Abstract parabolic smoothing for mild solutions

Homogeneous term with an arbitrary vertex. Let $\omega$ be a sectorial vertex for $A$ and put $B:=A-\omega I$. Since $R(\lambda ,B)=R(\lambda +\omega ,A)$, $B$ is sectorial with vertex $0$. The semigroup $S(t):=e^{-\omega t}T(t)$ has generator $B$ on $D(A)$: its difference quotient converges exactly when that of $T$ does, since $(S(h)y-y)\mathrm{/}h=e^{-\omega h}(T(h)y-y)\mathrm{/}h+(e^{-\omega h}-1)y\mathrm{/}h$. It is exponentially bounded by [L1], hence equals the contour semigroup of $B$ by [L1]. Induction using $B=A-\omega I$ gives $D(B^k)=D(A^k)$ and $A^k=(B+\omega I{)}^k={\sum }_{j=0}^k\left(\genfrac{}{}{0px}{}{k}{j}\right){\omega }^{k-j}{B}^j$ on this common domain: in the induction step, the lower powers $B^{j}y$ for $j<k$ already lie in $D(B)=D(A)$, so $A^{k}y\in D(A)$ is equivalent to $B^{k}y\in D(B)$. Put $K_0:={\sup }_{t\ge 0}\mathrm{\parallel }S(t)\mathrm{\parallel }$. For $0<t\le b$, [L1] now gives $T(t)X\subseteq D(A^m)$ and $\mathrm{\parallel }{A}^{m}T(t)\mathrm{\parallel }\le e^{\max \{\omega ,0\}b}{\sum }_{j=0}^m\left(\genfrac{}{}{0px}{}{m}{j}\right)\mathrm{\mid }\omega {\mathrm{\mid }}^{m-j}{K}_j{b}^{m-j}{t}^{-m}=:C_m{t}^{-m}$. In particular $c_0,c_1$ are finite. Differentiating $T(t)=e^{\omega t}S(t)$ gives $T^{(m)}(t)=A^{m}T(t)$, continuous in operator norm for $t>0$. This $C_m$ is a finite-interval smoothing constant; no global $t^{-m}$ bound for a nonzero vertex is asserted. Writing $u=T(\cdot )x+v$, with $v(t):={\int }_0^{t}T(t-s)f(s)ds$, reduces the remaining membership, continuity and estimate to those for $A^{m}v$.

The curves $h_j$ and their images. The graph norm on $D(A^{m-1})$ dominates $\mathrm{\parallel }\cdot \mathrm{\parallel }$ and $\mathrm{\parallel }{A}^j\cdot \mathrm{\parallel }$ for every $0\le j\le m-1$, so each $h_j:=A^{j}f$ is a continuous $X$-valued curve on $[0,b]$; the curve $g=h_{m-1}$ satisfies $\mathrm{\parallel }g(0)\mathrm{\parallel }\le \mathrm{\parallel }f(0){\mathrm{\parallel }}_{D(A^{m-1})}$ and, for $m\ge 2$, is differentiable in $X$ with $g^{\mathrm{\prime }}=A^{m-1}{f}^{\mathrm{\prime }}$ bounded, hence Lipschitz and α-Hölder, while for $m=1$ it equals $f$ and is α-Hölder by hypothesis; thus $g\in C^{\alpha }([0,b],X)$ with $\mathrm{\parallel }g(0)\mathrm{\parallel }$ and $[g{]}_{\alpha }$ controlled by $\mathrm{\parallel }f{\mathrm{\parallel }}_{C^{m-1,\alpha }([0,b],D(A^{m-1}))}$. Fix $0<t\le b$ and $0\le j\le m-2$ and put $u_j(s):=T(t-s)h_j(s)$: by strong continuity of $T$ from [L3] and continuity of $h_j$ the curve $u_j$ is continuous on $[0,t]$, and for every $s\in [0,t]$ one has $h_j(s)\in D(A)$, $u_j(s)\in D(A)$ and $A{u}_j(s)=T(t-s)h_{j+1}(s)=u_{j+1}(s)$ by [L2].

Domain induction by closedness. The claim is that for every $0\le j\le m-1$ one has $v(t)\in D(A^j)$ with $A^{j}v(t)=V_j(t):={\int }_0^{t}T(t-s)h_j(s)ds$; the case $j=0$ is the definition of $v$. Assume the claim for some $j\le m-2$ and take right-endpoint Riemann sums $S_n$ of the continuous curve $u_j$ along partitions of $[0,t]$ with mesh tending to $0$: then $S_n\to V_j(t)=A^{j}v(t)$, while each $S_n$ lies in $D(A)$ and $A{S}_n$ is the corresponding Riemann sum of $u_{j+1}$, so $A{S}_n\to V_{j+1}(t)$ by [step 1.2]; since $A$ is closed by [L3], $V_j(t)\in D(A)$ and $A{V}_j(t)=V_{j+1}(t)$, that is $v(t)\in D(A^{j+1})$ and $A^{j+1}v(t)=V_{j+1}(t)$. Induction up to $j=m-1$ gives $v(t)\in D(A^{m-1})$ and $A^{m-1}v(t)=w(t):={\int }_0^{t}T(t-s)g(s)ds$, which is exactly the function $U$ of [L4] with $x_0=0$ and $h=g$.

Classical regularity of $w$. By [step 1.2] the forcing $g$ lies in $C^{\alpha }([0,b],X)$, so [L4] applied to $x_0=0$ and $h=g$ makes $w$ a classical solution with $w(t)\in D(A)$ for every $t\in [0,b]$, $Aw\in C([0,b],X)$ and $Aw(0)=0$, and [L5] gives $Aw(t)=(T(t)-I)g(0)+R_g(t)$ with $\mathrm{\parallel }{R}_g(t)\mathrm{\parallel }\le (\frac{c_1}{\alpha }+c_0+1)[g{]}_{\alpha }{t}^{\alpha }$ for $0<t\le b$; since $A^{m-1}v(t)=w(t)$ by [step 2.1], the recursive definition of $D(A^m)$ yields $v(t)\in D(A^m)$ with $A^{m}v(t)=Aw(t)$.

Final estimate and continuity. Adding the homogeneous bound of [step 1.1] to the bound of [step 3.1] gives, for every $0<t\le b$, $\mathrm{\parallel }{A}^{m}u(t)\mathrm{\parallel }\le C_m{t}^{-m}\mathrm{\parallel }x\mathrm{\parallel }+(c_0+1)\mathrm{\parallel }g(0)\mathrm{\parallel }+(\frac{c_1}{\alpha }+c_0+1)[g{]}_{\alpha }{t}^{\alpha }$, and $A^{m}u$ is continuous on $(0,b]$ because both $A^{m}T(\cdot )x$ and $Aw$ are continuous there by [step 1.1] and [step 3.1]; since $t\le b$ and the norms of $g(0)$ and $[g{]}_{\alpha }$ are controlled by [step 1.2], this gives $\mathrm{\parallel }{A}^{m}u(t)\mathrm{\parallel }\le C_m^{\mathrm{\prime }}{t}^{-m}(\mathrm{\parallel }x\mathrm{\parallel }+\mathrm{\parallel }f{\mathrm{\parallel }}_{C^{m-1,\alpha }([0,b],D(A^{m-1}))})$ with $C_m^{\mathrm{\prime }}$ depending only on $m,\alpha ,b$ and the semigroup bounds. For $m=1$ the same argument runs with the single curve $h_0=f$ and the vacuous tower, and the splitting of [step 1.1] is what removes every requirement on $x\in X$; no choice principle beyond Dependent Choice is used.