Proof of The Transition-Rate Family and Aggregate Drift of Affine-Controlled Data

Throughout, $(\sigma,\gamma)$ denotes an ordered pair of distinct elements of $\{1,\dots,l\}$; when such a pair is fixed we abbreviate $\beta_0(\Sigma)=\beta_0(\sigma,\gamma,\Sigma)$, $\beta_1(\Sigma)=\beta_1(\sigma,\gamma,\Sigma)$ and $\beta(\Sigma,\alpha)=\beta(\sigma,\gamma,\Sigma,\alpha)$. We write $|\cdot|$ for the Euclidean norm and use without further comment the norm properties: clause 1 (the square of the norm is the sum of the squares of the coordinates), clause 4 (each coordinate satisfies $|x_i|\le|x|$) and clause 6 (the triangle inequality). By the definition of an affine-controlled transition-rate family, the control set $\mathcal{A}$ is nonempty, convex and compact, the maps $\beta_0(\sigma,\gamma,\cdot)$ and $\beta_1(\sigma,\gamma,\cdot)$ are Lipschitz with constant $\Lambda$ on $\Delta^l$, and $\beta(\Sigma,\alpha)\ge0$ for $\Sigma\in\Delta^l$ and $\alpha\in\mathcal{A}$.

Step 0. Two elementary bounds.

(0a) Let $n\ge1$ be a natural number, let $x\in\mathbb{R}^n$ and let $C$ be a real number with $C\ge0$ and $|x_i|\le C$ for every $i\in\{1,\dots,n\}$. Then $|x|\le\sqrt{n}\,C$, where $\sqrt{n}$ denotes the nonnegative square root of $n$. Indeed, applying monotonicity of squaring (clause 2) to the nonnegative reals $|x_i|$ and $C$ gives $x_i^2=|x_i|^2\le C^2$ for every $i$, whence

Since $|x|\ge0$ and $\sqrt{n}\,C\ge0$, clause 2 of the same lemma, read in the other direction, gives $|x|\le\sqrt{n}\,C$.

(0b) By compactness of the simplex and the control set (clause 1), $\Delta^l$ is nonempty and closed and satisfies $|\Sigma|\le1$ for every $\Sigma\in\Delta^l$; consequently $|\Sigma-\Sigma'|\le|\Sigma|+|\Sigma'|\le2$ for all $\Sigma,\Sigma'\in\Delta^l$. Moreover $\Delta^l$ is convex: if $\Sigma,\Sigma'\in\Delta^l$ and $t$ is a real number with $0\le t\le1$, then the point $t\Sigma+(1-t)\Sigma'$ has nonnegative coordinates, and those coordinates sum to $t\cdot1+(1-t)\cdot1=1$, so it lies in the probability simplex. Being nonempty, closed and convex, $\Delta^l$ admits the nearest-point projection $\pi_{\Delta^l}$ by clause 1 of that lemma. This proves the second assertion of clause 1 of the statement.

Step 1. Finiteness of $R$, $K_1$ and $B$ (clause 1).

The control set $\mathcal{A}$ is compact, hence bounded, so by the definition of a bounded subset there are a point $\alpha_*\in\mathbb{R}^m$ and a real number $r\ge0$ with $|\alpha-\alpha_*|\le r$ for every $\alpha\in\mathcal{A}$; the triangle inequality then gives $|\alpha|\le|\alpha_*|+r$ for every $\alpha\in\mathcal{A}$. The set $\{|\alpha|:\alpha\in\mathcal{A}\}$ is therefore nonempty and bounded above, so its supremum $R$ exists by the least upper bound property, and $0\le R$ because $\mathcal{A}$ is nonempty and norms are nonnegative.

Fix an ordered pair $(\sigma,\gamma)$ and a point $\Sigma_*\in\Delta^l$. For every $\Sigma\in\Delta^l$ the Lipschitz property and (0b) give

Since there are only finitely many ordered pairs $(\sigma,\gamma)$, the set of all values $|\beta_1(\sigma,\gamma,\Sigma)|$, taken over all such pairs and all $\Sigma\in\Delta^l$, is nonempty and bounded above, so $K_1$ exists by the least upper bound property and $K_1\ge0$. Writing $M_0$ for the supremum of the values $|\beta_0(\sigma,\gamma,\Sigma)|$ over the same index set, which exists and is nonnegative for the same reason, we obtain from the Cauchy-Schwarz inequality that

for all $\Sigma\in\Delta^l$ and $\alpha\in\mathcal{A}$, the first inequality being the nonnegativity requirement. Hence the set of values of $\beta$ is nonempty and bounded above, $B$ exists, and $0\le B\le M_0+K_1R$. This proves clause 1.

