I am reading the following note:
https://arxiv.org/pdf/1309.3112v1.pdf
Please see Ch. 1.3 (p.6~p.7).
On p.7, it says the infimum is not attained with a control law $u(t)$ belonging to the space of Lebesgue integrable functions.
But for $t\in [0,1]$, $$\int_0^1 |u_k(t)| dt < \infty,$$ for $k\rightarrow \infty $. I am confused why $u(t)$ may not be Lebesgue integrable.
The minimizing sequence $u_k$ is bounded in $L^1$, oscillatory, which means that it converges weakly to the zero function. However, the functional is not weakly lower semi-continuous with respect to $u$, so in the limit the function $u$ is not a minimizer.
This shows that the infimum of the functional is zero. If there is $(x,u)$ such that the functional is zero, then necessarily $x\equiv 0$, $|u|\equiv 1$, a contradiction to $\dot x = u$.