[proofplan] We proceed in three stages. First we show the convolution-type integral $v(t) := \int_0^t T(t-s)f(s)\, d\mathcal{L}^1(s)$ is well-defined as a Bochner integral, using the strong continuity of $T$ and the uniform bound $\|T(t)\|_{\mathcal{L}(X)} \le Me^{\omega t}$ on bounded intervals. Second, we prove $v \in C([0, T]; X)$ by splitting the difference $v(t+h) - v(t)$ into a "shift" piece controlled by strong continuity of $T$ and a "tail" piece controlled by integrability of $f$. Third, under the regularity hypothesis $g \in D(A)$ and $f \in C^1([0,T];X)$, we change variables to $v(t) = \int_0^t T(s)f(t-s)\, d\mathcal{L}^1(s)$, which puts the $t$-dependence on $f$ rather than on $T$. We then differentiate $v$ in $t$, verify $v(t) \in D(A)$ via a difference-quotient argument using the semigroup property, and combine these to obtain $u'(t) = Au(t) + f(t)$. [/proofplan] [step:Show the convolution $v(t) = \int_0^t T(t-s) f(s) \, d\mathcal{L}^1(s)$ is well-defined as a Bochner integral] Define \begin{align*} v: [0, T] &\to X, \\ t &\mapsto \int_0^t T(t - s) f(s) \, d\mathcal{L}^1(s). \end{align*} Fix $t \in [0, T]$. The integrand $\Phi_t: [0, t] \to X$, $s \mapsto T(t-s) f(s)$ is the composition of the strongly continuous family $\{T(\tau)\}_{\tau \ge 0}$ with the strongly measurable map $f$. Since $\{T(\tau)\}_{\tau \ge 0}$ is strongly continuous on $[0, T]$, by the [Closure and Density of the $C_0$-Semigroup Generator](/theorems/3144) there exist constants $M \ge 1$ and $\omega \in \mathbb{R}$ with $\|T(\tau)\|_{\mathcal{L}(X)} \le M e^{\omega \tau}$ for all $\tau \in [0, T]$. Set \begin{align*} M_T := \sup_{0 \le \tau \le T} \|T(\tau)\|_{\mathcal{L}(X)} \le Me^{\omega^+ T} < \infty, \end{align*} where $\omega^+ = \max(\omega, 0)$. **Strong measurability.** $\Phi_t$ is the product of two strongly measurable maps in the appropriate sense: $s \mapsto f(s)$ is strongly measurable (as $f \in L^1(0, T; X)$), and $s \mapsto T(t-s)$ is strongly continuous, hence strongly measurable. The map $(\tau, x) \mapsto T(\tau) x$ is jointly continuous on $[0, T] \times X$ in the strong sense (continuous in $\tau$ uniformly on compact sets in $x$, and bounded linear in $x$), so $s \mapsto T(t-s) f(s)$ is strongly measurable as the composition of strongly measurable inputs with a Carathéodory map. **Bochner integrability.** We have \begin{align*} \int_0^t \|T(t-s) f(s)\|_X \, d\mathcal{L}^1(s) \le M_T \int_0^t \|f(s)\|_X \, d\mathcal{L}^1(s) \le M_T \|f\|_{L^1(0, T; X)} < \infty. \end{align*} Hence $\Phi_t$ is Bochner integrable, and $v(t)$ is well-defined with the bound \begin{align*} \|v(t)\|_X \le M_T \|f\|_{L^1(0, T; X)} \quad \text{for all } t \in [0, T]. \end{align*} [/step] [step:Show $v \in C([0, T]; X)$ by splitting into shift and tail pieces] Let $0 \le t < t + h \le T$ with $h > 0$. Then \begin{align*} v(t + h) - v(t) &= \int_0^{t+h} T(t + h - s) f(s) \, d\mathcal{L}^1(s) - \int_0^t T(t - s) f(s) \, d\mathcal{L}^1(s) \\ &= \underbrace{\int_0^t [T(t + h - s) - T(t - s)] f(s) \, d\mathcal{L}^1(s)}_{=: I_1(h)} + \underbrace{\int_t^{t+h} T(t + h - s) f(s) \, d\mathcal{L}^1(s)}_{=: I_2(h)}. \end{align*} **Bounding $I_2(h)$.