Step 2. $\beta$ is a transition-rate family (clause 2).

Bounds. For all $\Sigma\in\Delta^l$ and $\alpha\in\mathcal{A}$ we have $0\le\beta(\Sigma,\alpha)$ as above, and $\beta(\Sigma,\alpha)\le B$ because $B$ is an upper bound for the set of values of $\beta$.

Lipschitz dependence on the state. Let $\Sigma,\Sigma'\in\Delta^l$ and $\alpha\in\mathcal{A}$. Using the triangle inequality, then Cauchy-Schwarz and $|\alpha|\le R$,

which is the asserted Lipschitz bound with constant $\Lambda_\beta=\Lambda(1+R)$.

Joint continuity. Let $(\Sigma_n,\alpha_n)_{n\in\mathbb{N}}$ be a sequence in $\Delta^l\times\mathcal{A}$ such that the Euclidean distances $d(\Sigma_n,\Sigma)$ and $d(\alpha_n,\alpha)$ converge to $0$, where $\Sigma\in\Delta^l$ and $\alpha\in\mathcal{A}$. Let $\varepsilon>0$ be real. Choose $n_0$ so large that $d(\Sigma_n,\Sigma)\le\varepsilon/(2\Lambda_\beta+2)$ and $d(\alpha_n,\alpha)\le\varepsilon/(2K_1+2)$ for every $n\ge n_0$; this is possible because both distance sequences converge to $0$ and both denominators are positive. For such $n$, the triangle inequality, the Lipschitz bound just proved and Cauchy-Schwarz give

where the middle step used $\beta(\Sigma,\alpha_n)-\beta(\Sigma,\alpha)=\beta_1(\Sigma)\cdot(\alpha_n-\alpha)$, which is immediate from the defining formula for $\beta$. As $\varepsilon$ was arbitrary, $\beta(\sigma,\gamma,\Sigma_n,\alpha_n)$ converges to $\beta(\sigma,\gamma,\Sigma,\alpha)$. Both requirements of a transition-rate family with control set $\mathcal{A}$ and rate bound $B$ are met, proving clause 2.

Step 3. Affine form of the drift (clause 3).

Fix $\Sigma\in\Delta^l$, $\alpha\in\mathcal{A}$ and $\gamma$. Substituting the defining formula for $\beta$ into the definition of the aggregate state drift gives

The first sum is $b^\gamma_0(\Sigma)$. In the second sum, the dot product is given coordinatewise by $v\cdot\alpha=\sum_{k=1}^mv^k\alpha^k$, so it is linear in $v$ for fixed $\alpha$; rearranging the resulting finite double sum therefore gives

Adding the two contributions proves clause 3.

Step 4. Bounds on the drift (clause 4).

Throughout this step we use $0\le\Sigma^\sigma\le1$, $\sum_{\sigma\neq\gamma}\Sigma^\sigma\le\sum_{\sigma=1}^l\Sigma^\sigma=1$, and the fact that the index set $\{\sigma:\sigma\neq\gamma\}$ has $l-1$ elements. We also use repeatedly that $l\le2(l-1)$, which holds because $l\ge2$.

(a) Norm bound. For every $\gamma$,

By (0a) applied in $\mathbb{R}^l$, $|b(\Sigma,\alpha)|\le\sqrt{l}\cdot2(l-1)B=2\sqrt{l}\,(l-1)B$.

