[proofplan]
The solution operator at time $t$ sends initial data to the value of the unique trajectory at time $t$. The identity law follows directly from the initial condition. The semigroup law follows by shifting a solution forward by time $s$ and using autonomy plus uniqueness to identify the shifted trajectory with the unique solution starting from the intermediate state. Bounded linearity and strong continuity are exactly the two remaining hypotheses translated into operator language.
[/proofplan]
[step:Use the initial condition to identify $T(0)$ with the identity operator]
Let $u_0 \in X$. By the definition of $T(0):X \to X$ and the assumed initial condition for the solution $u(\cdot;u_0):[0,\infty)\to X$,
\begin{align*}
T(0)u_0 = u(0;u_0)=u_0.
\end{align*}
Since this holds for every $u_0 \in X$, the operator $T(0)$ equals the identity map $I_X:X\to X$.
[/step]
[step:Shift a solution and apply uniqueness to prove the semigroup law]
Fix $s,t \ge 0$ and fix $u_0 \in X$. Define the shifted trajectory $v_s:[0,\infty)\to X$ by
\begin{align*}
v_s(r)=u(r+s;u_0)
\end{align*}
for every $r \ge 0$. By the assumed time-translation stability of the solution class and the autonomy of the Cauchy problem, $v_s$ is an admissible solution of the same problem with initial value
\begin{align*}
v_s(0)=u(s;u_0)=T(s)u_0.
\end{align*}
The uniqueness hypothesis applied to the initial datum $T(s)u_0 \in X$ therefore gives
\begin{align*}
v_s(r)=u(r;T(s)u_0)
\end{align*}
for every $r \ge 0$. Evaluating this identity at $r=t$ gives
\begin{align*}
u(t+s;u_0)=u(t;T(s)u_0).
\end{align*}
Using the definition of $T(t+s)$ and $T(t)$, this becomes
\begin{align*}
T(t+s)u_0=T(t)(T(s)u_0).
\end{align*}
Since $u_0 \in X$ was arbitrary, $T(t+s)=T(t)T(s)$ as maps from $X$ to $X$.
[guided]
Fix two times $s,t \ge 0$ and an initial state $u_0 \in X$. The goal is to compare two ways of reaching time $t+s$: first solve directly from $u_0$ up to time $t+s$, and second solve up to time $s$ and then restart the same autonomous problem from the intermediate state.
Define the shifted trajectory $v_s:[0,\infty)\to X$ by
\begin{align*}
v_s(r)=u(r+s;u_0)
\end{align*}
for every $r \ge 0$. This definition is meaningful because $r+s \ge 0$, so $u(r+s;u_0)$ lies in the domain of the original solution trajectory. Its initial value is
\begin{align*}
v_s(0)=u(s;u_0)=T(s)u_0.
\end{align*}
Now we use the two structural hypotheses on the Cauchy problem. Time-translation stability says that the shifted trajectory remains in the admissible solution class. Autonomy says that after shifting the time origin, the equation being solved is the same equation, not a different time-dependent equation. Therefore $v_s$ is an admissible solution of the same homogeneous autonomous Cauchy problem with initial value $T(s)u_0$.
There is also, by hypothesis, a unique admissible solution with initial value $T(s)u_0$. That unique solution is denoted $u(\cdot;T(s)u_0):[0,\infty)\to X$. Since both $v_s$ and $u(\cdot;T(s)u_0)$ are admissible solutions with the same initial value, uniqueness gives equality of trajectories:
\begin{align*}
v_s(r)=u(r;T(s)u_0)
\end{align*}
for every $r \ge 0$. Evaluating at $r=t$ yields
\begin{align*}
u(t+s;u_0)=u(t;T(s)u_0).
\end{align*}
Finally, translate this identity into the notation of solution operators. The left-hand side is $T(t+s)u_0$, while the right-hand side is $T(t)(T(s)u_0)$. Hence
\begin{align*}
T(t+s)u_0=T(t)(T(s)u_0).
\end{align*}
Because the initial state $u_0 \in X$ was arbitrary, the two maps $T(t+s):X\to X$ and $T(t)T(s):X\to X$ agree on all of $X$. Thus $T(t+s)=T(t)T(s)$.
[/guided]
[/step]
[step:Translate bounded linear dependence on the initial datum into bounded linear operators]
Let $t \ge 0$. By hypothesis, the map $u_0 \mapsto u(t;u_0)$ is a bounded [linear map](/page/Linear%20Map) from $X$ to $X$. Since $T(t):X\to X$ is defined by $T(t)u_0=u(t;u_0)$, it follows that $T(t)$ is bounded and linear. Thus $T(t)\in \mathcal{L}(X)$ for every $t \ge 0$.
[/step]
[step:Use trajectory continuity to obtain strong continuity]
Let $u_0 \in X$ and let $t_0 \ge 0$. By hypothesis, the map $t \mapsto u(t;u_0)$ is continuous from $[0,\infty)$ to $X$. Therefore,
\begin{align*}
\lim_{t\to t_0}\|u(t;u_0)-u(t_0;u_0)\|_X=0.
\end{align*}
Using the definition $T(t)u_0=u(t;u_0)$, this is exactly
\begin{align*}
\lim_{t\to t_0}\|T(t)u_0-T(t_0)u_0\|_X=0.
\end{align*}
Since $u_0 \in X$ and $t_0 \ge 0$ were arbitrary, $(T(t))_{t\ge 0}$ is strongly continuous. Together with $T(0)=I_X$, the semigroup identity, and $T(t)\in\mathcal{L}(X)$ for every $t \ge 0$, this proves that $(T(t))_{t \ge 0}$ is a strongly continuous semigroup on $X$.
[/step]
