Proof of Regularity and Derivative Bounds of the Extended Aggregate Observation Drift

Throughout fix $\upsilon\in\{1,\dots,\tilde{l}\}$ and recall from the extension definition its clauses 1-4. On $\Delta^l$ we use, from the probability simplex: $\Sigma^\sigma\ge0$ for all $\sigma$ and $\sum_{\sigma=1}^{l}\Sigma^\sigma=1$; and from clause 1 together with the observation-rate family bounds: $0\le\bar{\tilde{\beta}}(\sigma,\upsilon,\Sigma)\le\tilde{B}$ for $\Sigma\in\Delta^l$ and all admissible pairs. We write $\mathbf{1}_{\{\cdot\}}$ for the indicator equal to $1$ when the subscripted condition holds and $0$ otherwise. The argument is the same as, but logically independent of, the published proof of the regularity and derivative bounds of the extended aggregate state drift, with the control coordinates absent.

(i). For $\sigma\in\{1,\dots,l\}$ let $\pi_\sigma:\tilde{U}\to\mathbb{R}$ be the coordinate function $\pi_\sigma(\Sigma)=\Sigma^\sigma$. Directly from the definition of the partial derivative (the difference quotients being constant), $\partial_\gamma\pi_\sigma$ exists and equals the constant $\mathbf{1}_{\{\gamma=\sigma\}}$, so $\pi_\sigma$ is a $C^1$ map, and each constant function is again $C^1$ with vanishing partial derivatives. By the definition of the extended aggregate observation drift,

Fix $\gamma$ and a point of $\tilde{U}$. The set of slice parameters $s$ for which the point shifted by $s$ along the $\gamma$-th coordinate remains in the open set $\tilde{U}$ is an open subset of $\mathbb{R}$ containing the given parameter value; choose an open interval $I_0$ around that value contained in it. On $I_0$, every factor above restricts to a one-variable function whose derivative exists and is given by the corresponding partial derivative (this is the definition of the partial derivative), so the one-dimensional sum and product rules give that $\partial_\gamma\bar{\tilde{b}}^\upsilon$ exists at the given point and equals

the first displayed formula of the statement. Each term of that formula is continuous: $\bar{\tilde{\beta}}(\sigma,\upsilon,\cdot)$ is continuous because it is differentiable at every point (clause 2 with $C^1$ implies differentiable, and the inequality in the definition of differentiability at a point forces $f(a+h)\to f(a)$ as $h\to0$); its partial derivatives are continuous by clause 2 and the definition of a $C^1$ map; the coordinate functions are continuous; and finite sums and products of continuous real functions are continuous (immediate from the sequential formulation of continuity at a point). Hence $\partial_\gamma\bar{\tilde{b}}^\upsilon$ is continuous for every $\gamma$, so $\bar{\tilde{b}}^\upsilon$ is a $C^1$ map. Applying the same slice-interval argument to the first displayed formula --- a finite sum of products of coordinate functions, the functions $\bar{\tilde{\beta}}(\sigma,\upsilon,\cdot)$, and their first partials, all of which are $C^1$ by clause 2 --- gives that $\partial_\delta\partial_\gamma\bar{\tilde{b}}^\upsilon$ exists and equals

which is the second displayed formula after evaluating the indicator sum, and this expression is continuous by the same reasoning; hence each $\partial_\gamma\bar{\tilde{b}}^\upsilon$ is a $C^1$ map. Finally, for $\Sigma\in\Delta^l$, clause 1 allows replacing every $\bar{\tilde{\beta}}$ by $\tilde{\beta}$ in the defining formula of $\bar{\tilde{b}}^\upsilon(\Sigma)$, which then coincides with the defining formula of the aggregate observation drift of $\tilde{\beta}$ at $\Sigma$; hence $\bar{\tilde{b}}$ agrees with it on $\Delta^l$.

