[proofplan] The proof uses only the definition of the infinitesimal generator and the semigroup law. First we prove that each orbit starting in $D(A)$ remains in $D(A)$ and that $A$ commutes with $T(t)$ on $D(A)$. Then we compute right and left difference quotients of $t\mapsto T(t)u_0$ to obtain differentiability and the equation $u'(t)=Au(t)$. Finally, strong continuity gives continuity of both $u$ and $Au$, so the orbit is classical on every finite interval. [/proofplan]

[step:Record the generator limit at the initial datum] Since $u_0\in D(A)$ and $A$ is the infinitesimal generator of $(T(t))_{t\geq 0}$, define the right difference quotient family $(q_h)_{h>0}\subset X$ by \begin{align*} q_h:=\frac{T(h)u_0-u_0}{h}. \end{align*} It satisfies \begin{align*} \lim_{h\downarrow 0}\|q_h-Au_0\|_X=0. \end{align*} This is the defining property of membership in $D(A)$. [/step]

[step:Show that the semigroup preserves the generator domain]Fix $t\geq 0$. We prove that $T(t)u_0\in D(A)$ and compute its image under $A$. For $h>0$, the semigroup law gives \begin{align*} \frac{T(h)T(t)u_0-T(t)u_0}{h}=T(t)\frac{T(h)u_0-u_0}{h}=T(t)q_h. \end{align*} Because $T(t):X\to X$ is bounded and $q_h\to Au_0$ in $X$ as $h\downarrow 0$, we obtain \begin{align*} \lim_{h\downarrow 0}\left\|\frac{T(h)T(t)u_0-T(t)u_0}{h}-T(t)Au_0\right\|_X=0. \end{align*} By the definition of the generator domain, this proves $T(t)u_0\in D(A)$ and \begin{align*} A T(t)u_0 = T(t)Au_0. \end{align*}[/step]

[guided]The point of this step is to check domain regularity directly from the generator definition. To prove that $T(t)u_0$ belongs to $D(A)$, we must show that the right derivative at time $0$ of the orbit starting from $T(t)u_0$ exists in $X$. Fix $t\geq 0$. For $h>0$, define the original generator quotient $q_h\in X$ by \begin{align*} q_h:=\frac{T(h)u_0-u_0}{h}. \end{align*} Also define the shifted quotient $r_h\in X$ by \begin{align*} r_h:=\frac{T(h)T(t)u_0-T(t)u_0}{h}. \end{align*} Since $u_0\in D(A)$, the definition of the infinitesimal generator gives $q_h\to Au_0$ in $X$ as $h\downarrow 0$. Using the semigroup identity $T(h)T(t)=T(h+t)=T(t)T(h)$, we rewrite the shifted quotient as \begin{align*} r_h=\frac{T(t)T(h)u_0-T(t)u_0}{h}=T(t)\frac{T(h)u_0-u_0}{h}=T(t)q_h. \end{align*} The quotient $q_h$ satisfies $q_h\to Au_0$ in $X$ as $h\downarrow 0$. Since $T(t):X\to X$ is a [bounded linear operator](/page/Bounded%20Linear%20Operator), bounded linear maps preserve norm limits. Hence \begin{align*} \lim_{h\downarrow 0}\|r_h-T(t)Au_0\|_X=\lim_{h\downarrow 0}\|T(t)(q_h-Au_0)\|_X\leq \|T(t)\|_{\mathcal{L}(X)}\lim_{h\downarrow 0}\|q_h-Au_0\|_X=0. \end{align*} Therefore the generator limit for the point $T(t)u_0$ exists and equals $T(t)Au_0$. By the definition of the infinitesimal generator, this means \begin{align*} T(t)u_0\in D(A) \end{align*} and \begin{align*} A T(t)u_0=T(t)Au_0. \end{align*}[/guided]

