain (NG8) because $n/(\lambda-in)=i(1-\lambda/(\lambda-in))$ is bounded and $x_n\to0$. Thus the coordinate formula verifies the claimed range, as well as the inverse equation.

The source comparison is Takesaki, Theory of Operator Algebras II, XI.1, p.312 and Lemma1.19, pp.325–326. The norm argument above retains both time signs and both domain-sensitive inverse identities.