Proof of Forward Invariance of the Probability Simplex under the Projected Drift

For a real number $c$ write $c^+=\max(c,0)$ and $c^-=\max(-c,0)$, so that $c=c^+-c^-$, $|c|=c^++c^-$, and $-c\le c^-$. Put

$$u_t=\sum_{\gamma=1}^l\big(x^\gamma_t\big)^-\qquad(t\in[0,T]),\qquad K_b=2\sqrt{l}\,(l-1)B .$$

The maps $c\mapsto c^+$ and $c\mapsto c^-$ are continuous, so $u$ is continuous on $[0,T]$, and $u_0=0$ because $x_0\in\Delta^l$ has nonnegative coordinates.

Step 0: the integrands. By the lemma on affine-controlled data, $|\hat{b}(y,\alpha)|\le K_b$ for all $(y,\alpha)$, and

$$|\hat{b}(y,\alpha)-\hat{b}(y',\alpha')|\le|\hat{b}(y,\alpha)-\hat{b}(y',\alpha)|+|b(\pi_{\Delta^l}(y'),\alpha)-b(\pi_{\Delta^l}(y'),\alpha')|\le\Lambda_b|y-y'|+2\sqrt{l}(l-1)K_1|\alpha-\alpha'| ,$$

so each component $\hat{b}^\gamma$ is sequentially continuous on $E=\mathbb{R}^l\times\mathcal{A}$. The map $s\mapsto(x_s,A_s)$ takes values in $E$, and its components are measurable: each component $x^\gamma$ of $x$ is itself continuous, since the coordinate bound $|x^\gamma_t-x^\gamma_s|\le|x_t-x_s|$ of claim 4 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ transfers the defining $\varepsilon$--$\delta$ condition, and is therefore measurable by claim 4 of the measurability toolkit; the components of $A$ are measurable by hypothesis. Hence $s\mapsto\hat{b}^\gamma(x_s,A_s)$ is measurable by measurability of sequentially continuous functions of measurable maps, and it is bounded by $K_b$, so the integrals in the hypothesis exist. The same applies to $s\mapsto\big(\hat{b}^\gamma(x_s,A_s)\big)^-$, bounded by $K_b$, and to $s\mapsto u_s$, which is measurable by the same composition device and bounded: the hypothesis and $|\hat{b}^\gamma|\le K_b$ give $|x^\gamma_s|\le|x^\gamma_0|+K_bT$, so $0\le u_s\le\sum_{\gamma=1}^{l}\bigl(|x^\gamma_0|+K_bT\bigr)$.

Step 1: the coordinates sum to $1$. Summing the hypothesis over $\gamma$ and using linearity of the integral together with the conservation identity of the lemma on affine-controlled data, applied at the point $\pi_{\Delta^l}(x_s)\in\Delta^l$,

$$\sum_\gamma x^\gamma_t=\sum_\gamma x^\gamma_0+\int_{[0,t]}\sum_\gamma b^\gamma\big(\pi_{\Delta^l}(x_s),A_s\big)\,ds=\sum_\gamma x^\gamma_0=1 .$$

Step 2: distance to the simplex. We claim $|x_t-\pi_{\Delta^l}(x_t)|\le2u_t$ for every $t$. Note first that $|z|\le\sum_\gamma|z^\gamma|$ for $z\in\mathbb{R}^l$, since $|z|^2=\sum_\gamma(z^\gamma)^2\le\big(\sum_\gamma|z^\gamma|\big)^2$. Write $v^\gamma=(x^\gamma_t)^+$. By Step 1,

$$\sum_\gamma v^\gamma=\sum_\gamma x^\gamma_t+\sum_\gamma(x^\gamma_t)^-=1+u_t .$$

If $u_t=0$ then all coordinates of $x_t$ are nonnegative and, by Step 1, $x_t\in\Delta^l$, so the claim holds with both sides $0$. Otherwise put $y=v/(1+u_t)$, which has nonnegative coordinates summing to $1$, hence lies in $\Delta^l$. The coordinates of $x_t-v$ are $-(x^\gamma_t)^-$, so $|x_t-v|\le u_t$, and $|v|\le\sum_\gamma v^\gamma=1+u_t$, so

$$|v-y|=\Big(1-\frac{1}{1+u_t}\Big)|v|=\frac{u_t}{1+u_t}\,|v|\le u_t .$$

By the triangle inequality, $|x_t-y|\le2u_t$, and since $\pi_{\Delta^l}(x_t)$ is a nearest point of $\Delta^l$ to $x_t$ and $y\in\Delta^l$, the projection lemma gives $|x_t-\pi_{\Delta^l}(x_t)|\le|x_t-y|\le2u_t$.

