Proof of Boundedness and Uniform Continuity of Population Cost Data on a Compact Control Set

We use throughout the continuity clause of population cost data: if $\Sigma_k\to\Sigma$ in $\Delta^l$ and $\alpha_k\to\alpha$ in $\mathbb{R}^m$ in Euclidean distance, then $L(\Sigma_k,\alpha_k)\to L(\Sigma,\alpha)$ and $G(\Sigma_k)\to G(\Sigma)$.

Preliminaries. The simplex $\Delta^l$ is bounded, since $|\Sigma|^2=\sum_\gamma(\Sigma^\gamma)^2\le\big(\sum_\gamma\Sigma^\gamma\big)^2=1$ for $\Sigma\in\Delta^l$, the coordinates being nonnegative; and $\mathcal{A}$ is closed and bounded by the Heine-Borel theorem. Both sets contain the limits of their convergent sequences: for $\Delta^l$, if $\Sigma_k\to\Sigma$ with $\Sigma_k\in\Delta^l$, then by the coordinate bound of the elementary properties of the Euclidean norm each coordinate converges, so the coordinates of $\Sigma$ are nonnegative with sum $1$; for $\mathcal{A}$, if some convergent sequence in $\mathcal{A}$ had its limit $p$ outside $\mathcal{A}$, then, $\mathcal{A}$ being closed, some open ball around $p$ would miss $\mathcal{A}$, contradicting the convergence.

Claim 1. Suppose no bound $C$ works for $L$. Then for every natural number $k$ there are $\Sigma_k\in\Delta^l$ and $\alpha_k\in\mathcal{A}$ with $|L(\Sigma_k,\alpha_k)|>k$. The sequence of pairs $(\Sigma_k,\alpha_k)$ is bounded in $\mathbb{R}^{l+m}$, so by the Bolzano-Weierstrass theorem some subsequence converges, and by the coordinate bound its two blocks converge separately, say $\Sigma_{k_j}\to\Sigma$ and $\alpha_{k_j}\to\alpha$. By the preliminaries $\Sigma\in\Delta^l$ and $\alpha\in\mathcal{A}$, so the continuity clause gives $L(\Sigma_{k_j},\alpha_{k_j})\to L(\Sigma,\alpha)$; a convergent sequence of real numbers is bounded, contradicting $|L(\Sigma_{k_j},\alpha_{k_j})|>k_j$ for all $j$. Hence a bound $C_L$ exists for $L$, and the same argument with $G$ in place of $L$ gives a bound $C_G$; take $C=\max(C_L,C_G,0)$.

Claim 2. Suppose the assertion fails for $L$. Then there is $\varepsilon>0$ such that for every natural number $k$ the choice $\delta=1/k$ fails, that is there are $\Sigma_k,\Sigma'_k\in\Delta^l$ and $\alpha_k\in\mathcal{A}$ with

$$|\Sigma_k-\Sigma'_k|\le\tfrac{1}{k}\qquad\text{and}\qquad|L(\Sigma_k,\alpha_k)-L(\Sigma'_k,\alpha_k)|>\varepsilon .$$

The triples $(\Sigma_k,\Sigma'_k,\alpha_k)$ form a bounded sequence in $\mathbb{R}^{2l+m}$, so by Bolzano-Weierstrass and the coordinate bound some subsequence satisfies $\Sigma_{k_j}\to\Sigma$, $\Sigma'_{k_j}\to\Sigma'$ and $\alpha_{k_j}\to\alpha$, with $\Sigma,\Sigma'\in\Delta^l$ and $\alpha\in\mathcal{A}$. By the triangle inequality,

$$|\Sigma-\Sigma'|\le|\Sigma-\Sigma_{k_j}|+|\Sigma_{k_j}-\Sigma'_{k_j}|+|\Sigma'_{k_j}-\Sigma'|\longrightarrow0 ,$$

so $\Sigma=\Sigma'$. The continuity clause now gives $L(\Sigma_{k_j},\alpha_{k_j})\to L(\Sigma,\alpha)$ and $L(\Sigma'_{k_j},\alpha_{k_j})\to L(\Sigma',\alpha)=L(\Sigma,\alpha)$, so the difference of the two tends to $0$, contradicting that it exceeds $\varepsilon$ for every $j$. Hence a suitable $\delta_L$ exists for $L$; the same argument gives $\delta_G$ for $G$, and $\delta=\min(\delta_L,\delta_G)$ works for both. $\blacksquare$