** \begin{align*} \|I_2(h)\|_X \le M_T \int_t^{t+h} \|f(s)\|_X \, d\mathcal{L}^1(s). \end{align*} The right-hand side tends to $0$ as $h \to 0^+$ by absolute continuity of the Bochner integral (the integral over a set of measure $h$ tends to zero with $h$, since $\|f(\cdot)\|_X \in L^1(0, T)$). **Bounding $I_1(h)$.** For each fixed $s \in [0, t]$, $T(t + h - s) f(s) - T(t - s) f(s) \to 0$ in $X$ as $h \to 0^+$, by strong continuity of $\{T(\tau)\}_\tau$. Moreover the integrand is dominated: \begin{align*} \|T(t + h - s) f(s) - T(t - s) f(s)\|_X \le 2 M_T \|f(s)\|_X, \end{align*} which is integrable in $s$ on $[0, t]$. By the Dominated Convergence Theorem (Bochner version), $I_1(h) \to 0$ in $X$ as $h \to 0^+$. The case $h \to 0^-$ is analogous (swap the roles of $t$ and $t + h$). Hence $v$ is continuous on $[0, T]$. Since $T(\cdot)g \in C([0, T]; X)$ is also continuous (by strong continuity of the semigroup), $u(t) = T(t)g + v(t)$ is continuous, i.e., $u \in C([0, T]; X)$. [guided] We want to prove that $v(t) := \int_0^t T(t-s) f(s)\, d\mathcal{L}^1(s)$ is continuous in $t$. The naive temptation is to differentiate under the integral or "check the integrand is continuous in $t$" — but $t$ enters $v(t)$ in two different ways: through the upper limit of integration and through the integrand $T(t-s)$. We isolate these two effects with a **shift-plus-tail split**. Fix $0 \le t < t+h \le T$ with $h > 0$. Write \begin{align*} v(t+h) - v(t) &= \int_0^{t+h} T(t+h-s) f(s)\, d\mathcal{L}^1(s) - \int_0^t T(t-s) f(s)\, d\mathcal{L}^1(s). \end{align*} Add and subtract $\int_0^t T(t+h-s) f(s)\, d\mathcal{L}^1(s)$ to separate the two effects: \begin{align*} v(t+h) - v(t) = I_1(h) + I_2(h), \end{align*} where \begin{align*} I_1(h) &:= \int_0^t \big[T(t+h-s) - T(t-s)\big] f(s)\, d\mathcal{L}^1(s), \\ I_2(h) &:= \int_t^{t+h} T(t+h-s) f(s)\, d\mathcal{L}^1(s). \end{align*} The piece $I_1(h)$ holds the integration interval $[0, t]$ fixed and only changes the integrand: this is the "shift" piece, controlled by **strong continuity of $T$** through dominated convergence. The piece $I_2(h)$ has the new integration interval $[t, t+h]$ that did not exist at time $t$: this is the "tail" piece, controlled by **absolute continuity of the Bochner integral**. **Bounding the tail $I_2(h)$.** Using $\|T(\tau)\|_{\mathcal{L}(X)} \le M_T$ for $\tau \in [0, T]$, \begin{align*} \|I_2(h)\|_X \le \int_t^{t+h} \|T(t+h-s)\|_{\mathcal{L}(X)} \|f(s)\|_X\, d\mathcal{L}^1(s) \le M_T \int_t^{t+h} \|f(s)\|_X\, d\mathcal{L}^1(s). \end{align*} Since $\|f(\cdot)\|_X \in L^1(0, T)$, absolute continuity of the Lebesgue integral gives \begin{align*} \int_t^{t+h} \|f(s)\|_X\, d\mathcal{L}^1(s) \to 0 \quad \text{as } h \to 0^+. \end{align*} Why does absolute continuity hold? It is a property of every $L^1$ function: the integral over a set of Lebesgue measure $h$ tends to $0$ as $h \to 0$. Hence $\|I_2(h)\|_X \to 0$. **Bounding the shift $I_1(h)$ via dominated convergence.