[proofplan] For existence with $g \in D(A)$, we apply the differentiation formula from the [Closure and Density of the $C_0$-Semigroup Generator](/theorems/3144), which gives $T(t)g \in D(A)$ and $\frac{d}{dt}T(t)g = AT(t)g = T(t)Ag$ — exactly the Cauchy problem with continuous derivative $T(\cdot)Ag$. For uniqueness, given any classical solution $v$, the function $w(s) := T(t - s)v(s)$ has derivative zero on $[0, t]$ by a Leibniz computation that combines the semigroup differentiation rule with the equation $v'(s) = Av(s)$; constancy on $[0, t]$ then forces $v(t) = T(t)g$. For $g \notin D(A)$, $u(t) = T(t)g$ is continuous by strong continuity but cannot be differentiable at $t = 0$ (since differentiability there is exactly the definition of $D(A)$). [/proofplan]

[step:Verify that $u(t) := T(t)g$ solves the Cauchy problem when $g \in D(A)$] Let $g \in D(A)$ and define $u: [0, \infty) \to X$ by $u(t) := T(t) g$. By part 3 of the [Closure and Density of the $C_0$-Semigroup Generator](/theorems/3144), we have $T(t) g \in D(A)$ for every $t \ge 0$ and \begin{align*} \frac{d}{dt} T(t) g &= A T(t) g = T(t) A g. \end{align*} This shows that $u$ is differentiable on $[0, \infty)$ with $u'(t) = A u(t)$, that $u(t) \in D(A)$ for every $t$, and that \begin{align*} u'(t) &= T(t) A g. \end{align*} Strong continuity of the semigroup applied to the vector $A g \in X$ yields $t \mapsto T(t) A g \in C([0, \infty); X)$, hence $u \in C^1([0, \infty); X)$. Finally $u(0) = T(0) g = g$, so the initial condition holds. [/step]

[step:Show $u \in C([0, \infty); D(A))$ with the graph norm] Recall that the graph norm on $D(A)$ is $\|x\|_{D(A)} := \|x\|_X + \|Ax\|_X$, and $D(A)$ is a Banach space under this norm because $A$ is closed (part 2 of the [Closure and Density of the $C_0$-Semigroup Generator](/theorems/3144)). For $g \in D(A)$ and $t \ge 0$, the previous step gives $u(t) = T(t)g \in D(A)$ and $A u(t) = T(t) A g$. Therefore \begin{align*} \|u(t) - u(s)\|_{D(A)} &= \|T(t) g - T(s) g\|_X + \|T(t) A g - T(s) A g\|_X. \end{align*} Each of the two terms tends to $0$ as $s \to t$ by strong continuity of the semigroup applied to $g$ and to $Ag$, respectively. Hence $u: [0, \infty) \to D(A)$ is continuous in the graph norm, i.e. $u \in C([0, \infty); D(A))$. [/step]

