Given $T > 0$, standard theory implies there is a unique $u_T \in L^2(0,T;V)$ with $u_T' \in L^2(0,T;V^*)$ such that $$u_T' + Au_T = f\quad\text{as an equality in $L^2(0,T;V^*)$}$$ $$u_T(0) = u_0$$ where $f \in L^2(0,T;V^*)$, $u_0 \in H$ and $A$ is some smooth elliptic operator. Here $V \subset H \subset V^*$ is Gelfand triple.

Given any $T_\infty > T$, it follows by uniqueness of solutions that $u_{T_\infty}$ restriced to $[0,T]$ is $u_T:$ $$u_{T_\infty}|_{[0,T]} = u_T$$.

So then this means that there is a global solution to the PDE in $L^2_{loc}(0,\infty;V)$, right? Or is the $loc$ not necessary?

What is wrong with this idea?

For any $n > 0$, standard theory implies there is a unique $u_n \in L^2(0,n;V)$ with $u_n' \in L^2(0,n;V^*)$ such that $$u_n' + Au_n = f\quad\text{as an equality in $L^2(0,n;V^*)$}$$ $$u_n(0) = u_0$$ where $f \in L^2(0,n;V^*)$, $u_0 \in H$ and $A$ is some smooth elliptic operator. Here $V \subset H \subset V^*$ is Gelfand triple.

If $n > m$, then we see that $u_n|_{[0,m]} = u_m$ by uniqueness of solutions.

Now define $v(t) = u_n(t)$ if $t \leq n$.

Then isn't $v$ in some sense a global solution of the PDE (assuming we have $f \in L^2(0,\infty;V^*)$)?

My question is this is sense of a global solution useful or not; what is the usual sense of a global solution? Because I have read many times "if the norm of the solution stays bounded then we can extend the solution globally" but never got any details.