Classical regularity for Holder-continuous forcing under initial compatibility

Finite-interval smoothing. Choose a sectorial vertex $\omega$ for $A$ and set $B=A-\omega I$ on $D(A)$. The identity $R(\lambda ,B)=R(\lambda +\omega ,A)$ makes $B$ sectorial with vertex $0$. The strongly continuous semigroup $S(t)=e^{-\omega t}T(t)$ has generator $B$ on exactly $D(A)$, because $(S(h)y-y)\mathrm{/}h=e^{-\omega h}(T(h)y-y)\mathrm{/}h+(e^{-\omega h}-1)y\mathrm{/}h$. By [L4] it is exponentially bounded and equals the contour semigroup of $B$. Writing $K_0={\sup }_{t\ge 0}\mathrm{\parallel }S(t)\mathrm{\parallel }$ and using $AT(t)=e^{\omega t}(BS(t)+\omega S(t))$ gives $T(t)X\subseteq D(A)$ for $t>0$, $c_0\le e^{\max \{\omega ,0\}b}{K}_0$ and $c_1\le e^{\max \{\omega ,0\}b}(K_1+\mathrm{\mid }\omega \mathrm{\mid }b{K}_0)<\mathrm{\infty }$. These finite-interval bounds meet the Duhamel cancellation hypotheses and supply the positive-time domain inclusion for the given vertex.

The Duhamel data. By [L1] the function $v$ is continuous in the graph norm and $Av$ is continuous on $[0,b]$ with $Av(0)=0$, so $s\mapsto Av(s)+f(s)$ is a continuous $X$-valued curve; moreover by [L1] the decomposition $Av=A{v}_1+A{v}_2$ holds with $\mathrm{\parallel }A{v}_1(t)\mathrm{\parallel }\le \frac{c_1}{\alpha }[f{]}_{\alpha }{t}^{\alpha }$ and $\mathrm{\parallel }A{v}_2(t)\mathrm{\parallel }\le (c_0+1)\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$, and $\mathrm{\parallel }(T(t)-I)(f(t)-f(0))\mathrm{\parallel }\le (c_0+1)[f{]}_{\alpha }{t}^{\alpha }$.

The integral identity. By [L2] one has $v(t)=A{\int }_0^{t}v(s)ds+{\int }_0^{t}f(s)ds$; since $v$ and $Av$ are continuous on $[0,b]$, the graph-norm integral ${\int }_0^{t}v(s)ds$ lies in $D(A)$ with $A{\int }_0^{t}v(s)ds={\int }_0^{t}Av(s)ds$, because $A$ is closed and the Riemann sums of the $D(A)$-valued continuous curve $s\mapsto v(s)$ converge in the graph norm. Hence $v(t)={\int }_0^t(Av(s)+f(s))ds$ with a continuous integrand.

Classicality. The fundamental theorem [L3] applied to the continuous curve $s\mapsto Av(s)+f(s)$ shows $v\in C^1([0,b],X)$ with $v^{\mathrm{\prime }}=Av+f$ and $v(0)=0$. For $t>0$ the orbit $T(t)x$ has derivative $AT(t)x=T(t)Ax$ by [L2] and this derivative extends continuously to $0$ with value $Ax$ because $x\in D(A)$ and $T$ is strongly continuous; hence $u=T(\cdot )x+v\in C^1([0,b],X)$ with $u(0)=x$ and $u^{\mathrm{\prime }}(t)=AT(t)x+Av(t)+f(t)=Au(t)+f(t)$ for every $t\in [0,b]$, and $u(t)\in D(A)$ because both $T(t)x$ and $v(t)$ lie in the domain. Thus $u$ is a classical solution in the sense of the cited definition, and $Au\in C([0,b],X)$ with $Au(0)=Ax$.

Endpoint identity, modulus and caveat. Subtracting $Ax$ from $Au(t)=AT(t)x+A{v}_1(t)+A{v}_2(t)$ and writing $A{v}_2(t)=(T(t)-I)f(t)=(T(t)-I)f(0)+(T(t)-I)(f(t)-f(0))$ gives $Au(t)-Ax=(T(t)-I)(Ax+f(0))+R_f(t)$ with $R_f$ as displayed, and the bounds of [step 2.1] give $\mathrm{\parallel }{R}_f(t)\mathrm{\parallel }\le (\frac{c_1}{\alpha }+c_0+1)[f{]}_{\alpha }{t}^{\alpha }$. The term $(T(t)-I)(Ax+f(0))$ tends to $0$ by strong continuity but admits no uniform power modulus as stated, so the hypotheses give continuity and classicality of $u$ but not Hölder continuity of $Au$ in terms of $[f{]}_{\alpha }$ alone. The argument used only the Duhamel cancellation, the variation-of-constants identity and the fundamental theorem, so no choice principle beyond Dependent Choice was used.