(ii). Let $\Sigma\in\Delta^l$ and $\gamma\in\{1,\dots,l\}$. In the first displayed formula, $|\bar{\tilde{\beta}}(\gamma,\upsilon,\Sigma)|\le\tilde{B}$ as recorded above, and $\sum_{\sigma}\Sigma^\sigma|\partial_\gamma\bar{\tilde{\beta}}(\sigma,\upsilon,\Sigma)|\le\tilde{K}\sum_{\sigma}\Sigma^\sigma=\tilde{K}$ by clause 3; hence $|\partial_\gamma\bar{\tilde{b}}^\upsilon(\Sigma)|\le\tilde{B}+\tilde{K}$. For the Lipschitz estimate let $\Sigma,\Sigma'\in\Delta^l$. The segment from $\Sigma$ to $\Sigma'$ stays in $\Delta^l$: a convex combination of two points of the simplex has nonnegative entries summing to $1$, hence lies in the simplex. The segment therefore lies in the open set $\tilde{U}$, and part (i) of the Taylor expansion lemma, applied to the $C^1$ map $\bar{\tilde{b}}^\upsilon$ with $n=l$ and $M_1=\tilde{B}+\tilde{K}$, gives $|\bar{\tilde{b}}^\upsilon(\Sigma)-\bar{\tilde{b}}^\upsilon(\Sigma')|\le\sqrt{l}\,(\tilde{B}+\tilde{K})\,d(\Sigma,\Sigma')$.

(iii). For $\Sigma\in\Delta^l$, estimating the three groups of the second displayed formula with clause 3 and $\sum_{\sigma}\Sigma^\sigma=1$: the two first-derivative terms contribute at most $\tilde{K}$ each, and the sum $\sum_\sigma\Sigma^\sigma\,\partial_\delta\partial_\gamma\bar{\tilde{\beta}}(\sigma,\upsilon,\Sigma)$ contributes at most $\tilde{K}$ in total; hence $|\partial_\delta\partial_\gamma\bar{\tilde{b}}^\upsilon(\Sigma)|\le3\tilde{K}$.

For the uniform continuity claim, let $\varepsilon>0$ and set $C^*=2\sqrt{l}\,\tilde{K}+l\,\tilde{K}$. By clause 4 there is $\delta_1>0$ such that $|\partial_\delta\partial_\gamma\bar{\tilde{\beta}}(\sigma,\upsilon,\Sigma)-\partial_\delta\partial_\gamma\bar{\tilde{\beta}}(\sigma,\upsilon,\Sigma')|\le\varepsilon/2$ for all admissible indices whenever $d(\Sigma,\Sigma')\le\delta_1$. Set $\delta^\circ=\delta_1$ if $C^*=0$ and $\delta^\circ=\min(\delta_1,\varepsilon/(2C^*))$ otherwise. Let $\Sigma,\Sigma'\in\Delta^l$ with $d(\Sigma,\Sigma')\le\delta^\circ$, and take the difference of the second displayed formula at $\Sigma$ and at $\Sigma'$ term by term. For the two first-derivative terms: each of the two functions $\partial_\delta\bar{\tilde{\beta}}(\gamma,\upsilon,\cdot)$ and $\partial_\gamma\bar{\tilde{\beta}}(\delta,\upsilon,\cdot)$ is a $C^1$ map (clause 2) whose partial derivatives are bounded by $\tilde{K}$ on $\tilde{U}$ (clause 3), so part (i) of the Taylor expansion lemma along the segment from $\Sigma$ to $\Sigma'$ (which lies in the simplex, as above) bounds each difference by $\sqrt{l}\,\tilde{K}\,d(\Sigma,\Sigma')$. For the product terms,

and summing over $\sigma$, using $|\Sigma^\sigma-\Sigma'^\sigma|\le d(\Sigma,\Sigma')$ for every $\sigma$ and $\sum_\sigma\Sigma'^\sigma=1$, the product group contributes at most $l\,\tilde{K}\,d(\Sigma,\Sigma')+\varepsilon/2$. Altogether the difference is at most $C^*d(\Sigma,\Sigma')+\varepsilon/2\le\varepsilon/2+\varepsilon/2=\varepsilon$, uniformly over $\gamma,\delta,\upsilon$, as claimed. $\blacksquare$