Laplace resolvents of a unitary group

For either sign put $F_x(t)=e^{-\lambda t}U(\pm t)x$ on $[0,\mathrm{\infty })$. It is continuous. For integer $n\ge 1$, approximate it on $[0,n)$ by its values at the left endpoints of intervals of length $2^{-n}$ and put zero elsewhere. These are measurable simple functions with finite support measure, converging at every fixed $t\ge 0$ to $F_x(t)$ by continuity. Thus $F_x$ is strongly measurable. Compact Newton--Leibniz with primitive $-e^{-\lambda t}\mathrm{/}\lambda$ and the bridge in [A4] give ${\int }_0^N{e}^{-\lambda t}dt=(1-e^{-\lambda N})\mathrm{/}\lambda$. Monotone convergence as integer $N\to \mathrm{\infty }$ gives ${\int }_0^{\mathrm{\infty }}{e}^{-\lambda t}dt=1\mathrm{/}\lambda$. Since $\mathrm{\parallel }{F}_x(t)\mathrm{\parallel }=e^{-\lambda t}\mathrm{\parallel }x\mathrm{\parallel }$, [A1] defines $Q_{\pm }x$ and gives $\mathrm{\parallel }{Q}_{\pm }x\mathrm{\parallel }\le \mathrm{\parallel }x\mathrm{\parallel }\mathrm{/}\lambda$. Linearity follows first for simple integrals and then by adding their approximating sequences in integral norm. Therefore these are bounded linear operators.

For $x\in D(G)$ write $v_h=(U(h)x-x)\mathrm{/}h\to Gx$ for real $h\mathrm{\ne }0$. Norm preservation and the inner-product expansion give $0=2\mathrm{Re}⟨v_h,x⟩+h\mathrm{\parallel }{v}_h{\mathrm{\parallel }}^2$. Since a convergent family is bounded near zero, the limit gives $\mathrm{Re}⟨Gx,x⟩=0$. Hence $\mathrm{Re}⟨(\lambda \mp G)x,x⟩=\lambda \mathrm{\parallel }x{\mathrm{\parallel }}^2$. If $(\lambda \mp G)x=0$, positivity of $\lambda$ gives $x=0$; both shifts are injective. This argument requires no density or closedness theorem for $G$.

Translation of a Bochner integral is legitimate here: for simple integrable functions it follows termwise from [A5]; for their integral-norm limits the scalar change-of-variables identity follows first for nonnegative simple functions, then by monotone convergence, and shows that translation preserves the approximation error. Thus the simple identities pass to the Bochner integral by [A1]. Subdivision and linearity follow in the same way from simple integrals. Also $\mathrm{\parallel }{h}^{-1}{\int }_0^h{F}_x(s)ds-x\mathrm{\parallel }\le {\sup }_{0\le s\le h}\mathrm{\parallel }{F}_x(s)-x\mathrm{\parallel }\to 0$ as $h\downarrow 0$. The tail ${\int }_h^{\mathrm{\infty }}{F}_x$ tends to $Q_{\pm }x$, since the norm of the omitted integral is at most $h\mathrm{\parallel }x\mathrm{\parallel }$.

For the plus sign and $h>0$, commuting $U(h)$ with integration and translating gives $\frac{U(h)-I}{h}{Q}_{+}x=\frac{e^{\lambda h}-1}{h}{\int }_h^{\mathrm{\infty }}{e}^{-\lambda s}U(s)xds-\frac{1}{h}{\int }_0^h{e}^{-\lambda s}U(s)xds\to \lambda Q_{+}x-x$ by step 2.1. To obtain the required two-sided derivative, if $y\in H$ has right quotient $v_h=(U(h)y-y)\mathrm{/}h\to v$, then $\frac{U(-h)y-y}{-h}=U(-h)v_h\to v$: its error is bounded by $\mathrm{\parallel }{v}_h-v\mathrm{\parallel }+\mathrm{\parallel }U(-h)v-v\mathrm{\parallel }$. Thus $Q_{+}x\in D(G)$ and $(\lambda -G)Q_{+}x=x$. For $V(t)=U(-t)$, substitution $s=-t$ in the two-sided derivative definition gives $D(G_V)=D(G)$ and $G_V=-G$. Applying the proved plus-sign argument to $V$ gives $Q_{-}x\in D(G)$ and $(\lambda +G)Q_{-}x=x$.

For $x\in D(G)$ let $y=Q_{\pm }(\lambda \mp G)x$. Step 3.1 places $y$ in $D(G)$ and gives $(\lambda \mp G)y=(\lambda \mp G)x$. Injectivity from step 1.2 implies $y=x$, proving $Q_{\pm }(\lambda \mp G)=I$ on $D(G)$. Together with step 3.1 this shows $\mathrm{ran}{Q}_{\pm }=D(G)$ and both inverse identities with their stated domains.

From $(\lambda +G)Q_{-}=I$ obtain $G{Q}_{-}=I-\lambda Q_{-}$ and $(\lambda -G)Q_{-}=2\lambda Q_{-}-I$. Multiplication on the left by $Q_{+}$ is legitimate on every vector because $\mathrm{ran}{Q}_{-}\subseteq D(G)$. Step 4.1 gives $Q_{-}=2\lambda Q_{+}{Q}_{-}-Q_{+}$, hence $Q_{+}+Q_{-}=2\lambda Q_{+}{Q}_{-}$.

The norm bound, range and inverse claims are steps 1.1, 3.1 and 4.1, and the sum identity is step 5.1. For $H=\{0\}$ the same formulas directly concern its unique full-domain operator; zero vectors give zero integrals. The strict condition $\lambda >0$ ensures integrability and injectivity. Countable Choice supplies the compact integration bridge and the Bochner framework; the declared Dependent Choice is not additionally needed by this proof. No half-line fundamental theorem for a merely bounded derivative is invoked.