by property [**LS4**]{}, belongs to the funnel $S(t_0, a)$. Since $u$ maximizes $\zeta$ over $S(t_0, a)$, $\zeta(u) \ge \zeta(w)$, i.e., $$\int_0^{T(t_0, a)} e^{-\lambda t} \vp(u(t_0 + t; t_0, a))\,dt \ge \int_0^{T(t_0, a)} e^{-\lambda t} \vp(w(t_0 + t; t_0, a))\,dt\,.$$ Now compute $$\begin{aligned}
& \int_0^{T(t_0, a)} e^{-\lambda t} \vp(w(t_0 + t; t_0, a))\,dt = \\ 
& \int_0^\tau e^{-\lambda t} \vp(u(t_0 + t; t_0, a))\,dt + 
\int_\tau^{T(t_0, a)} e^{-\lambda t} \vp(v(t_0 + t; t_0 + \tau, u(t_0 + \tau; t_0, a))\,dt = \\ 
& \int_0^\tau e^{-\lambda t} \vp(u(t_0 + t; t_0, a))\,dt + e^{-\lambda \tau}\,\int_0^{T(t_0, a) 
- \tau} e^{-\lambda t} \vp(v(t_0 + \tau + t; t_0 + \tau, u(t_0 + \tau; t_0, a) ))\,dt\,,
\end{aligned}$$ and $$\begin{aligned}
& \int_0^{T(t_0, a)} e^{-\lambda t} \vp(u(t_0 + t; t_0, a))\,dt = \\ 
& \int_0^\tau e^{-\lambda t} \vp(u(t_0 + t; t_0, a))\,dt + e^{-\lambda \tau}\,\int_0^{T(t_0, a) - \tau} e^{-\lambda t} \vp(u(t_0 + \tau + t; t_0, a))\,dt\,.
\end{aligned}$$ Hence, $$\begin{aligned}
& \zeta(\sigma_\tau u) = \int_0^{T(t_0, a) - \tau} e^{-\lambda t} \vp(u(t_0 + \tau + t; t_0, a))\,dt \ge 
 \\ 
& \int_0^{T(t_0, a) 
- \tau} e^{-\lambda t} \vp(v(t_0 + \tau + t; t_0 + \tau, u(t_0 + \tau; t_0, a) ))\,dt  = \zeta(v),
\end{aligned}$$ which means $\sigma_\tau 