** We apply the Dominated Convergence Theorem (Bochner version). We need (a) pointwise convergence of the integrand to $0$ in $X$ a.e. on $[0, t]$, and (b) a measurable real-valued majorant integrable on $[0, t]$. For (a): fix $s \in [0, t]$. Then $t - s \ge 0$ and $t + h - s \to t - s$ as $h \to 0^+$. By strong continuity of $\{T(\tau)\}_{\tau \ge 0}$ at $\tau = t - s$, \begin{align*} T(t+h-s) f(s) - T(t-s) f(s) \to 0 \quad \text{in } X \text{ as } h \to 0^+. \end{align*} (At the boundary case $s = t$: $t + h - s = h \to 0^+$, and strong continuity of $T$ at $0$ gives $T(h) f(t) \to f(t) = T(0) f(t)$, so the difference still tends to $0$.) For (b): for every $s \in [0, t]$ and $h$ small, \begin{align*} \|T(t+h-s) f(s) - T(t-s) f(s)\|_X \le \big(\|T(t+h-s)\|_{\mathcal{L}(X)} + \|T(t-s)\|_{\mathcal{L}(X)}\big) \|f(s)\|_X \le 2 M_T \|f(s)\|_X, \end{align*} and the function $s \mapsto 2 M_T \|f(s)\|_X$ is in $L^1(0, t)$ since $f \in L^1(0, T; X)$. The dominated convergence theorem applies and yields \begin{align*} I_1(h) = \int_0^t \big[T(t+h-s) - T(t-s)\big] f(s)\, d\mathcal{L}^1(s) \to 0 \quad \text{in } X \text{ as } h \to 0^+. \end{align*} **Combining.** Since $\|v(t+h) - v(t)\|_X \le \|I_1(h)\|_X + \|I_2(h)\|_X \to 0$ as $h \to 0^+$, the function $v$ is right-continuous at $t$. **Left continuity.** For $h \to 0^-$, the same argument applies after swapping the roles of $t$ and $t+h$: write $v(t) - v(t+h)$ in place of $v(t+h) - v(t)$, with the new "tail" interval being $[t+h, t]$ of length $|h|$, and the same "shift" piece $I_1(h)$ on $[0, t+h]$. Both pieces vanish as $h \to 0^-$ by the same reasoning. Hence $v: [0, T] \to X$ is continuous. Combined with $T(\cdot)g \in C([0, T]; X)$ (which is the strong continuity of the semigroup applied to the fixed vector $g$), we conclude $u = T(\cdot)g + v \in C([0, T]; X)$. [/guided] [/step] [step:Under $g \in D(A)$ and $f \in C^1([0, T]; X)$, rewrite the convolution as $v(t) = \int_0^t T(s) f(t - s)\, d\mathcal{L}^1(s)$] We change variables in the convolution. Let $\sigma = t - s$, so $s = t - \sigma$, $d\mathcal{L}^1(s) = d\mathcal{L}^1(\sigma)$, and the limits become $s = 0 \mapsto \sigma = t$, $s = t \mapsto \sigma = 0$. Thus \begin{align*} v(t) = \int_0^t T(t-s) f(s) \, d\mathcal{L}^1(s) = \int_0^t T(\sigma) f(t - \sigma) \, d\mathcal{L}^1(\sigma). \end{align*} Renaming the dummy variable $\sigma \to s$, \begin{align*} v(t) = \int_0^t T(s) f(t - s) \, d\mathcal{L}^1(s). \end{align*} This form is preferable for differentiation: now $T(s)$ does not depend on $t$, so when we differentiate in $t$ the derivative falls on $f$ — and $f$ is $C^1$ by hypothesis. [/step] [step:Differentiate $v$ in $t$ using the $C^1$ hypothesis on $f$] Compute the difference quotient at $t \in (0, T)$ for small $h \ne 0$ with $t + h \in (0, T)$: \begin{align*} \frac{v(t + h) - v(t)}{h} = \frac{1}{h}\left[\int_0^{t+h} T(s) f(t + h - s)\, d\mathcal{L}^1(s) - \int_0^t T(s) f(t - s) \, d\mathcal{L}^1(s)\right]. \end{align*} Split: \begin{align*} \frac{v(t + h) - v(t)}{h} &= \int_0^t T(s) \cdot \frac{f(t + h - s) - f(t - s)}{h}\, d\mathcal{L}^1(s) \\ &\qquad + \frac{1}{h}\int_t^{t+h} T(s) f(t + h - s) \, d\mathcal{L}^1(s). \end{align*} **First term.** Since $f \in C^1([0, T]; X)$, the difference quotient $(f(t + h - s) - f(t - s))/h \to f'(t - s)$ uniformly in $s \in [0, t]$ as $h \to 0$ (by uniform continuity of $f'$ on $[0, T]$ — a continuous function on a compact interval is uniformly continuous). Combined with the uniform bound $\|T(s)\|_{\mathcal{L}(X)} \le M_T$, the first term converges to \begin{align*} \int_0^t T(s) f'(t - s)\, d\mathcal{L}^1(s) \quad \text{in } X \text{ as } h \to 0. \end{align*} **Second term.