[step:Prove uniqueness via the auxiliary function $w(s) := T(t - s) v(s)$]Suppose $v: [0, \infty) \to X$ is another classical solution: $v \in C^1([0, \infty); X) \cap C([0, \infty); D(A))$, $v'(s) = A v(s)$ for $s \ge 0$, and $v(0) = g$. We show $v(t) = T(t) g$ for every $t \ge 0$. Fix $t > 0$ and define \begin{align*} w: [0, t] &\to X, & s &\mapsto T(t - s) v(s). \end{align*} [claim:The function $w$ is differentiable on $[0, t]$ with $w'(s) = 0$ identically.] [/claim] [proof] Fix $s \in (0, t)$ and let $h \in \mathbb{R}$ with $|h|$ small enough that $s + h \in (0, t)$. We expand the difference \begin{align*} w(s + h) - w(s) &= T(t - s - h) v(s + h) - T(t - s) v(s) \\ &= T(t - s - h) [v(s + h) - v(s)] + [T(t - s - h) - T(t - s)] v(s). \end{align*} Dividing by $h$: \begin{align*} \frac{w(s + h) - w(s)}{h} &= T(t - s - h)\, \frac{v(s + h) - v(s)}{h} + \frac{T(t - s - h) - T(t - s)}{h} v(s). \end{align*} Examine the two terms separately as $h \to 0$. First term. Since $v$ is differentiable at $s$ with $v'(s) = A v(s)$, the difference quotient $\frac{v(s + h) - v(s)}{h} \to A v(s)$ in $X$. By the local exponential bound, $\|T(t - s - h)\|_{\mathcal{L}(X)} \le M e^{|\omega| t}$ for $|h|$ small, and $T(t - s - h) \to T(t - s)$ strongly as $h \to 0$. Together with the convergence of the difference quotient, the operator $\times$ vector limit gives \begin{align*} T(t - s - h)\, \frac{v(s + h) - v(s)}{h} &\xrightarrow{h \to 0} T(t - s)\, A v(s). \end{align*} Second term. Since $v(s) \in D(A)$, the differentiation formula from the [Closure and Density of the $C_0$-Semigroup Generator](/theorems/3144) applied to the orbit $\tau \mapsto T(\tau) v(s)$ at $\tau = t - s$ gives \begin{align*} \lim_{h \to 0} \frac{T(t - s - h) - T(t - s)}{h} v(s) &= \lim_{h \to 0} \frac{T(t - s - h) v(s) - T(t - s) v(s)}{h} = -\, A T(t - s) v(s). \end{align*} Using $A T(t - s) v(s) = T(t - s) A v(s)$ (the same theorem): \begin{align*} \lim_{h \to 0} \frac{T(t - s - h) - T(t - s)}{h} v(s) &= -\, T(t - s) A v(s). \end{align*} Adding the two limits: \begin{align*} w'(s) &= T(t - s) A v(s) - T(t - s) A v(s) = 0. \end{align*} The same computation with one-sided difference quotients gives $w'(0) = 0$ and $w'(t) = 0$, so $w' \equiv 0$ on $[0, t]$. [/proof] Since $w \in C([0, t]; X)$ (composition of continuous maps and the locally bounded semigroup) and $w' \equiv 0$ in $X$, the Mean Value Inequality for Banach-Valued Functions implies $w$ is constant on $[0, t]$. Comparing endpoints: \begin{align*} w(0) &= T(t) v(0) = T(t) g, & w(t) &= T(0) v(t) = v(t). \end{align*} Constancy gives $v(t) = T(t) g$. Since $t > 0$ was arbitrary and $v(0) = g = T(0) g$, we conclude $v = u$ on $[0, \infty)$.[/step]

