## Section4Asymptotic behavior of controls

In this section, let $u\colon [0,\infty)\to V_1$ be a fixed control satisfying the PMP 3.1. Let $a\colon[0,\infty)\to V_1^*$ be the associated curve of subdifferentials and let $B\colon V_1\times V_1\to\RR$ be the associated bilinear form.

Fix an arbitrary vector $X\in V_1\text{.}$ Bilinearity of the map $B$ implies that

$$B\left(\intavg_{0}^Tu(t)\,dt,X\right) = \frac{1}{T}\int_{0}^TB\left(u(t),X\right)\,dt\text{.}\label{eq-B-integral-average}\tag{4.1}$$

Since the curve $a$ is absolutely continuous, the derivative condition PMP 3.1i implies that

$$\int_{0}^{T}B(u(t),X) = \int_{0}^{T}\frac{d}{dt}a(t)X = a(T)X-a(0)X\text{.}\label{eq-B-integral-bound-in-T}\tag{4.2}$$

By the subdifferential condition PMP 3.1ii, for almost every $T\text{,}$ the linear map $a(T)$ is a subdifferential of the squared norm $\frac{1}{2}\norm{\cdot}^2$ at the point $u(T)\text{.}$ Since $\norm{u(T)}\equiv 1$ is constant, continuity of the curve $a$ and Lemma 2.19 imply the bound $\abs{a(T)X} \leq \norm{X}$ for every $T\in[0,\infty)\text{.}$ The identities (4.1) and (4.2) then imply the desired conclusion that

\begin{equation*} \lim\limits_{T\to\infty}\abs{B\left(\intavg_{0}^Tu(t)\,dt,X\right)} \leq \lim\limits_{T\to\infty}\frac{2}{T}\norm{X} = 0\text{.} \end{equation*}

By the Lebesgue differentiation theorem it suffices to prove that $\intavg_{a}^b\tilde{u}(t)\,dt\in \ker B$ for any $0\leq a\lt b\lt \infty\text{.}$

Fix $0\leq a\lt b\lt \infty\text{.}$ By assumption $u_{\dilationfactor_k}\to \tilde{u}$ in $\mathrm{L}^2([a,b];V_1)\text{,}$ so there exists some error term $\epsilon\colon \NN\to V_1$ with $\lim\limits_{k\to\infty}\epsilon(k)=0$ such that

$$\intavg_{a}^b\tilde{u}(t)\,dt = \intavg_a^bu(\dilationfactor_kt)\,dt +\epsilon(k) = \intavg_{a\dilationfactor_k}^{b\dilationfactor_k}u(t)\,dt+\epsilon(k)\text{.}\label{eq-limit-intavg}\tag{4.3}$$

The right-hand integral average can further be expressed as a difference of integral averages as

$$\intavg_{a\dilationfactor_k}^{b\dilationfactor_k}u(t)\,dt = \frac{b}{b-a}\cdot\intavg_{0}^{b\dilationfactor_k}u(t)\,dt-\frac{a}{b-a}\cdot\intavg_{0}^{a\dilationfactor_k}u(t)\,dt\text{.}\label{eq-difference-of-intavg}\tag{4.4}$$

Lemma 4.1 implies that for any $X\in V_1$

\begin{equation*} \lim\limits_{k\to\infty}B\left(\intavg_{0}^{b\dilationfactor_k}u(t)\,dt,X\right) = \lim\limits_{k\to\infty}B\left(\intavg_{0}^{a\dilationfactor_k}u(t)\,dt,X\right) = 0\text{.} \end{equation*}

Combining the identities (4.3) and (4.4) and using bilinearity of $B$ then implies that $B\left(\intavg_a^b\tilde{u}(t)\,dt,X\right)=0\text{.}$ Since the vector $X\in V_1$ was arbitrary, this proves the desired claim that $\intavg_{a}^b\tilde{u}(t)\,dt\in\ker B\text{.}$