Asymptotic behavior of controls

Section 4 Asymptotic 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.

Lemma 4.1.

For every vector \(X\in V_1\text{,}\)

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

Proof.

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

\begin{equation} 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} \end{equation}

Since the curve \(a\) is absolutely continuous, the derivative condition PMP 3.1 i implies that

\begin{equation} \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} \end{equation}

By the subdifferential condition PMP 3.1 ii, 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*}

Lemma 4.2.

Let \(\dilationfactor_k\to \infty\) be a diverging sequence and let \(u_{\dilationfactor_k}(t)=u(\dilationfactor_kt)\) be the corresponding dilated controls. If \(u_{\dilationfactor_k}\to \tilde{u}\) in \(\Lloc([0,\infty);V_1)\text{,}\) then \(\tilde{u}(t)\in\ker B\) for almost every \(t\in[0,\infty)\text{.}\)

Proof.

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

\begin{equation} \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} \end{equation}

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

\begin{equation} \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} \end{equation}

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 \(\intavg_{a}^b\tilde{u}(t)\,dt\in\ker B\text{.}\)