[step:Compute the derivative of the orbit] We next prove differentiability of $u:(0,\infty)\to X$. Fix $t>0$. For $h>0$, the semigroup law gives \begin{align*} \frac{u(t+h)-u(t)}{h}=T(t)\frac{T(h)u_0-u_0}{h}=T(t)q_h. \end{align*} Letting $h\downarrow 0$ yields \begin{align*} \lim_{h\downarrow 0}\left\|\frac{u(t+h)-u(t)}{h}-T(t)Au_0\right\|_X=0. \end{align*} For the left difference quotient, let $0<k<t$. Then \begin{align*} \frac{u(t-k)-u(t)}{-k}=T(t-k)\frac{T(k)u_0-u_0}{k}=T(t-k)q_k. \end{align*} Since $q_k\to Au_0$ in $X$ and $T(t-k)Au_0\to T(t)Au_0$ in $X$ by strong continuity, it remains to control $T(t-k)(q_k-Au_0)$ uniformly for $0<k<t$. Define the family of bounded linear maps $\mathcal{T}_t\subset\mathcal{L}(X)$ by \begin{align*} \mathcal{T}_t:=\{T(s):0\leq s\leq t\}. \end{align*} For each fixed $x\in X$, the orbit map $s\mapsto T(s)x$ is continuous on the compact interval $[0,t]$, so its range is bounded in $X$. Hence \begin{align*} \sup_{0\leq s\leq t}\|T(s)x\|_X<\infty \end{align*} for every $x\in X$. By the [Uniform Boundedness Principle](/theorems/549) applied to the pointwise bounded family $\mathcal{T}_t\subset\mathcal{L}(X)$ on the [Banach space](/page/Banach%20Space) $X$, the constant \begin{align*} M_t:=\sup_{0\leq s\leq t}\|T(s)\|_{\mathcal{L}(X)} \end{align*} is finite. Therefore \begin{align*} \|T(s)x\|_X\leq M_t\|x\|_X \end{align*} for every $s\in[0,t]$ and every $x\in X$. Therefore \begin{align*} \|T(t-k)q_k-T(t)Au_0\|_X\leq M_t\|q_k-Au_0\|_X+\|T(t-k)Au_0-T(t)Au_0\|_X, \end{align*} and the right-hand side tends to $0$ as $k\downarrow 0$. Thus the two one-sided difference quotients at $t>0$ have the same limit, and \begin{align*} u'(t)=T(t)Au_0. \end{align*} By the domain invariance already proved, \begin{align*} T(t)Au_0=A T(t)u_0=A u(t). \end{align*} Hence \begin{align*} u'(t)=Au(t) \end{align*} for every $t>0$. The same right-quotient computation at $t=0$ gives \begin{align*} \lim_{h\downarrow 0}\left\|\frac{u(h)-u(0)}{h}-Au_0\right\|_X=0. \end{align*} [/step]

[step:Verify continuity and conclude classical regularity on each finite interval] Let $S>0$ be fixed. Strong continuity of the semigroup gives continuity of the map $u:[0,S]\to X$ defined by \begin{align*} u(t)=T(t)u_0. \end{align*} The previous steps show that $u(t)\in D(A)$ for every $t\in[0,S]$, that $u$ is differentiable on $(0,S)$, and that \begin{align*} u'(t)=Au(t) \end{align*} for every $t\in(0,S)$. It remains to check the continuity of the derivative in the usual classical-solution sense. Since \begin{align*} Au(t)=A T(t)u_0=T(t)Au_0, \end{align*} and $Au_0\in X$, strong continuity of $(T(t))_{t\geq 0}$ gives continuity of the map $[0,S]\to X$ defined by \begin{align*} t\mapsto Au(t)=T(t)Au_0. \end{align*} Finally, \begin{align*} u(0)=T(0)u_0=u_0. \end{align*} Thus $u|_{[0,S]}$ satisfies the initial condition, remains in $D(A)$, is continuous as an $X$-valued map, is differentiable on $(0,S)$ with right derivative at $0$, and satisfies $u'(t)=Au(t)$. This is precisely the claimed classical solution property on $[0,S]$. [/step]