[guided]We are given a second classical solution $v$ and must show $v(t) = T(t) g$. The standard trick is to introduce an auxiliary function combining the candidate $T(t - s)$ (which moves backward in time) with the unknown $v(s)$ (which moves forward in time): \begin{align*} w: [0, t] &\to X, & s &\mapsto T(t - s) v(s). \end{align*} The motivation is that the two time-derivatives should cancel: differentiating the semigroup factor with respect to $s$ produces $-A T(t - s) v(s)$, while differentiating $v(s)$ produces $T(t - s) A v(s)$, and these should match up to a sign because $A T(t - s) = T(t - s) A$ on $D(A)$. If $w' = 0$ then $w$ is constant, and comparing the endpoints $w(0) = T(t) g$ and $w(t) = v(t)$ forces $v(t) = T(t) g$. To make this rigorous, fix $s \in (0, t)$ and a small $h \in \mathbb{R}$ with $s + h \in (0, t)$. We expand the difference \begin{align*} w(s + h) - w(s) &= T(t - s - h) v(s + h) - T(t - s) v(s) \\ &= T(t - s - h) [v(s + h) - v(s)] + [T(t - s - h) - T(t - s)] v(s). \end{align*} This algebraic regrouping isolates the $v$-derivative in the first term and the $T$-derivative in the second. Dividing by $h$: \begin{align*} \frac{w(s + h) - w(s)}{h} &= T(t - s - h)\, \frac{v(s + h) - v(s)}{h} + \frac{T(t - s - h) - T(t - s)}{h} v(s). \end{align*} We analyse the two terms separately. For the first term, $v$ is differentiable at $s$ with $v'(s) = A v(s)$ by hypothesis, so the difference quotient $\frac{v(s + h) - v(s)}{h}$ converges to $A v(s)$ in $X$. The operator $T(t - s - h)$ is uniformly bounded for $|h|$ small (by the local exponential bound $M e^{|\omega| t}$) and converges strongly to $T(t - s)$ by strong continuity. Combining strong convergence of the operator with norm convergence of the vector, \begin{align*} T(t - s - h)\, \frac{v(s + h) - v(s)}{h} &\xrightarrow{h \to 0} T(t - s)\, A v(s). \end{align*} For the second term, we are differentiating the semigroup orbit $\tau \mapsto T(\tau) v(s)$ at $\tau = t - s$. This is exactly where we use that $v(s) \in D(A)$ — without this, $T(\tau) v(s)$ need not be differentiable in $\tau$. The differentiation formula from the [Closure and Density of the $C_0$-Semigroup Generator](/theorems/3144) gives \begin{align*} \lim_{h \to 0} \frac{T(t - s - h) v(s) - T(t - s) v(s)}{h} &= -\, A T(t - s) v(s) = -\, T(t - s) A v(s). \end{align*} The minus sign comes from the chain rule: $\frac{d}{ds} T(t - s) = - \frac{d}{d\tau} T(\tau) \big|_{\tau = t - s}$. The second equality uses the commutation $A T(\tau) = T(\tau) A$ on $D(A)$. Adding the two limits cancels them: \begin{align*} w'(s) &= T(t - s) A v(s) - T(t - s) A v(s) = 0. \end{align*} The boundary cases $s = 0$ and $s = t$ work the same way with one-sided difference quotients. Thus $w' \equiv 0$ on $[0, t]$. Now $w \in C([0, t]; X)$ because $s \mapsto v(s)$ is continuous (in fact $C^1$) and $(s, x) \mapsto T(t - s) x$ is jointly strongly continuous on $[0, t] \times X$ (using the local boundedness $\|T(t-s)\|_{\mathcal{L}(X)} \le M e^{|\omega| t}$). Since $w$ is continuous with $w' = 0$ in the Banach space $X$, the Mean Value Inequality for Banach-Valued Functions implies $w$ is constant. Evaluating at the endpoints: \begin{align*} w(0) &= T(t - 0) v(0) = T(t) g, & w(t) &= T(0) v(t) = v(t), \end{align*} and constancy gives $v(t) = T(t) g$. Since $t > 0$ was arbitrary and $v(0) = g$, the two solutions agree on $[0, \infty)$.[/guided]

[step:Treat the case $g \in X \setminus D(A)$: $u(t) := T(t)g$ is continuous but not classical] For arbitrary $g \in X$, the map $u: [0, \infty) \to X$, $t \mapsto T(t) g$, is continuous by the strong continuity of the semigroup. Indeed, for $s, t \ge 0$ with $|t - s|$ small, \begin{align*} \|u(t) - u(s)\|_X &= \|T(t) g - T(s) g\|_X \xrightarrow{|t - s| \to 0} 0, \end{align*} so $u \in C([0, \infty); X)$. We call this the **mild solution** with initial datum $g$. (Equivalently, $u(t) = T(t) g$ is the [Duhamel mild solution](/theorems/3141) of the homogeneous problem.) Suppose now that $g \in X \setminus D(A)$. We claim $u$ is not differentiable at $t = 0$ in $X$. By definition, \begin{align*} u'(0^+) &= \lim_{h \to 0^+} \frac{u(h) - u(0)}{h} = \lim_{h \to 0^+} \frac{T(h) g - g}{h} \end{align*} exists in $X$ if and only if $g \in D(A)$. Since $g \notin D(A)$, no such limit exists, and $u$ fails to be differentiable at $0$. In particular, $u$ is not a classical solution. This completes the proof. [/step]