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Weakly convergent bounded controls have uniformly convergent states

Statement

For the data of Finite-horizon bounded-control linear-quadratic ODE problem, if $u_{k}, u \in U_{c}$ and $u_{k} ⇀ u$ weakly in $L^{2} (0, T; R^{m})$ , then $x_{u_{k}} \to x_{u}$ uniformly on $[0, T]$ .

Facts & Assumptions

Given: The bounded measurable linear state equation, the compact control set $U_{c}$ , and a sequence $u_{k} ⇀ u$ in the pointwise admissible set.

[F1]

The state map has a unique continuous solution and its zero-initial-state part is bounded and linear from $L^{2}$ to $C ([0, T]; R^{n})$ (The bounded measurable linear ODE has a unique affine state map).

[F2]

Every control in $U_{c}$ satisfies $∥ u_{k} (t) ∥_{2} \leq M_{c} : = \sup_{v \in U_{c}} ∥ v ∥_{2} < \infty$ almost everywhere (Finite-horizon bounded-control linear-quadratic ODE problem and compactness of $U_{c}$ ).

[F3]

Weak convergence means convergence under every bounded linear functional on $L^{2}$ (Weak convergence of nets and sequences).

[F4]

For a Lebesgue-integrable vector function $h$ , the componentwise integral obeys $∥ \int h ∥_{2} \leq \int ∥ h ∥_{2}$ : if $I = \int h \neq 0$ , test the integral against $I / ∥ I ∥_{2}$ and use componentwise linearity, Euclidean Cauchy--Schwarz, scalar integral monotonicity, and $∣ \int q^{T} h ∣ \leq \int ∣ q^{T} h ∣$ (The Lebesgue integral is linear on $L^{1} (μ)$ , Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation, The modulus of an integral is bounded by the integral of the modulus, Monotonicity and nonnegative homogeneity of the nonnegative integral); if $I = 0$ the bound is immediate.

Proof

1.1F1F2F4algebra

By [F1] and [F2], the states have a common bound $M_{x}$ in supremum norm. For $s, t \in [0, T]$ , the integral equation and [F4] give $∥ x_{u_{k}} (t) - x_{u_{k}} (s) ∥_{2} \leq L ∣ t - s ∣$ , where $L : = ∥ A ∥_{\infty} M_{x} + ∥ B ∥_{\infty} M_{c}$ . The same bound holds for $x_{u}$ .

1.2F1F3

For each fixed $t$ , evaluation at $t$ composed with the bounded linear zero-initial-state map in [F1] is a bounded linear functional on $L^{2}$ . Thus [F3] gives $x_{u_{k}} (t) \to x_{u} (t)$ coordinatewise.

2.1step 1.1step 1.2choosealgebra∎

If $L = 0$ , step 1.1 makes each state constant, and step 1.2 gives uniform convergence. If $L > 0$ , fix $ε > 0$ and choose a finite partition of $[0, T]$ with mesh smaller than $ε / (3 L)$ . By step 1.2 and finiteness of its nodes, there is an index after which $∥ x_{u_{k}} - x_{u} ∥_{2} < ε / 3$ at every node. Any $t$ is within the mesh of a node, so step 1.1 gives $∥ x_{u_{k}} (t) - x_{u} (t) ∥_{2} \leq L ∣ t - t_{j} ∣ + ∥ x_{u_{k}} (t_{j}) - x_{u} (t_{j}) ∥_{2} + L ∣ t - t_{j} ∣ < ε$ . Hence the convergence is uniform.

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Sources

J. Monnier, Optimal Control of ODEs: The Linear-Quadratic (LQ) Case, course notes, Chapter 2 §2.5.1 (standard reference, not scraped)