Proof of Weak Lower Semicontinuity of the Running-Cost Integral along a State Path

Preliminaries. Since $\mathcal{A}$ is compact it is closed by Compact Subset of $\mathbb{R}^n$ is Closed and bounded by Compact Subset of $\mathbb{R}^n$ is Bounded, so there are $x\in\mathbb{R}^{m}$ and a real $R'\ge0$ with $d_{E}(x,a)\le R'$ for every $a\in\mathcal{A}$, where $d_{E}$ is the Euclidean distance. Putting $R=|x|+R'$ and using the triangle inequality for the Euclidean norm (Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$) gives $|a|\le|a-x|+|x|=d_{E}(x,a)+|x|\le R$ for every $a\in\mathcal{A}$. Hence The Set of Controls with Values in a Closed Bounded Set is Nonempty, Bounded, Convex and Closed applies to $\mathcal{A}$ and $R$: $\mathcal{U}_{\mathcal{A}}$ is nonempty, bounded in norm by $RT^{1/2}$, convex and closed. By claim 1 of Boundedness and Uniform Continuity of Population Cost Data on a Compact Control Set there is a real $C\ge0$ with $|L(\Sigma,\alpha)|\le C$ for all $\Sigma\in\Delta^{l}$ and $\alpha\in\mathcal{A}$. Next, every element of $\mathcal{U}_{\mathcal{A}}$ has a representative all of whose values lie in $\mathcal{A}$; such representatives are called normalized below. Indeed, let $\xi\in\mathcal{U}_{\mathcal{A}}$. By the definition of the set of $\mathcal{A}$-valued controls there are a representative $u$ of $\xi$ and a set $N\in\mathcal{B}$ with $\lambda(N)=0$ such that $u(t)\in\mathcal{A}$ for every $t\in[0,T]\setminus N$. Since $\mathcal{A}$ is nonempty, fix $a_{0}\in\mathcal{A}$ and define $u'(t)=u(t)$ for $t\in[0,T]\setminus N$ and $u'(t)=a_{0}$ for $t\in N$. Then $u'(t)\in\mathcal{A}$ for every $t\in[0,T]$. Each component of $u'$ is measurable: for a Borel subset $B$ of $\mathbb{R}$, the preimage of $B$ under the $i$-th component of $u'$ is $\bigl((u^{i})^{-1}(B)\setminus N\bigr)\cup N_{B}$, where $N_{B}=N$ if the $i$-th component of $a_{0}$ lies in $B$ and $N_{B}=\emptyset$ otherwise, and both parts lie in $\mathcal{B}$. Since every value of $u'$ lies in $\mathcal{A}$ we have $|u'(t)|^{2}\le R^{2}$ for every $t$, and $\lambda([0,T])=T<\infty$, so $u'$ is square-integrable by claim 1 of Linearity and Monotonicity of the Lebesgue Integral applied to the nonnegative functions $|u'|^{2}$ and the constant $R^{2}$. Finally $u'$ agrees with $u$ off the null set $N$, so $u'$ represents the same element $\xi$. Finally, $\lambda([0,T])=T$ by the interval toolkit, and modifying an integrand on a set of $\lambda$-measure zero changes neither its measurability nor its integral, again by that lemma.

Claim 1. Let $S$ be an admissible state path and let $\xi\in\mathcal{U}_{\mathcal{A}}$. By the Preliminaries, $\xi$ has a normalized representative, that is, a representative $u$ with $u(t)\in\mathcal{A}$ for every $t\in[0,T]$; this is the existence assertion of claim 1. Let $u$ be such a representative. Put

$$E=\bigl\{(\Sigma,\alpha)\in\mathbb{R}^{l+m}\;:\;\Sigma\in\Delta^{l},\ \alpha\in\mathcal{A}\bigr\},$$

a nonempty subset of $\mathbb{R}^{l+m}$, and let $g:E\to\mathbb{R}$ be given by $g(\Sigma,\alpha)=L(\Sigma,\alpha)$. The map $g$ is sequentially continuous on $E$: if $(\Sigma_{n},\alpha_{n})\in E$ and $(\Sigma,\alpha)\in E$ with the Euclidean distances $d\bigl((\Sigma_{n},\alpha_{n}),(\Sigma,\alpha)\bigr)$ converging to $0$, then, since the squared distance is the sum of the squared distances of the two blocks of coordinates, $d(\Sigma_{n},\Sigma)$ and $d(\alpha_{n},\alpha)$ converge to $0$ as well, so claim 1 of Population Cost Data gives $L(\Sigma_{n},\alpha_{n})\to L(\Sigma,\alpha)$.

