Resolvent power estimates for semigroup generators

For $\lambda >\omega$ the integrand $t\mapsto e^{-\lambda t}T(t)x$ is dominated in norm by $M{e}^{(\omega -\lambda )t}\mathrm{\parallel }x\mathrm{\parallel }$; for a fixed ${\lambda }_0\in (\omega ,\lambda )$ the tails satisfy ${\int }_R^{\mathrm{\infty }}t{e}^{(\omega -{\lambda }_0)t}dt\to 0$ as $R\to \mathrm{\infty }$, which is the uniform-in-$h$ domination used below.

The function $\mu \mapsto R(\mu ,A)x$ is differentiable on $(\omega ,\mathrm{\infty })$ with derivative $-R(\lambda ,A{)}^{2}x$: by [F2], $\frac{R(\lambda +h,A)-R(\lambda ,A)}{h}x=-R(\lambda +h,A)R(\lambda ,A)x$, and $R(\lambda +h,A)\to R(\lambda ,A)$ in operator norm as $h\to 0$ because $\mathrm{\parallel }R(\lambda +h,A)-R(\lambda ,A)\mathrm{\parallel }\le \mathrm{\mid }h\mathrm{\mid }\mathrm{\parallel }R(\lambda +h,A)\mathrm{\parallel }\mathrm{\parallel }R(\lambda ,A)\mathrm{\parallel }$ and $\mathrm{\parallel }R(\lambda +h,A)\mathrm{\parallel }$ is bounded near $\lambda$ by [F1].

The same derivative computed from the integral is $-{\int }_0^{\mathrm{\infty }}t{e}^{-\lambda t}T(t)xdt$: the difference quotient is ${\int }_0^{\mathrm{\infty }}\frac{e^{-ht}-1}{h}{e}^{-\lambda t}T(t)xdt$, whose integrands converge pointwise to $-t{e}^{-\lambda t}T(t)x$ and are dominated by $t{e}^{(\omega -{\lambda }_0)t}M\mathrm{\parallel }x\mathrm{\parallel }$ for $\mathrm{\mid }h\mathrm{\mid }$ small and ${\lambda }_0\in (\omega ,\lambda ]$; splitting the integral at $R$ and using uniform convergence on $[0,R]$ for the mean-value estimate $\mathrm{\mid }(e^{-ht}-1)\mathrm{/}h\mathrm{\mid }\le t{e}^{\mathrm{\mid }h\mathrm{\mid }t}$ and the tail estimate of [step 1.1] passes the limit through the improper integral.

Comparing [step 1.2] and [step 2.1]: $R(\lambda ,A{)}^{2}x={\int }_0^{\mathrm{\infty }}t{e}^{-\lambda t}T(t)xdt$ for every $x$, and the integral converges absolutely.

Induction on $m$ gives $R(\lambda ,A{)}^{m}x=\frac{1}{(m-1)!}{\int }_0^{\mathrm{\infty }}{s}^{m-1}{e}^{-\lambda s}T(s)xds$: the case $m=1$ is [F1] and the case $m=2$ is [step 3.1]. Assume the formula for $m$; since the resolvents commute, $\frac{d}{d\lambda }R(\lambda ,A{)}^m=-mR(\lambda ,A{)}^{m+1}$ by [step 1.2], while differentiating the integral representation in $\lambda$ (the same tail-splitting argument as [step 2.1], with domination $s^m{e}^{(\omega -{\lambda }_0)s}M\mathrm{\parallel }x\mathrm{\parallel }$ with $\omega <{\lambda }_0<\lambda$ and $\mathrm{\mid }h\mathrm{\mid }<\lambda -{\lambda }_0$) gives $-\frac{1}{(m-1)!}{\int }_0^{\mathrm{\infty }}{s}^m{e}^{-\lambda s}T(s)xds$. Equating the two expressions yields the formula for $m+1$.

Norm bound: by [F3] and $M\ge 1$, $\mathrm{\parallel }R(\lambda ,A{)}^{m}x\mathrm{\parallel }\le \frac{1}{(m-1)!}{\int }_0^{\mathrm{\infty }}{s}^{m-1}{e}^{(\omega -\lambda )s}M\mathrm{\parallel }x\mathrm{\parallel }ds=\frac{M\mathrm{\parallel }x\mathrm{\parallel }}{(\lambda -\omega {)}^m}$, the scalar integral is $(m-1)!\mathrm{/}(\lambda -\omega {)}^m$: integration by parts on $[0,R]$ gives $I_m=(m-1)I_{m-1}\mathrm{/}(\lambda -\omega )$ after $R\to \mathrm{\infty }$, with $I_1=1\mathrm{/}(\lambda -\omega )$ and vanishing polynomial-exponential boundary terms. For all nearby difference quotients use a strictly smaller parameter ${\lambda }_0\in (\omega ,\lambda )$; polynomial times $e^{-({\lambda }_0-\omega )s}$ is integrable by the same recurrence. Thus taking the supremum over $\mathrm{\parallel }x\mathrm{\parallel }\le 1$ gives $\mathrm{\parallel }R(\lambda ,A{)}^m\mathrm{\parallel }\le M\mathrm{/}(\lambda -\omega {)}^m$.