Smoothing estimates for the semigroup generated by a sectorial operator

The differentiated contour formula. For $t>0$ and every $k\ge 0$ the integral $\frac{1}{2\pi i}{\int }_{\mathrm{\Gamma }}{\lambda }^k{e}^{\lambda t}R(\lambda ,A)d\lambda$ converges absolutely, because on the rays $\mathrm{\parallel }{\lambda }^k{e}^{\lambda t}R(\lambda ,A)\mathrm{\parallel }\le M_{\epsilon }{s}^{k-1}{e}^{-cst}$ with $c>0$ (the admissible angle gives $\mathrm{Re}(\lambda t)\le -cst$) and ${\int }_r^{\mathrm{\infty }}{s}^{k-1}{e}^{-cst}ds<\mathrm{\infty }$ for $t>0$; differentiating $k$ times under the integral sign is justified by the same integrable majorant on compact time intervals bounded away from $0$, so $\frac{d^k}{d{t}^k}T(t)=\frac{1}{2\pi i}{\int }_{\mathrm{\Gamma }}{\lambda }^k{e}^{\lambda t}R(\lambda ,A)d\lambda$ and $T\in C^{\mathrm{\infty }}((0,\mathrm{\infty }),\mathcal{B}(X))$ in operator norm.

The truncated integral lies in the domain. For $k\ge 0$ the truncated integral $I_{k,R}:=\frac{1}{2\pi i}{\int }_{{\mathrm{\Gamma }}_R}{\lambda }^k{e}^{\lambda t}R(\lambda ,A)d\lambda$ is the norm limit of Riemann sums of elements of $D(A)$, and by [L2] $A$ of each such sum equals the corresponding sum of ${\lambda }^{k+1}{e}^{\lambda t}R(\lambda ,A)-{\lambda }^k{e}^{\lambda t}I$; since $A$ is closed, the limit $I_{k,R}$ lies in $D(A)$ with $A{I}_{k,R}=\frac{1}{2\pi i}{\int }_{{\mathrm{\Gamma }}_R}{\lambda }^{k+1}{e}^{\lambda t}R(\lambda ,A)d\lambda -\frac{1}{2\pi i}\left({\int }_{{\mathrm{\Gamma }}_R}{\lambda }^k{e}^{\lambda t}d\lambda \right)I$.

The scalar contour integrals vanish. For every $k\ge 0$ one has ${\int }_{\mathrm{\Gamma }}{\lambda }^k{e}^{\lambda t}d\lambda =0$: close the truncated contour by the arc at radius $R$ through the left, on which $\mathrm{Re}(\lambda t)\le -cRt$; the integral of the entire function ${\lambda }^k{e}^{\lambda t}$ over the closed truncated curve vanishes (it has the global primitive ${\int }_0^{\lambda }{w}^k{e}^{wt}dw$), while the closing arc contribution is at most $2\pi R^{k+1}{e}^{-cRt}\to 0$.

First order: $AT(t)=T^{\mathrm{\prime }}(t)$. Apply [step 1.2] with $k=0$: the truncated integral $I_{0,R}$ converges in operator norm to $T(t)$, while its $A$-image is $I_{1,R}-\frac{1}{2\pi i}({\int }_{{\mathrm{\Gamma }}_R}{e}^{\lambda t}d\lambda )I$. The first term converges in operator norm to $T^{\mathrm{\prime }}(t)$ by [step 1.1], and the scalar term tends to zero by [step 1.3]. Since $A$ is closed by [L2], for each $x\in X$ the convergence of $I_{0,R}x$ and $A{I}_{0,R}x$ gives $T(t)x\in D(A)$ and $AT(t)x=T^{\mathrm{\prime }}(t)x$. Thus $T(t)X\subseteq D(A)$ and $AT(t)=T^{\mathrm{\prime }}(t)$.

Higher orders by the same argument. Suppose $T(t)X\subseteq D(A^m)$ and $A^{m}T(t)=T^{(m)}(t)$. Apply [step 1.2] with $k=m$ to the truncated contour integral $I_{m,R}$ for $T^{(m)}$: its $A$-image is $I_{m+1,R}$ minus the scalar term in [step 1.2]. As $R\to \mathrm{\infty }$, $I_{m,R}\to T^{(m)}(t)=A^{m}T(t)$ and $I_{m+1,R}\to T^{(m+1)}(t)$ in operator norm by [step 1.1], while the scalar term tends to zero by [step 1.3]. Closedness of $A$ then gives $A^{m+1}T(t)x=T^{(m+1)}(t)x$ for every $x\in X$, and $T(t)X\subseteq D(A^{m+1})$. Induction proves the claim for every $m\ge 1$.

The bound. By the independence of the inner radius [L1], compute $T^{(m)}(t)$ with $r=1\mathrm{/}t$. On the rays, $\mathrm{\parallel }{\lambda }^m{e}^{\lambda t}R(\lambda ,A)\mathrm{\parallel }\le M_{\epsilon }{s}^{m-1}{e}^{-cst}$ and ${\int }_{1\mathrm{/}t}^{\mathrm{\infty }}{s}^{m-1}{e}^{-cst}ds=t^{-m}{\int }_1^{\mathrm{\infty }}{\sigma }^{m-1}{e}^{-c\sigma }d\sigma$ after $\sigma =st$; on the arc $\mathrm{\mid }\lambda \mathrm{\mid }=1\mathrm{/}t$ one has $\mathrm{\mid }\lambda {\mathrm{\mid }}^m=t^{-m}$, length at most $2\theta \mathrm{/}t$, $\mathrm{\mid }{e}^{\lambda t}\mathrm{\mid }\le e$ and $\mathrm{\parallel }R(\lambda ,A)\mathrm{\parallel }\le M_{\epsilon }t$, so the arc contributes at most $2\theta e{M}_{\epsilon }{t}^{-m}$. Hence $\mathrm{\parallel }{A}^{m}T(t)\mathrm{\parallel }=\mathrm{\parallel }{T}^{(m)}(t)\mathrm{\parallel }\le C_m{t}^{-m}$ with $C_m$ depending only on $m,\delta ,M_{\delta \mathrm{/}4}$ through the prescribed $\theta =\pi \mathrm{/}2+\delta \mathrm{/}2$ and $c=-\cos \theta$, and [step 3.1] supplies the domain membership and the derivative identity for every $m$.