The map $t\mapsto(S_{t},u(t))$ takes values in $E$ and each of its $l+m$ components is measurable, by the definition of an admissible state path and because $u\in\mathcal{L}^{2}([0,T];\mathbb{R}^{m})$. Hence $t\mapsto L(S_{t},u(t))$ is measurable by Sequentially Continuous Functions of Measurable Euclidean Maps are Measurable. It satisfies $|L(S_{t},u(t))|\le C$ for every $t$, so by monotonicity of the integral (claim 1 of Linearity and Monotonicity of the Lebesgue Integral) $\int_{[0,T]}|L(S_{t},u(t))|\,d\lambda\le CT<\infty$ and the map is integrable.

If $u'$ is another normalized representative of $\xi$, then $u=u'$ off a set of measure zero, so the two integrands agree off a set of measure zero and the two integrals coincide. Hence $\Phi_{S}(\xi)$ is well defined.

Finally, by claim 2 of Linearity and Monotonicity of the Lebesgue Integral and monotonicity,

$$|\Phi_{S}(\xi)|\le\int_{[0,T]}\bigl|L(S_{t},u(t))\bigr|\,d\lambda\le CT,$$

with $C$ as in the preliminaries. This bound does not depend on $S$ or on $\xi$, so for sequences $(S^{n})$ of admissible state paths and $(\xi_{n})$ in $\mathcal{U}_{\mathcal{A}}$ one has $|\Phi_{S^{n}}(\xi_{n})|\le CT<CT+1$ for every $n$, with $CT+1>0$; hence the real sequence $\bigl(\Phi_{S^{n}}(\xi_{n})\bigr)_{n}$ is a bounded sequence, and the limit inferiors appearing in claims 3 and 4 are defined.

Claim 2, convexity. Let $\xi,\xi'\in C_{M}$ and let $s$ be real with $0\le s\le1$; let $u,u'$ be normalized representatives. For every $t\in[0,T]$ the values $u(t)$ and $u'(t)$ lie in $\mathcal{A}$, which is convex, so $su(t)+(1-s)u'(t)\in\mathcal{A}$. The map sending a pair $(a,b)\in\mathbb{R}^{2m}$ to $sa+(1-s)b$ is sequentially continuous, and $t\mapsto(u(t),u'(t))$ is componentwise measurable, so $su+(1-s)u'$ is measurable by Sequentially Continuous Functions of Measurable Euclidean Maps are Measurable; all its values lie in $\mathcal{A}$, hence $|su(t)+(1-s)u'(t)|^{2}\le R^{2}$ for every $t$, and it is square-integrable by claim 1 of Linearity and Monotonicity of the Lebesgue Integral as above. It represents $s\xi+(1-s)\xi'$, which lies in $\mathcal{U}_{\mathcal{A}}$ because $\mathcal{U}_{\mathcal{A}}$ is convex by the Preliminaries. Hence $su+(1-s)u'$ is a normalized representative of $s\xi+(1-s)\xi'$. By the convexity of $L$ in the control, for every $t$

$$L\bigl(S_{t},su(t)+(1-s)u'(t)\bigr)\le s\,L\bigl(S_{t},u(t)\bigr)+(1-s)\,L\bigl(S_{t},u'(t)\bigr),$$

and integrating, using claim 2 of Linearity and Monotonicity of the Lebesgue Integral for both the monotonicity and the linearity of the integral of signed integrands, gives

$$\Phi_{S}\bigl(s\xi+(1-s)\xi'\bigr)\le s\,\Phi_{S}(\xi)+(1-s)\,\Phi_{S}(\xi')\le sM+(1-s)M=M .$$

Hence $s\xi+(1-s)\xi'\in C_{M}$.

Claim 2, closedness. The metric open subsets of $H$ form a topology by Metric Open Sets Form a Topology. Let $\zeta$ belong to the closure of $C_{M}$. Since $C_{M}\subseteq\mathcal{U}_{\mathcal{A}}$ and $\mathcal{U}_{\mathcal{A}}$ is closed, $\zeta\in\mathcal{U}_{\mathcal{A}}$; let $u$ be a normalized representative of $\zeta$.

