Let $F:H^s\rightarrow H^s$ and suppose I have proved LWP for some PDE
\begin{align*} \partial_t u &= Fu \qquad \text{on }\, \mathbf{R}\times [0,\infty) \\ u(x,0)&=u_0(x) \qquad \text{on }\, \mathbf{R} \end{align*}
and in particular, given $u_0\in H^s$, there is a time $T$ such that $u\in C([0,T);H^s(\mathbf{R}))$. Furthermore, I have proven that
\begin{align*} \sup_{0\leqslant t < T}\lvert\lvert u(\cdot,t)\rvert\rvert_{H^s} < \infty \end{align*}
Is it the case that $u(x,T)\in H^s(\mathbf{R)}$?
I have argued yes, since letting $u(x,T)=:\lim_{t\rightarrow T}u(x,t)$ means that by Fatou's
\begin{align*} \lvert\lvert u(\cdot,T)\rvert\rvert_{H^s} \leqslant \liminf_{t\rightarrow T}\, \lvert\lvert u(\cdot,t)\rvert\rvert_{H^s} \leqslant \sup_{0\leqslant t < T}\lvert\lvert u(\cdot,t)\rvert\rvert_{H^s} < \infty \end{align*}
However, this doesn't make much sense to me, because in this case couldn't we easily get LWP for $u\in C_t([0,T];H^s(\mathbf{R}))$ and by a similar argument keep extending the regularity to the rest of the space? I always thought we lost LWP regularity since we could no longer control the $H^s$ norm at $T$ (via Grönwall or some similar argument).
If it helps I have also proved that there are solutions with $u_0\in H^s$ which lose their $H^s$ regularity in finite time.
Where have I made a mistake? And is my intuition about losing control on a Sobolev norm giving us the endpoint for LWP correct?
Solved:
I have abused Fatou's lemma in a subtle way. Fatou's lemma requires setting $u(x,T):=\liminf_{t\rightarrow T}u(x,t)$ which does not equal the limit in general. All I have proved is that the limit inferior is finite... which is true. The pointwise convergence in this case is suspect.
It is also instructive to see why this fails to be true in general. The condition that $\sup_{t<T}\lvert\lvert u(\cdot, t)\rvert\rvert_{H^s}$ says that $u(x, t)$ is a bounded sequence in the space $C_t H^s$, which of course does not imply convergence. In fact, since I know there are solutions which lose their regularity at time $T$, this set is guaranteed not to be compact; hence we cannot conclude that $u(x,T)\in H^s$.