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LQR minimizers satisfy a feasible-set variational inequality

Statement

Assume the Axioms of Countable Choice and Dependent Choice.

For the LQR data of Finite-horizon bounded-control linear-quadratic ODE problem, let $F$ be nonempty and convex and let $u \in F$ . Put $g (t) = B (t)^{T} p (t) + R (t) u (t)$ , with the adjoint from Continuous adjoint equation for the linear-quadratic ODE. If $u$ minimizes $J$ on $F$ , then $\int_{0}^{T} g (t)^{T} (v (t) - u (t)) d t \geq 0 (v \in F) .$ Conversely, if this variational inequality holds at $u$ , then $u$ is the unique minimizer on $F$ .

If $F = U_{c}$ , the variational inequality is equivalent (under Dependent Choice, hence Countable Choice) to the existence of one null set $N$ such that $g (t)^{T} (c - u (t)) \geq 0$ for every $t \notin N$ and every $c \in U_{c}$ . The pointwise condition is not asserted for additional state or terminal constraints.

Facts & Assumptions

Given: The LQR data, a nonempty convex feasible set $F$ , a control $u \in F$ , its continuous adjoint, and the variational-inequality hypothesis when proving sufficiency or the reverse pointwise implication.

[F1]

The objective has bounded first-order map $D J (u) [w] = \int_{0}^{T} (B^{T} p + R u)^{T} w d t$ , with remainder $O (∥ w ∥_{2}^{2})$ (The LQR adjoint identity gives the reduced gradient).

[F2]

The state map is affine, and for every $u, v \in L^{2}$ the quadratic expansion has the exact remainder $J (v) - J (u) - D J (u) [v - u] = \frac{1}{2} \int_{0}^{T} (δ x^{T} Q δ x + (v - u)^{T} R (v - u)) d t + \frac{1}{2} δ x (T)^{T} P δ x (T)$ , where $δ x = x_{v} - x_{u}$ (Finite-horizon bounded-control linear-quadratic ODE problem, The bounded measurable linear ODE has a unique affine state map).

[F3]

Cauchy–Schwarz makes the integral pairing finite for $L^{2}$ factors (Cauchy-Schwarz inequality for $L^{2}$ ).

[F4]

Rational open boxes form a countable basis of the product topology, and on $R^{m}$ that topology is the Euclidean metric topology ( $Q^{n}$ is a countable dense subset of $R^{n}$ , and rational open boxes form a countable basis, For $n \geq 1$ the product topology on $n$ copies of the usual topology of $R$ is the metric topology of $d_{\infty}$ on $R^{n}$ , and hence also of $d_{1}$ and $d_{2}$ , so $R^{n}$ as a product and $R^{n}$ as a metric space are one space).

[F5]

Under Countable Choice, Lebesgue measurable sets form a sigma-algebra and countable unions of measurable null sets are null by countable subadditivity (Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume, Finite and countable subadditivity of measures).

[F6]

The Lebesgue integral is linear on $L^{1}$ , and the nonnegative integral is monotone (The Lebesgue integral is linear on $L^{1} (μ)$ , Monotonicity and nonnegative homogeneity of the nonnegative integral).

[A1]

Dependent Choice and Countable Choice are available for the countable dense subset of $U_{c}$ used in the pointwise equivalence (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain, The Axiom of Countable Choice ( ${A C}_{ω}$ ), Dependent choice implies countable choice).

Proof

1.1F1givenchoosealgebra

Suppose $u$ minimizes $J$ on $F$ and fix $v \in F$ . Convexity gives $u_{θ} = u + θ (v - u) \in F$ for $0 \leq θ \leq 1$ . Hence $(J (u_{θ}) - J (u)) / θ \geq 0$ for $0 < θ \leq 1$ . The first-order expansion [F1] with increment $θ (v - u)$ has quotient $D J (u) [v - u] + O (θ ∥ v - u ∥_{2}^{2})$ , so its limit is $D J (u) [v - u]$ . This proves the displayed variational inequality.

1.2F3givenalgebra

Suppose the pointwise inequalities hold outside one null set and $F = U_{c}$ . For any $v \in U_{c}$ , substitute $c = v (t)$ outside that null set and the null set on which $v (t) \notin U_{c}$ . The vector $g = B^{T} p + R u$ lies in $L^{2}$ because $B, R$ are essentially bounded, $p$ is continuous on the compact horizon, and $u \in L^{2}$ ; [F3] makes $g^{T} (v - u)$ integrable. Integrating proves the global variational inequality.

1.3F2givenalgebra

Assume the variational inequality at $u$ . For any $v \in F$ , [F2] and $Q ⪰ 0$ , $R ⪰ r I$ give $J (v) - J (u) \geq D J (u) [v - u] + \frac{r}{2} ∥ v - u ∥_{2}^{2}$ . The linear term is nonnegative by the assumed inequality; hence $J (v) \geq J (u)$ , with strict inequality if $v \neq u$ . Thus $u$ is the unique minimizer.

2.1A1F3F4F5F6givenchoosealgebra∎

Assume the global variational inequality and suppose no common null set makes all pointwise inequalities true. By [F4] and [A1], choose one point from each rational open box meeting $U_{c}$ ; the resulting countable set $D \subseteq U_{c}$ is dense. Define $M = ⋃_{d \in D, k \geq 1} E_{d, k}$ , where $E_{d, k} = {t : g (t)^{T} (d - u (t)) \leq - 1 / k}$ . These sets are measurable, and $M$ is measurable by [F5]. Continuity in $c$ and density show that $M$ is exactly the set where some $c \in U_{c}$ violates the pointwise inequality, so the absence of a common null set means $∣ M ∣ > 0$ . By [F5], some $E_{d, k}$ has positive measure. Put $v (t) = d$ on that set and $v (t) = u (t)$ elsewhere. This measurable control lies in $U_{c}$ . On $E_{d, k}$ , $- g (t)^{T} (d - u (t)) \geq 1 / k$ ; [F6] and monotonicity give $\int_{0}^{T} g^{T} (v - u) d t \leq - ∣ E_{d, k} ∣ / k < 0$ , contradicting the global variational inequality. Therefore one null set suffices for every $c \in U_{c}$ .

Remarks

For arbitrary active state or terminal constraints the valid first-order condition is the integral inequality over the feasible set. The multiplier $p$ in the unconstrained adjoint equation alone does not encode normal multipliers for those additional constraints.

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Sources

J. Monnier, Optimal Control of ODEs: The Linear-Quadratic (LQ) Case, course notes, Chapter 2 §2.5.2 (standard reference, not scraped)
W. W. Hager, Runge-Kutta methods in optimal control and the transformed adjoint system, Numerische Mathematik 87 (2000) (standard reference, not scraped)