For $n\in\mathbb{N}$ let $B_{n}$ be the set of normalized representatives $v$ of elements of $C_{M}$ with $d_{L^{2}}(\zeta,[v])<\iota(n)^{-1}$, where $\iota$ is the canonical map of $\mathbb{R}$, so $\iota(n)^{-1}>0$ by claim 3 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field. Each $B_{n}$ is nonempty by Characterization of the Closure in a Metric Space by Open Balls, so Axiom of Countable Choice yields a sequence $(u_{n})_{n\in\mathbb{N}}$ with $u_{n}\in B_{n}$. The real sequence $\bigl(\lVert[u_{n}]-\zeta\rVert_{L^{2}}\bigr)_{n}$ has limit $0$: its $n$-th term is nonnegative and at most $\iota(n)^{-1}$, the sequence $\bigl(\iota(n)^{-1}\bigr)_{n}$ has limit $0$ by claim 3 of The Archimedean Property of the Real Numbers, and null domination, claim 3 of Order Properties of Limits of Real Sequences, applies.

By claim 2 of Completeness of the Lebesgue Space of Square-Integrable Vector-Valued Functions there are natural numbers $n_{1}<n_{2}<\dots$ and a set $N\in\mathcal{B}$ with $\lambda(N)=0$ such that $\bigl(u_{n_{j}}(t)\bigr)_{j}$ converges to $u(t)$ in $\mathbb{R}^{m}$ for every $t\in[0,T]\setminus N$. Define $\tilde u_{j}(t)=u_{n_{j}}(t)$ for $t\notin N$ and $\tilde u_{j}(t)=u(t)$ for $t\in N$; each $\tilde u_{j}$ is a normalized representative of $[u_{n_{j}}]$ (it differs from $u_{n_{j}}$ only on $N$, and all its values lie in $\mathcal{A}$), and now $\tilde u_{j}(t)\to u(t)$ for every $t\in[0,T]$.

By the sequential continuity established in claim 1, $L\bigl(S_{t},\tilde u_{j}(t)\bigr)\to L\bigl(S_{t},u(t)\bigr)$ for every $t\in[0,T]$; all these functions are measurable by claim 1 and satisfy $\bigl|L(S_{t},\tilde u_{j}(t))\bigr|\le C$, and the constant function $C$ is integrable on $([0,T],\mathcal{B},\lambda)$ because $\lambda([0,T])=T<\infty$. By Dominated Convergence Theorem,

$$\Phi_{S}\bigl([u_{n_{j}}]\bigr)=\int_{[0,T]}L\bigl(S_{t},\tilde u_{j}(t)\bigr)\,d\lambda\ \longrightarrow\ \int_{[0,T]}L\bigl(S_{t},u(t)\bigr)\,d\lambda=\Phi_{S}(\zeta).$$

Each term satisfies $\Phi_{S}([u_{n_{j}}])\le M$, so $\Phi_{S}(\zeta)\le M$ by Order Properties of Limits of Real Sequences, comparing with the constant sequence $M$. Hence $\zeta\in C_{M}$. So the closure of $C_{M}$ is contained in $C_{M}$; as it also contains $C_{M}$ and is closed, $C_{M}$ is closed.

Claim 3. We first record a level form of the assertion, which is where claim 2 is used. Let $(\zeta_{i})_{i\in\mathbb{N}}$ be a sequence in $\mathcal{U}_{\mathcal{A}}$ with $\zeta_{i}\rightharpoonup\zeta$ for some $\zeta\in H$, and let $M$ be a real number with $\Phi_{S}(\zeta_{i})\le M$ for every $i$. Then $\zeta\in\mathcal{U}_{\mathcal{A}}$ and $\Phi_{S}(\zeta)\le M$. Indeed $\zeta_{i}\in C_{M}$ for every $i$, and $C_{M}$ is convex and closed by claim 2, so Closed Convex Subsets of the Lebesgue Space of Square-Integrable Vector-Valued Functions are Weakly Sequentially Closed gives $\zeta\in C_{M}$, which is the assertion.

Now let $(\xi_{n})_{n}$ and $\xi$ be as in claim 3, and put $\ell=\liminf_{n}\Phi_{S}(\xi_{n})$, which is defined because $\bigl(\Phi_{S}(\xi_{n})\bigr)_{n}$ is a bounded real sequence by claim 1. Let $\varepsilon>0$ be real. By claim 4 of Basic Properties of the Limit Inferior and Limit Superior of a Bounded Real Sequence there are natural numbers $n_{1}<n_{2}<\dots$ with $\Phi_{S}(\xi_{n_{j}})<\ell+\varepsilon$ for every $j$. For every $v\in H$ the real sequence $\bigl(\langle\xi_{n_{j}},v\rangle_{L^{2}}\bigr)_{j}$ is a subsequence of $\bigl(\langle\xi_{n},v\rangle_{L^{2}}\bigr)_{n}$ and hence has the same limit $\langle\xi,v\rangle_{L^{2}}$, by the fact that a subsequence of a convergent sequence has the same limit; so $\xi_{n_{j}}\rightharpoonup\xi$. The level form, applied to $(\xi_{n_{j}})_{j}$ with the bound $\ell+\varepsilon$, gives $\xi\in\mathcal{U}_{\mathcal{A}}$ and $\Phi_{S}(\xi)\le\ell+\varepsilon$.