** The map $s \mapsto T(s) f(t+h-s)$ is continuous in $s$ on $[t, t+h]$, so by the mean-value theorem for Bochner integrals (the integral of a continuous function over an interval of length $h$ divided by $h$ tends to the value at the endpoint as $h \to 0$): \begin{align*} \frac{1}{h}\int_t^{t+h} T(s) f(t + h - s)\, d\mathcal{L}^1(s) \to T(t) f(0) \quad \text{in } X \text{ as } h \to 0^+. \end{align*} For $h \to 0^-$ the corresponding limit is the same, after writing the integral as $-\frac{1}{|h|}\int_{t+h}^t T(s) f(t+h-s)\, d\mathcal{L}^1(s)$. Therefore $v$ is differentiable on $(0, T)$ with \begin{align*} v'(t) = T(t) f(0) + \int_0^t T(s) f'(t - s)\, d\mathcal{L}^1(s). \end{align*} [/step] [step:Show $v(t) \in D(A)$ and compute $Av(t)$ via the difference quotient] We verify $v(t) \in D(A)$ by showing the difference quotient $(T(h) v(t) - v(t))/h$ converges in $X$ as $h \to 0^+$, then identify the limit as $Av(t)$ using the [Closure and Density of the $C_0$-Semigroup Generator](/theorems/3144). Fix $t \in (0, T)$. Using the change-of-variable form $v(t) = \int_0^t T(s) f(t-s)\, d\mathcal{L}^1(s)$ from Step 3 and the semigroup property $T(h) T(s) = T(s + h)$: \begin{align*} T(h) v(t) &= \int_0^t T(h) T(s) f(t-s)\, d\mathcal{L}^1(s) = \int_0^t T(s+h) f(t-s)\, d\mathcal{L}^1(s) \\ &= \int_h^{t+h} T(\sigma) f(t - \sigma + h)\, d\mathcal{L}^1(\sigma) \quad (\sigma = s + h). \end{align*} Subtracting $v(t) = \int_0^t T(\sigma) f(t-\sigma)\, d\mathcal{L}^1(\sigma)$ and dividing by $h$: \begin{align*} \frac{T(h) v(t) - v(t)}{h} &= \frac{1}{h}\int_h^{t+h} T(\sigma) f(t-\sigma+h)\, d\mathcal{L}^1(\sigma) - \frac{1}{h}\int_0^t T(\sigma) f(t-\sigma)\, d\mathcal{L}^1(\sigma). \end{align*} Splitting the first integral as $\int_h^{t+h} = \int_h^t + \int_t^{t+h}$ and the second as $\int_0^t = \int_0^h + \int_h^t$, then collecting the matching $[h, t]$ pieces: \begin{align*} \frac{T(h) v(t) - v(t)}{h} &= \frac{1}{h}\int_t^{t+h} T(\sigma) f(t-\sigma+h)\, d\mathcal{L}^1(\sigma) - \frac{1}{h}\int_0^h T(\sigma) f(t-\sigma)\, d\mathcal{L}^1(\sigma) \\ &\qquad + \int_h^t T(\sigma) \cdot \frac{f(t-\sigma+h) - f(t-\sigma)}{h}\, d\mathcal{L}^1(\sigma). \end{align*} We now compute the limit of each term as $h \to 0^+$. - **First term.** The integrand $\sigma \mapsto T(\sigma) f(t-\sigma+h)$ is continuous in $\sigma$ on $[t, t+h]$, and at $\sigma = t$ it equals $T(t) f(h) \to T(t) f(0)$ as $h \to 0^+$ (using continuity of $f$ and boundedness of $T(t)$). By the mean-value property of Bochner integrals, \begin{align*} \frac{1}{h}\int_t^{t+h} T(\sigma) f(t-\sigma+h)\, d\mathcal{L}^1(\sigma) \to T(t) f(0) \quad \text{in } X \text{ as } h \to 0^+. \end{align*} - **Second term.** Similarly, the integrand $\sigma \mapsto T(\sigma) f(t-\sigma)$ is continuous on $[0, h]$ and at $\sigma = 0$ equals $T(0) f(t) = f(t)$. Hence \begin{align*} \frac{1}{h}\int_0^h T(\sigma) f(t-\sigma)\, d\mathcal{L}^1(\sigma) \to f(t) \quad \text{in } X \text{ as } h \to 0^+. \end{align*} - **Third term.