\; \left\langle \alpha,x\right\rangle < 0\\
-\bar \alpha(t) & if \; \left\langle \alpha,x\right\rangle > 0
\end{array}
\right.$$ and is not differentiable whenever $\left\langle
\alpha,x\right\rangle = 0$. However we can apply Theorem \[thdeterministicverification\] and prove the following result.

\[propqui\] The feedback map given by $$u^{op}(t,x)
= \left\lbrace
\begin{array}{ll}
- \left\langle \beta,\bar\alpha(t) \right\rangle  & if \; \left\langle \alpha,x\right\rangle \le 0\\
\left\langle \beta,\bar\alpha(t) \right\rangle & if \; \left\langle
\alpha,x\right\rangle > 0
\end{array}
\right.$$ is optimal. Similarly, also the feedback map $$\bar u^{op}(t,x)
= \left\lbrace
\begin{array}{ll}
- \left\langle \beta,\bar\alpha(t) \right\rangle  & if \; \left\langle \alpha,x\right\rangle < 0\\
\left\langle \beta,\bar\alpha(t) \right\rangle & if \; \left\langle
\alpha,x\right\rangle \ge 0
\end{array}
\right.$$ is optimal.

Let $(t,x)\in (0,T]\times \mathcal{H}$ be the initial datum. If $\left\langle \alpha,x\right\rangle \le 0$, taking the control $-\left\langle \beta,\bar\alpha(t) \right\rangle$ the associated state trajectory is $$x^{op}(s)= e^{(s-t)A}x - \int_t^{s}e^{(s-r)A} R(\left\langle
\beta,\bar\alpha(r) \right\rangle) {\mathrm{d}}r$$ and it easy to check that it satisfies $\left\langle
\alpha,x^{op}(s) \right\rangle \le 0 $ for every $s\ge t$. Indeed, using the form of $R$ and the fact that $\alpha $ is eigenvector of $A^*$ we get $$\left\langle \alpha,x^{op}(s)  \right\rangle = e^{\lambda(s-t)}
\left\langle \alpha, x \right\rangle
 - \left\langle \alpha, \beta \right\rangle
\int_t^{s}e^{\lambda(s-r)} \left\langle \beta,\bar\alpha(r)
\right\rangle {\mathrm{d}}r$$ $$=e^{\lambda(s-t)}\left\langle \alpha, x \right\rangle
 - \left\langle \alpha, 