[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*}
[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]
[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]