Step 3: an increment inequality. We first record: for all real numbers $c,e$,

$$c^--e^-\le(e-c)^+\,\mathbf{1}\{c<0\} ,$$

where $\mathbf{1}\{\cdot\}$ is $1$ if the condition holds and $0$ otherwise. Indeed, if $c\ge0$ the left side is $-e^-\le0$ and the right side is $0$; if $c<0$ and $e\ge0$ the left side is $-c\le e-c=(e-c)^+$; and if $c<0$ and $e<0$ the left side is $e-c\le(e-c)^+$.

Fix $0\le r\le t\le T$ and $\gamma$, and apply this with $c=x^\gamma_t$ and $e=x^\gamma_r$. Suppose $x^\gamma_t<0$. Using $-c\le c^-$ pointwise and monotonicity of the integral,

$$x^\gamma_r-x^\gamma_t=-\int_{[r,t]}\hat{b}^\gamma(x_s,A_s)\,ds\le\int_{[r,t]}\big(\hat{b}^\gamma(x_s,A_s)\big)^-\,ds .$$

By the inflow bound of the lemma on affine-controlled data, applied at $\pi_{\Delta^l}(x_s)\in\Delta^l$,

$$\hat{b}^\gamma(x_s,A_s)=b^\gamma\big(\pi_{\Delta^l}(x_s),A_s\big)\ge-(l-1)B\,\pi_{\Delta^l}(x_s)^\gamma ,$$

and $\pi_{\Delta^l}(x_s)^\gamma\ge0$, so $\big(\hat{b}^\gamma(x_s,A_s)\big)^-\le(l-1)B\,\pi_{\Delta^l}(x_s)^\gamma$. By the coordinate bound and Step 2,

$$\pi_{\Delta^l}(x_s)^\gamma\le x^\gamma_s+\big|\pi_{\Delta^l}(x_s)-x_s\big|\le x^\gamma_s+2u_s .$$

Moreover, for $s\in[r,t]$, the hypothesis and $|\hat{b}^\gamma|\le K_b$ give $x^\gamma_s=x^\gamma_t-\int_{[s,t]}\hat{b}^\gamma(x_q,A_q)\,dq\le x^\gamma_t+K_b(t-r)<K_b(t-r)$, because $x^\gamma_t<0$. Combining the last three displays,

$$x^\gamma_r-x^\gamma_t\le(l-1)B\int_{[r,t]}\big(K_b(t-r)+2u_s\big)ds=(l-1)B\,K_b(t-r)^2+2(l-1)B\int_{[r,t]}u_s\,ds .$$

When $x^\gamma_t\ge0$ the elementary inequality has right-hand side $0$, so in all cases

$$\big(x^\gamma_t\big)^--\big(x^\gamma_r\big)^-\le(l-1)B\,K_b(t-r)^2+2(l-1)B\int_{[r,t]}u_s\,ds .$$

Summing over the $l$ values of $\gamma$,

$$u_t-u_r\le l(l-1)B\,K_b\,(t-r)^2+2l(l-1)B\int_{[r,t]}u_s\,ds .$$

Step 4: telescoping. Fix $t\in(0,T]$ and a natural number $n$, and put $s_k=kt/n$ for $k\in\{0,\dots,n\}$. Applying Step 3 on each $[s_{k-1},s_k]$ and adding, the left sides telescope to $u_t-u_0=u_t$, and, since the intervals $[s_{k-1},s_k]$ overlap only in single points, which have Lebesgue measure zero, linearity of the integral gives

$$u_t\le l(l-1)B\,K_b\,n\Big(\frac{t}{n}\Big)^2+2l(l-1)B\int_{[0,t]}u_s\,ds=\frac{l(l-1)B\,K_b\,t^2}{n}+2l(l-1)B\int_{[0,t]}u_s\,ds .$$

The left side does not depend on $n$, so letting $n$ grow gives

$$u_t\le2l(l-1)B\int_{[0,t]}u_s\,ds ,$$

an inequality which also holds trivially at $t=0$.

Step 5: conclusion. The function $u$ is bounded and measurable on $[0,T]$, as recorded in Step 0, and satisfies the hypothesis of Gronwall's lemma for bounded measurable functions with $a=0$ and $c=2l(l-1)B$. Therefore $u_t\le0$ for every $t$. Since $u_t\ge0$ by construction, $u_t=0$, so every coordinate of $x_t$ is nonnegative; with Step 1 this gives $x_t\in\Delta^l$. Finally $\hat{b}(x_t,A_t)=b(\pi_{\Delta^l}(x_t),A_t)=b(x_t,A_t)$ by the fixed-point clause of the projection lemma. $\blacksquare$