Symmetric BAOAB one-step weak expansion

Choose an integer $p\ge 0$ and $K<\mathrm{\infty }$ such that $\mathrm{\mid }{D}^j\varphi (z)\mathrm{\mid }\le K(1+\mathrm{\mid }z{\mathrm{\mid }}^p)$ for $0\le j\le 6$, and put $W_r(z)=(1+\mathrm{\mid }z{\mathrm{\mid }}^2{)}^{r\mathrm{/}2}$. For a bounded $C_b^6$ family choose $p=0$ and a common $K$. The explicit A and B maps are $(q,p_0)\mapsto (q+s{M}^{-1}{p}_0,p_0)$ and $(q,p_0)\mapsto (q,p_0+sF(q))$, where $F=-\mathrm{\nabla }V$. They have linear growth and uniformly bounded spatial derivatives of orders one through six for $0\le s\le 1$, because $D^{j}F$ is bounded for $1\le j\le 6$. The O map is affine with centered Gaussian noise of uniformly bounded moments of every fixed order by [F2]. Differentiating these explicit maps and taking Gaussian expectations therefore shows that each $P_s^C$, $C\in \{A,B,O\}$, preserves bounds $\mathrm{\mid }{D}^j\psi (z)\mathrm{\mid }\le K_0{W}_r(z)$ through any stated order $k\le 6$, uniformly for $s\le 1$. Each generator maps such bounds through order $k$ to bounds through order $k-2$ with weight $W_{r+1}$: its drift coefficients grow at most linearly, their positive-order derivatives are bounded, and its second-order coefficients are constant. Thus a composition with at most three generator applications, even with subflow semigroups between them, uses at most six test derivatives and is bounded by $C{W}_{p+3}(z)$.

For any $r\ge 0$, direct differentiation of $W_r$ and the linear-growth drift and constant diffusion give $\mathrm{\mid }L{W}_r\mathrm{\mid }\le C_r{W}_r$. Itô's formula [F3] stopped on a ball, the integral Gronwall estimate, and Fatou [F5] as the stopping radius increases give $P_u{W}_r(z)\le C_r{W}_r(z)$ for $0\le u\le 1$. These are estimates from a fixed deterministic starting point; higher weights can be used to justify the localization limits. In particular, $\mathrm{\mid }{L}^3\varphi (z)\mathrm{\mid }\le C{W}_{p+3}(z)$ by the same coefficient and derivative count as in step 1.1, and $\mathrm{\mid }{P}_u{L}^3\varphi (z)\mathrm{\mid }\le C{W}_{p+3}(z)$. For the bounded-test branch this is the cubic weight $W_3$, using the same Lyapunov function with exponent $3\mathrm{/}2$, not a higher initial-moment assumption.

Put $(C_1,\dots ,C_5)=(B,A,O,A,B)$, $(c_1,\dots ,c_5)=(1\mathrm{/}2,1\mathrm{/}2,1,1\mathrm{/}2,1\mathrm{/}2)$, and $\mathrm{\Psi }(u,z)=(P_{c_{1}u}^{C_1}\cdots P_{c_{5}u}^{C_5}\varphi )(z)$. The subflow integral identity $P_s^C\psi -\psi ={\int }_0^s{P}_v^{C}C\psi dv$ follows from the explicit A/B maps and Itô's formula for O. It and the weighted derivative bounds in step 1.1 justify differentiating this five-factor product three times, with right derivatives at zero. Explicitly, its third derivative is the finite ordered sum. [F1, F2, F3, given, step 1.1, algebra] ${\mathrm{\partial }}_u^3\mathrm{\Psi }(u)={\sum }_{{\nu }_1+\cdots +{\nu }_5=3}\frac{3!{\prod }_{i=1}^5{c}_i^{{\nu }_i}}{{\prod }_{i=1}^5{\nu }_i!}(P_{c_{1}u}^{C_1}{C}_1^{{\nu }_1}\cdots P_{c_{5}u}^{C_5}{C}_5^{{\nu }_5}\varphi )\mathrm{.}$ Every summand has total generator length three, so step 1.1 bounds it by $C{W}_{p+3}(z)$; no term with more than three generators is formed. The first derivative at zero is $L\varphi$. Grouping the second derivative at zero gives the pure squares $A^2+B^2+O^2$ and each ordered mixed pair $AB,BA,AO,OA,BO,OB$ once, hence ${\mathrm{\partial }}_u^2\mathrm{\Psi }(0)=L^2\varphi$. Taylor's formula with its third-derivative integral remainder now gives $K_h\varphi =\varphi +hL\varphi +\frac{1}{2}{h}^2{L}^2\varphi +\frac{1}{2}{\int }_0^h(h-u{)}^2{\mathrm{\partial }}_u^3\mathrm{\Psi }(u)du,$ whose remainder is bounded by $C{h}^3{W}_{p+3}(z)$. This is a second-order Taylor expansion of the product itself, rather than multiplication of five truncated polynomials with higher differential cross-products.

Applying the localized Itô integral identity successively to $\varphi$, $L\varphi$ and $L^2\varphi$ in which the three test functions have respectively six, four and two continuous spatial derivatives, gives the following identity. [A1, F3, F4, F5, given, step 1.1, step 2.1, algebra] $P_h\varphi =\varphi +hL\varphi +\frac{1}{2}{h}^2{L}^2\varphi +\frac{1}{2}{\int }_0^h(h-u{)}^2{P}_u{L}^3\varphi du\mathrm{.}$ The polynomial bounds from steps 1.1 and 2.1 justify taking expectations and removing the stopping times in these three identities. Step 2.1 bounds the remainder by $C{h}^3{W}_{p+3}(z)$.

Subtracting steps 2.2 and 3.1 cancels all terms through order two and gives $\mathrm{\mid }{K}_h\varphi (z)-P_h\varphi (z)\mathrm{\mid }\le C{h}^3{W}_{p+3}(z)\le C^{\mathrm{\prime }}{h}^3(1+\mathrm{\mid }z{\mathrm{\mid }}^{p+3})$. This proves the original polynomial-growth claim with some finite exponent. When the test family has a common $C_b^6$ bound, $p=0$ in every preceding estimate, so the same argument proves the uniform cubic-growth bound $C{h}^3(1+\mathrm{\mid }z{\mathrm{\mid }}^3)$. Neither conclusion is an invariant-measure statement.