In [Li, Xungjing, and Jiongmin Yong. Optimal control theory for infinite dimensional systems.1995] at page 241 it is claimed that: \begin{equation} \lim _{s \downarrow t}\left|e^{A^{*}(s-t)} \frac{y_{t, x}(s)}{\left|y_{t, x}(s)\right|}-\frac{x}{|x|}\right|=0 \end{equation}
uniformly in $u(\cdot) \in \mathcal{U}[t, T]$. Here the state equation is $$y_{t, x}(s)=e^{A(s-t)} x+\int_{t}^{s} e^{A(s-r)} f\left(r, y_{t, x}(r), u(r)\right) d r, \quad s \in[t, T], x \in X$$ where $X$ is Hilbert, $A : D(A) \subset X \to X$ linear operator is the generator of a strongly continuous semigroup of contractions $|e^{At}|\leq 1$, $U$ a metric space in which the control $u(\cdot)$ takes values $u(\cdot) \in \mathcal{U}[0, T] \equiv$ $\{u:[0, T] \rightarrow U \mid u(\cdot)$ measurable $\}$, $f : [0,T] \times X \times U \to X$ satisfies $$|f(t, x, u)-f(\bar{t}, \bar{x}, u)| \leq L|x-\bar{x}|+\omega(|t-\bar{t}|)$$ $\forall t, \bar{t} \in[0, T], x, \bar{x} \in X, u \in U$ and modulus of continuity $\omega$. $$|f(t, 0, u)| \leq L, \quad \forall(t, u) \in[0, T] \times U$$. Note that $A^*$ is the adjoint of $A$ and $e^{A^* t}$ is a strongly continuous semigroup of contractions with generator $A^*$.

I don't manage to prove the claim, how do you show that?

  • $\begingroup$ As a start, can you prove $y_{t,x}(s)\to x$ as $s\searrow t$? $\endgroup$
    – MaoWao
    Jul 11 at 10:03
  • $\begingroup$ yes of course @MaoWao $\endgroup$
    – carlos85
    Jul 11 at 10:07
  • $\begingroup$ Well, if $\xi_\alpha\to \xi$, then $\xi_\alpha/\|\xi_\alpha\|\to \xi/\|\xi\|$ (provided $\xi\neq 0$). And if $(T_\alpha)$ is uniformly bounded and converges strongly to $T$ and $\eta_\alpha\to\eta$, then $T_\alpha \eta_\alpha\to T\eta$. These two facts are not hard to prove and combined they give the desired convergence. $\endgroup$
    – MaoWao
    Jul 11 at 10:52
  • $\begingroup$ @MaoWao for $T_\alpha$ I think you mean $e^{A*(t-s)}$, ok it is uniformly bounded but does it converge strongly to the identity $I$ for $t \to s$? $\endgroup$
    – carlos85
    Jul 12 at 9:50
  • $\begingroup$ It does, but this is indeed not obvious (depending on your definition of $e^{(s-t)A^\ast}$. I provided a reference in my answer. $\endgroup$
    – MaoWao
    Jul 12 at 10:26

As discussed in the comments, $y_{t,x}(s)\to x$ as $s\searrow t$. Thus $y_{t,x}(s)/|y_{t,x}(s)|\to x/|x|$.

Moreover, $(e^{rA^\ast})_{r\geq 0}$ is a strongly continuous semigroup. I would even say this is implicit in the definition, i.e., $(e^{rA^\ast})$ is defined to be the $C_0$-semigroup generated by $A^\ast$. In any case, the fact that the adjoint of a generator of a $C_0$-semigroup on a Hilbert space (or more generally on a reflexive Banach space) is again a generator of a $C_0$-semigroup is discussed in 5.14 of Engel, Nagel: One-Parameter Semigroups for Linear Evolution Equations.

This result relies on two ingredients: First, it is not hard to see that the adjoint of a $C_0$-semigroup is weak$^\ast$-continuous, hence weakly continuous if the underlying Banach space is reflexive. Then, and that's the harder part, one uses the uniform boundedness principle and the Hahn-Banach theorem to show that every weakly continuous semigroup is in fact strongly continuous.

Therefore, $e^{(s-t)A^\ast}\to \mathrm{id}_X$ strongly as $s\searrow t$ and $s\mapsto e^{(s-t)A^\ast}$ is norm bounded on compact intervals. Elementary estimates then show that $$ \lim_{s\searrow t}e^{(s-t) A^\ast}\frac{y_{t,x}(s)}{|y_{t,x}(s)|}= \frac{x}{|x|}. $$

  • $\begingroup$ I'm not seeing again why $e^{(t-s)A^*} \to Id$ strongly: you said that $e^{rA^*}$ is a strongly continuous semigroup. But then we can say that $e^{(t-s)A^*} x \to x$ for every fixed $x \in X$ not that the operator norm tends to zero: $|e^{(t-s)A^*} - Id| \to 0$. The rest is clear $\endgroup$
    – carlos85
    Jul 12 at 22:40
  • $\begingroup$ Strong convergence of operators does not mean that the operator norm of the difference goes to zero (which is indeed not necessarily true in this setting). It means that the operators converge pointwise, i.e., exactly what you get from strong continuity of the semigroup. $\endgroup$
    – MaoWao
    Jul 13 at 6:47
  • $\begingroup$ Ah ok thanks for clarifying! $\endgroup$
    – carlos85
    Jul 13 at 9:22
  • $\begingroup$ to prove this result we don't need a contraction semigroup but just a uniform bounded semigroup $|e^{At}|\leq M$ right? $\endgroup$
    – carlos85
    Jul 18 at 22:20
  • $\begingroup$ @carlos85 You don't even need that, you only need that $t\mapsto e^{tA}$ is norm bounded on every compact interval, which holds for every strongly continuous semigroup (in fact you always have a bound of the type $|e^{tA}|\leq M e^{\omega t}$). $\endgroup$
    – MaoWao
    Jul 19 at 6:38

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