Analytic Duhamel cancellation removes the generator singularity

Truncated first term. Fix $0<t\le b$ and $0<\epsilon <t$, and set $v_{1,\epsilon }(t):={\int }_0^{t-\epsilon }T(t-s)(f(s)-f(t))ds$. For $s\le t-\epsilon$ the semigroup law gives $T(t-s)=T(\epsilon )T(t-s-\epsilon )$, so the integrand lies in $D(A)$ and, since $AT(\epsilon )$ is bounded by [L1], Riemann sums and the norm inequality [L3] give $v_{1,\epsilon }(t)\in D(A)$ and $A{v}_{1,\epsilon }(t)={\int }_0^{t-\epsilon }AT(t-s)(f(s)-f(t))ds$.

The constant-endpoint term. Since $f(t)$ does not depend on the integration variable, $v_2(t)={\int }_0^{t}T(\tau )f(t)d\tau =J_{t}f(t)$, so [L2] gives $v_2(t)\in D(A)$ and $A{v}_2(t)=T(t)f(t)-f(t)$, whence $\mathrm{\parallel }A{v}_2(t)\mathrm{\parallel }\le (c_0+1)\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$.

The truncated first term is Cauchy in the graph norm. For $0<\epsilon <\eta <t$ the difference of the truncated $A$-images is the integral over $[t-\eta ,t-\epsilon ]$ of $AT(t-s)(f(s)-f(t))$, whose norm is at most $c_1[f{]}_{\alpha }(t-s{)}^{\alpha -1}$ by [L1] and Hölder continuity of $f$; integrating gives $\mathrm{\parallel }A{v}_{1,\epsilon }(t)-A{v}_{1,\eta }(t)\mathrm{\parallel }\le \frac{c_1}{\alpha }[f{]}_{\alpha }({\eta }^{\alpha }-{\epsilon }^{\alpha })\to 0$ as $\epsilon ,\eta \downarrow 0$. Likewise $\mathrm{\parallel }{v}_{1,\epsilon }(t)-v_{1,\eta }(t)\mathrm{\parallel }\le c_0[f{]}_{\alpha }{\int }_{t-\eta }^{t-\epsilon }(t-s{)}^{\alpha }ds\to 0$, so both $v_{1,\epsilon }(t)$ and $A{v}_{1,\epsilon }(t)$ converge; since $A$ is closed, $v_1(t)\in D(A)$ and $A{v}_1(t)={\int }_0^{t}AT(t-s)(f(s)-f(t))ds$, with $\mathrm{\parallel }A{v}_1(t)\mathrm{\parallel }\le \frac{c_1}{\alpha }[f{]}_{\alpha }{t}^{\alpha }$.

Continuity in the graph norm. The bounds just obtained give $\mathrm{\parallel }{v}_1(t)\mathrm{\parallel }\le c_0[f{]}_{\alpha }{t}^{1+\alpha }\mathrm{/}(1+\alpha )$ and $\mathrm{\parallel }A{v}_1(t)\mathrm{\parallel }\le \frac{c_1}{\alpha }[f{]}_{\alpha }{t}^{\alpha }$, so $v_1$ and $A{v}_1$ extend continuously to $t=0$ with value $0$; on every $[a,b]$ with $a>0$, the truncated expressions are continuous for $0<\epsilon <a$, and their tails are bounded uniformly in $t$ by $c_0[f{]}_{\alpha }{\epsilon }^{1+\alpha }\mathrm{/}(1+\alpha )$ and $c_1[f{]}_{\alpha }{\epsilon }^{\alpha }\mathrm{/}\alpha$, respectively. They therefore converge uniformly on $[a,b]$, proving continuity of $v_1$ and $A{v}_1$ at positive times. For the constant-endpoint term, $\mathrm{\parallel }A{v}_2(t)\mathrm{\parallel }\le (c_0+1)\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$ and $\mathrm{\parallel }A{v}_2(t)-A{v}_2(0)\mathrm{\parallel }\le (c_0+1)\mathrm{\parallel }f(t)-f(0)\mathrm{\parallel }+\mathrm{\parallel }(T(t)-I)f(0)\mathrm{\parallel }\to 0$ by continuity of $f$ and strong continuity of $T$; finally $v$ is continuous on $[0,b]$ by [L3]. Hence $v\in C([0,b],D(A))$ in the graph norm and $Av\in C([0,b],X)$ with $Av(0)=0$.

The final assertion. For $x\in D(A)$, [L2] gives $AT(t)x=T(t)Ax\to Ax$ as $t\downarrow 0$ by strong continuity, so $Au(t)=AT(t)x+Av(t)\to Ax+0=Ax$; the decomposition $u=T(\cdot )x+v$ therefore removes the singularity of $AT(t)x$ at the endpoint, and no choice principle beyond Dependent Choice was used.