This holds for every real $\varepsilon>0$, so $\Phi_{S}(\xi)\le\ell$: otherwise $\varepsilon=\bigl(\Phi_{S}(\xi)-\ell\bigr)/2$ is positive and yields $\Phi_{S}(\xi)\le\ell+\tfrac{1}{2}\bigl(\Phi_{S}(\xi)-\ell\bigr)<\Phi_{S}(\xi)$, which is false.

Claim 4. Put $\ell'=\liminf_{n}\Phi_{S^{n}}(\xi_{n})$, which is defined by claim 1. Let $\varepsilon>0$ be real. By claim 2 of Boundedness and Uniform Continuity of Population Cost Data on a Compact Control Set there is a real $\delta>0$ such that $|L(\Sigma,\alpha)-L(\Sigma',\alpha)|\le\varepsilon$ whenever $\Sigma,\Sigma'\in\Delta^{l}$ satisfy $|\Sigma-\Sigma'|\le\delta$ and $\alpha\in\mathcal{A}$. By hypothesis there is $N_{0}\in\mathbb{N}$ with $|S^{n}_{t}-S_{t}|\le\delta$ for every $t\in[0,T]$ and every $n\ge N_{0}$. Fix $n\ge N_{0}$ and a normalized representative $u_{n}$ of $\xi_{n}$. Then $L(S_{t},u_{n}(t))\le L(S^{n}_{t},u_{n}(t))+\varepsilon$ for every $t$, so by claim 2 of Linearity and Monotonicity of the Lebesgue Integral, which supplies both the monotonicity and the linearity of the integral for signed integrable integrands, together with $\lambda([0,T])=T$,

$$\Phi_{S}(\xi_{n})\le\Phi_{S^{n}}(\xi_{n})+\varepsilon T\qquad(n\ge N_{0}).$$

By claim 4 of Basic Properties of the Limit Inferior and Limit Superior of a Bounded Real Sequence there are natural numbers $n_{1}<n_{2}<\dots$ with $\Phi_{S^{n_{j}}}(\xi_{n_{j}})<\ell'+\varepsilon$ for every $j$. Since $n_{j}\ge j$ by the growth bound for strictly increasing sequences of natural numbers, we have $n_{j}\ge N_{0}$ whenever $j\ge N_{0}$, and for such $j$ the displayed inequality gives

$$\Phi_{S}(\xi_{n_{j}})\le\Phi_{S^{n_{j}}}(\xi_{n_{j}})+\varepsilon T<\ell'+\varepsilon(1+T).$$

Consider the sequence $\zeta_{i}=\xi_{n_{N_{0}+i-1}}$ ($i\in\mathbb{N}$), which lies in $\mathcal{U}_{\mathcal{A}}$ and satisfies $\Phi_{S}(\zeta_{i})<\ell'+\varepsilon(1+T)$ for every $i$. For every $v\in H$ the real sequence $\bigl(\langle\zeta_{i},v\rangle_{L^{2}}\bigr)_{i}$ is a subsequence of $\bigl(\langle\xi_{n},v\rangle_{L^{2}}\bigr)_{n}$, the index map $i\mapsto n_{N_{0}+i-1}$ being strictly increasing, hence has the same limit $\langle\xi,v\rangle_{L^{2}}$ by the fact that a subsequence of a convergent sequence has the same limit; so $\zeta_{i}\rightharpoonup\xi$. The level form established in claim 3, applied to $(\zeta_{i})_{i}$ with the bound $\ell'+\varepsilon(1+T)$, gives $\xi\in\mathcal{U}_{\mathcal{A}}$ and $\Phi_{S}(\xi)\le\ell'+\varepsilon(1+T)$.

This holds for every real $\varepsilon>0$. If $\Phi_{S}(\xi)>\ell'$, then $\varepsilon=\bigl(\Phi_{S}(\xi)-\ell'\bigr)\bigl(2(1+T)\bigr)^{-1}$ is positive and yields $\Phi_{S}(\xi)\le\ell'+\tfrac{1}{2}\bigl(\Phi_{S}(\xi)-\ell'\bigr)<\Phi_{S}(\xi)$, which is false. Hence $\Phi_{S}(\xi)\le\ell'=\liminf_{n}\Phi_{S^{n}}(\xi_{n})$.