= Solution
A sufficient set of assumptions is that $L$ is a densely defined <closed operator> on the <Hilbert space> $H$, generating an <analytic semigroup> $S(t)$, and that $f\in C^{0,\theta}([0,T],H)$ for some $0<\theta\leq1$. For an operator criterion, one may assume that $A=\omega I-L$ is a <sectorial operator> of angle less than $\pi/2$, with dense domain. The <analytic semigroup> then satisfies, on a fixed finite time interval,
$$
\|S(t)\|\leq M_T,\qquad \|LS(t)\|\leq C_T/t,\qquad S(t)H\subset D(L)\quad(t>0).
$$
The <Hölder continuity> assumption means $\|f(t)-f(s)\|_H\leq K|t-s|^\theta$. It supplies time regularity of the forcing, not spatial membership of $f(t)$ in $D(L)$.
With initial datum $u_0\in H$, a <classical solution of an abstract Cauchy problem> means
$$
u\in C([0,T],H)\cap C^1((0,T],H),\qquad u(t)\in D(L),\qquad Lu\in C((0,T],H),
$$
with the evolution identity holding in $H$ at positive times and $u(t)\to u_0$ in $H$. Equivalently, the positive-time domain continuity is in the <graph norm>. If the definition requires differentiability also at zero, assume additionally $u_0\in D(L)$; the proof below then gives $u\in C^1([0,T],H)\cap C([0,T],D(L))$.
The <variation-of-constants formula> gives the candidate
$$
\boxed{u(t)=S(t)u_0+\int_0^tS(t-s)f(s)\,ds.}
$$
The integral is a <Bochner integral>. Boundedness of $S$ and continuity of $f$ make the expression continuous into $H$, and its integral term tends to zero with $t$. Thus it is a <mild solution of an abstract Cauchy problem> with the correct initial value. The issue is showing that its integral term lies in the domain of the unbounded <linear operator> $L$.
Write $v(t)=\int_0^tS(r)f(t-r)\,dr$. For $\varepsilon>0$, the truncated integral $v_\varepsilon(t)=\int_\varepsilon^tS(r)f(t-r)\,dr$ lies in $D(L)$. Subtract the current forcing before applying $L$:
$$
Lv_\varepsilon(t)=\int_\varepsilon^tLS(r)[f(t-r)-f(t)]\,dr+[S(t)-S(\varepsilon)]f(t).
$$
Here the last term follows by integrating $LS(r)f(t)=\frac d{dr}S(r)f(t)$ away from zero; it does not assume that $f(t)\in D(L)$. The first integrand satisfies
$$
\|LS(r)[f(t-r)-f(t)]\|_H\leq C_TK r^{\theta-1},\qquad
\int_0^t r^{\theta-1}\,dr=t^\theta/\theta<\infty.
$$
Since $v_\varepsilon(t)\to v(t)$ in $H$ and the displayed images also converge, the <closed operator> property gives
$$
v(t)\in D(L),\qquad Lv(t)=\int_0^tLS(r)[f(t-r)-f(t)]\,dr+[S(t)-I]f(t).
$$
This is the reason for the <Hölder continuity> assumption: the unsplit estimate would have the nonintegrable factor $r^{-1}$. Merely knowing that $f$ is continuous does not provide the integrable bound needed here. More generally, a modulus of continuity $\eta$ satisfying $\int_0^1\eta(r)\,dr/r<\infty$ suffices for this argument; a positive <Hölder exponent> is a convenient sufficient hypothesis.
The formula also proves continuity of $Lv$ at positive times. On compact positive-time intervals, split the integral at a small $r=\varepsilon$: its short part is uniformly bounded by $C\varepsilon^\theta$, while the remaining part is continuous by the <strongly continuous semigroup> property and ordinary dominated convergence. Together with analytic smoothing of $S(t)u_0$, this makes $Lu$ continuous at positive times. The <semigroup property> gives
$$
u(t+h)=S(h)u(t)+\int_0^hS(h-r)f(t+r)\,dr.
$$
Because $u(t)\in D(L)$, division of this identity by $h$ after subtracting $u(t)$ gives the continuous right derivative $Lu(t)+f(t)$. The left derivative agrees by the same identity and continuity of $Lu+f$. Thus $u'=Lu+f$, proving existence of the <classical solution of an abstract Cauchy problem> on the entire given finite interval, and in particular local existence.
For uniqueness, the difference $h$ of two such <classical solutions of an abstract Cauchy problem> satisfies $h'=Lh$ and $h(0)=0$. For $0<\varepsilon<t$, differentiation of $S(t-s)h(s)$ on $[\varepsilon,t]$ gives zero, so $h(t)=S(t-\varepsilon)h(\varepsilon)$. Letting $\varepsilon\downarrow0$ yields $h(t)=0$. Finally, if $u_0\in D(L)$, then $LS(t)u_0=S(t)Lu_0\to Lu_0$, and the expression for $Lv(t)$ tends to zero. Hence $u'(t)\to Lu_0+f(0)$ and the stronger time-zero regularity follows. \b[The solution is unique; the forcing regularity controls the singularity in the differentiated integral.]
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