Consider a system of ODEs
$$\dot{x}(t)=f(x(t))+g(x(t))u(t),$$
where $f:\mathbb{R}^n\to\mathbb{R}^n$ and $g:\mathbb{R}^n\to\mathbb{R}^{n\times m}$ are smooth. Let $L:\mathbb{R}^n\times \mathbb{R}^n\to [0,\infty)$ denote the controllability function, that is
$$L(x_o,x_f):=\text{The minimum amount of control energy (that is, $||u||^2_{L_2}$) required to }$$ $$\text{drive $x$ from $x(t_0)=x_0$ to $x(t_f)=x_f$, for any $t_f\geq t_0$.}$$
I would formalise the above as
$$L(x_0,x_f):=\inf_{t_0\leq t_f}\left(\inf_{u\in A_{t_0,t_f}} \int_{t_0}^{t_f}||u(t)||^2dt\right),\quad\quad (*)$$
where
$$A_{t_0,t_f}:=\{u\in L_2([t_0,t_f],\mathbb{R}^m) : \quad x(t_0)=x_0,\quad x(t_f)=x_f\}.$$
However, usually its simply stated (for example, here) that
$$L(x_0,x_f):=\inf_{u\in B} \int_{-\infty}^{0}||u(t)||^2dt,\quad\quad (**)$$
where
$$B:=\{u\in L_2((-\infty,0],\mathbb{R}^m): \quad \lim_{t\to-\infty}x(t)=x_0,\quad x(0)=x_f\}.$$
Is it true that $(*)=(**)$? If so why? Furthermore, if there is some $u_1$ and $u_2$ that achieve $(*)$ and $(**)$, respectively, is there any relationship between $u_1$ and $u_2$ (and similarly, between the corresponding trajectories $x_1$ and $x_2$)?