** The lower limit of integration is $h$ rather than $0$. We rewrite \begin{align*} \int_h^t T(\sigma) \cdot \frac{f(t-\sigma+h) - f(t-\sigma)}{h}\, d\mathcal{L}^1(\sigma) = \int_0^t T(\sigma) \cdot \frac{f(t-\sigma+h) - f(t-\sigma)}{h}\, d\mathcal{L}^1(\sigma) - R(h), \end{align*} where \begin{align*} R(h) := \int_0^h T(\sigma) \cdot \frac{f(t-\sigma+h) - f(t-\sigma)}{h}\, d\mathcal{L}^1(\sigma). \end{align*} Since $f \in C^1([0, T]; X)$, $\sup_{0 \le \tau \le T}\|f'(\tau)\|_X =: \|f'\|_\infty < \infty$, and the difference quotient is uniformly bounded by $\|f'\|_\infty$. Hence \begin{align*} \|R(h)\|_X \le M_T \cdot \|f'\|_\infty \cdot h \to 0 \quad \text{as } h \to 0^+, \end{align*} which justifies replacing the lower limit $h$ by $0$ in the limit. For the corrected integral $\int_0^t$: by uniform continuity of $f'$ on $[0, T]$, \begin{align*} \frac{f(t-\sigma+h) - f(t-\sigma)}{h} \to f'(t-\sigma) \quad \text{uniformly in } \sigma \in [0, t] \text{ as } h \to 0^+, \end{align*} and combined with $\|T(\sigma)\|_{\mathcal{L}(X)} \le M_T$, we obtain \begin{align*} \int_0^t T(\sigma) \cdot \frac{f(t-\sigma+h) - f(t-\sigma)}{h}\, d\mathcal{L}^1(\sigma) \to \int_0^t T(\sigma) f'(t - \sigma)\, d\mathcal{L}^1(\sigma) \quad \text{in } X. \end{align*} Reverting the substitution $\sigma \to t - s$: \begin{align*} \int_0^t T(\sigma) f'(t-\sigma)\, d\mathcal{L}^1(\sigma) = \int_0^t T(t-s) f'(s)\, d\mathcal{L}^1(s). \end{align*} Combining the three limits: \begin{align*} \lim_{h \to 0^+}\frac{T(h)v(t) - v(t)}{h} = T(t) f(0) - f(t) + \int_0^t T(t-s) f'(s)\, d\mathcal{L}^1(s). \end{align*} This limit exists in $X$, so by the [Closure and Density of the $C_0$-Semigroup Generator](/theorems/3144), $v(t) \in D(A)$ and \begin{align*} Av(t) = T(t) f(0) - f(t) + \int_0^t T(t-s) f'(s)\, d\mathcal{L}^1(s). \end{align*} [/step] [step:Combine to show $u'(t) = Au(t) + f(t)$ and conclude $u$ is a classical solution] From Steps 4-5, comparing \begin{align*} v'(t) &= T(t)f(0) + \int_0^t T(t-s) f'(s)\, d\mathcal{L}^1(s), \\ Av(t) &= T(t)f(0) - f(t) + \int_0^t T(t-s) f'(s)\, d\mathcal{L}^1(s), \end{align*} we read off \begin{align*} v'(t) = Av(t) + f(t). \end{align*} (Note: in Step 4 we wrote $v'(t) = T(t)f(0) + \int_0^t T(s) f'(t-s)\, d\mathcal{L}^1(s)$, which equals $T(t)f(0) + \int_0^t T(t-s) f'(s)\, d\mathcal{L}^1(s)$ by the same change of variable $\sigma = t - s$ used in Step 3.) Now consider $u(t) = T(t) g + v(t)$ with $g \in D(A)$. By the [Closure and Density of the $C_0$-Semigroup Generator](/theorems/3144), $T(t)g \in D(A)$ for all $t \ge 0$ and $\frac{d}{dt}T(t) g = A T(t) g$. Therefore $u(t) = T(t) g + v(t) \in D(A)$ (sum of $D(A)$ elements), and \begin{align*} u'(t) = AT(t)g + v'(t) = AT(t) g + Av(t) + f(t) = A(T(t)g + v(t)) + f(t) = Au(t) + f(t), \end{align*} where in the third equality we used linearity of $A$ on $D(A)$. Also $u(0) = T(0) g + v(0) = g + 0 = g$. So $u$ is a classical solution: $u \in C([0,T]; X) \cap C^1((0, T); X)$, $u(t) \in D(A)$ for $t \in (0, T)$, $u'(t) = Au(t) + f(t)$, and $u(0) = g$. The continuity at $t = 0$ of $u$ follows from $T(\cdot)g \in C([0, T]; X)$ and $v \in C([0, T]; X)$ (Step 2), and $u'$ extends continuously to $t = 0$ as $Ag + f(0)$. This completes the proof. [/step]