(b) Lipschitz dependence on the state. Let $\Sigma,\Sigma'\in\Delta^l$ and $\alpha\in\mathcal{A}$, and write $\beta(\cdot)=\beta(\sigma,\gamma,\cdot,\alpha)$ for the pair under consideration. For each $\sigma\neq\gamma$,

whose absolute value is at most $|\Sigma^\sigma-\Sigma'^\sigma|B+\Sigma'^\sigma\Lambda_\beta|\Sigma-\Sigma'|$ by Step 2; the same estimate with $\gamma$ in place of $\sigma$ bounds $|\Sigma^\gamma\beta(\gamma,\sigma,\Sigma,\alpha)-\Sigma'^\gamma\beta(\gamma,\sigma,\Sigma',\alpha)|$ by $|\Sigma^\gamma-\Sigma'^\gamma|B+\Sigma'^\gamma\Lambda_\beta|\Sigma-\Sigma'|$. Summing over the $l-1$ indices $\sigma\neq\gamma$ and using $|\Sigma^\sigma-\Sigma'^\sigma|\le|\Sigma-\Sigma'|$, $|\Sigma^\gamma-\Sigma'^\gamma|\le|\Sigma-\Sigma'|$, $\sum_{\sigma\neq\gamma}\Sigma'^\sigma\le1$ and $\Sigma'^\gamma\le1$, we obtain

By (0a), $|b(\Sigma,\alpha)-b(\Sigma',\alpha)|\le\sqrt{l}\cdot2(l-1)(B+\Lambda_\beta)|\Sigma-\Sigma'|=\Lambda_b|\Sigma-\Sigma'|$.

(c) Lipschitz dependence on the control. Let $\alpha,\alpha'\in\mathcal{A}$ and $\Sigma\in\Delta^l$. From the defining formula for $\beta$ and Cauchy-Schwarz, $|\beta(\sigma,\gamma,\Sigma,\alpha)-\beta(\sigma,\gamma,\Sigma,\alpha')|=|\beta_1(\sigma,\gamma,\Sigma)\cdot(\alpha-\alpha')|\le K_1|\alpha-\alpha'|$ for every pair. Hence

and (0a) gives $|b(\Sigma,\alpha)-b(\Sigma,\alpha')|\le2\sqrt{l}\,(l-1)K_1|\alpha-\alpha'|$. This proves clause 4.

Step 5. Conservation and the inflow bound (clause 5).

Let $P=\{(\sigma,\gamma):\sigma,\gamma\in\{1,\dots,l\},\ \sigma\neq\gamma\}$. Summing the definition of $b^\gamma$ over $\gamma$ and regrouping the resulting finite sum over $P$,

The map $(\sigma,\gamma)\mapsto(\gamma,\sigma)$ is a bijection of $P$ onto itself which carries the summand $\Sigma^\gamma\beta(\gamma,\sigma,\Sigma,\alpha)$ of the second sum to the summand $\Sigma^\sigma\beta(\sigma,\gamma,\Sigma,\alpha)$ of the first, so the two sums are equal and the difference vanishes.

For the inflow bound, the first group of terms in $b^\gamma(\Sigma,\alpha)$ is nonnegative because $\Sigma^\sigma\ge0$ and $\beta\ge0$, so

using $\beta\le B$ and $\Sigma^\gamma\ge0$. This proves clause 5.

Step 6. The projected drift (clause 6).

By Step 0(b) the projection $\pi_{\Delta^l}$ is defined, so $\hat{b}$ is well defined on $\mathbb{R}^l\times\mathcal{A}$, and $\pi_{\Delta^l}(x)\in\Delta^l$ for every $x\in\mathbb{R}^l$. If $x\in\Delta^l$ then $\pi_{\Delta^l}(x)=x$ by clause 3 of the projection lemma, so $\hat{b}(x,\alpha)=b(x,\alpha)$. For arbitrary $x\in\mathbb{R}^l$ and $\alpha\in\mathcal{A}$, clause 4(a) applied at the point $\pi_{\Delta^l}(x)\in\Delta^l$ gives $|\hat{b}(x,\alpha)|\le2\sqrt{l}\,(l-1)B$. Finally, for $x,x'\in\mathbb{R}^l$ and $\alpha\in\mathcal{A}$, clause 4(b) applied at the points $\pi_{\Delta^l}(x),\pi_{\Delta^l}(x')\in\Delta^l$ together with the nonexpansiveness of the projection (clause 4 of the projection lemma) gives

This proves clause 6 and completes the proof.