Forward Itô SDE Well-Posedness

For $Z\in {\mathcal{H}}_T$, the path supremum is measurable by [F5], [F3] makes each coefficient composition predictable, and [F1] and Tonelli's theorem [F2] give $E{\int }_0^T(\mathrm{\mid }b(s,Z_s){\mathrm{\mid }}^2+\mathrm{\parallel }\sigma (s,Z_s){\mathrm{\parallel }}_F^2)ds\le 2K^2{\int }_0^T(1+E\mathrm{\mid }{Z}_s{\mathrm{\mid }}^2)ds\le 2K^{2}T(1+\mathrm{\parallel }Z{\mathrm{\parallel }}_T^2)<\mathrm{\infty }\mathrm{.}$ This nonnegative energy is finite almost surely; on each such path, applying $0\le {\int }_0^T(g-\bar{g}{)}^{2}ds$ to $g(s)=\mathrm{\mid }b(s,Z_s)\mathrm{\mid }$ and its average gives ${\int }_0^T\mathrm{\mid }b(s,Z_s)\mathrm{\mid }ds\le \sqrt{T}({\int }_0^T\mathrm{\mid }b(s,Z_s){\mathrm{\mid }}^{2}ds{)}^{1\mathrm{/}2}<\mathrm{\infty }$. Every scalar diffusion coordinate has finite energy. By [F4] each scalar integral has a continuous adapted version; their finite sum and the drift integral define $\mathrm{\Phi }(Z)$ continuously and adaptedly.

Write $S_t=({\sum }_k{\int }_0^t{\sigma }^{ik}(s,Z_s)d{W}_s^k{)}_{i=1}^d$. For vectors $z_1,\dots ,z_m$, expansion gives $m{\sum }_k\mathrm{\mid }{z}_k{\mathrm{\mid }}^2-\mathrm{\mid }{\sum }_k{z}_k{\mathrm{\mid }}^2={\sum }_{k<\mathrm{\ell }}\mathrm{\mid }{z}_k-z_{\mathrm{\ell }}{\mathrm{\mid }}^2\ge 0$. Hence ${\sup }_{u\le t}\mathrm{\mid }{S}_u{\mathrm{\mid }}^2\le m{\sum }_k{\sum }_i{\sup }_{u\le t}\mathrm{\mid }{M}_u^{ik}{\mathrm{\mid }}^2$ for the scalar coordinate integrals $M^{ik}$. Applying [F4] to each gives $E{\sup }_{u\le t}\mathrm{\mid }{S}_u{\mathrm{\mid }}^2\le 4mE{\int }_0^t\mathrm{\parallel }\sigma (s,Z_s){\mathrm{\parallel }}_F^{2}ds$.

For a square-integrable vector path $h$ on $[0,u]$, put $\bar{h}=u^{-1}{\int }_0^{u}h(s)ds$ when $u>0$. The identity $0\le {\int }_0^u\mathrm{\mid }h-\bar{h}{\mathrm{\mid }}^{2}ds={\int }_0^u\mathrm{\mid }h{\mathrm{\mid }}^{2}ds-u^{-1}\mathrm{\mid }{\int }_0^{u}h(s)ds{\mathrm{\mid }}^2$ gives $\mathrm{\mid }{\int }_0^{u}h(s)ds{\mathrm{\mid }}^2\le u{\int }_0^u\mathrm{\mid }h(s){\mathrm{\mid }}^{2}ds$; for $u=0$ both sides vanish. Applying this pathwise to the drift and using $\mathrm{\mid }{X}_0+D+S{\mathrm{\mid }}^2\le 3(\mathrm{\mid }{X}_0{\mathrm{\mid }}^2+\mathrm{\mid }D{\mathrm{\mid }}^2+\mathrm{\mid }S{\mathrm{\mid }}^2)$ give $E{\sup }_{u\le t}\mathrm{\mid }\mathrm{\Phi }(Z{)}_u{\mathrm{\mid }}^2\le 3E\mathrm{\mid }{X}_0{\mathrm{\mid }}^2+3tE{\int }_0^t\mathrm{\mid }b(s,Z_s){\mathrm{\mid }}^{2}ds+12mE{\int }_0^t\mathrm{\parallel }\sigma (s,Z_s){\mathrm{\parallel }}_F^{2}ds\mathrm{.}$ By [F1], each squared coefficient is at most $2K^2(1+\mathrm{\mid }{Z}_s{\mathrm{\mid }}^2)$. Since $t\le T$, the last display is at most $3E\mathrm{\mid }{X}_0{\mathrm{\mid }}^2+(6K^{2}T+24m{K}^2){\int }_0^t(1+\mathrm{\parallel }Z{\mathrm{\parallel }}_s^2)ds$. Hence $\mathrm{\Phi }(Z)\in {\mathcal{H}}_T$ and the first bound holds.

For $Z,V\in {\mathcal{H}}_T$, the initial terms cancel. The two-term square inequality, the drift Cauchy–Schwarz bound, the coordinate estimate of step 1.2, and the Lipschitz part of [F1] give $E{\sup }_{u\le t}\mathrm{\mid }\mathrm{\Phi }(Z{)}_u-\mathrm{\Phi }(V{)}_u{\mathrm{\mid }}^2\le 2t{L}^2{\int }_0^{t}E\mathrm{\mid }{Z}_s-V_s{\mathrm{\mid }}^{2}ds+8m{L}^2{\int }_0^{t}E\mathrm{\mid }{Z}_s-V_s{\mathrm{\mid }}^{2}ds\le (2T{L}^2+8m{L}^2){\int }_0^{t}E{\sup }_{r\le s}\mathrm{\mid }{Z}_r-V_r{\mathrm{\mid }}^{2}ds\mathrm{.}$ This is the second bound.

For the constant input $Z_t=X_0$, apply $\mathrm{\mid }D+S{\mathrm{\mid }}^2\le 2\mathrm{\mid }D{\mathrm{\mid }}^2+2\mathrm{\mid }S{\mathrm{\mid }}^2$ to $\mathrm{\Phi }(Z)-X_0$. Cauchy–Schwarz bounds the drift term by $2T{\int }_0^{T}E\mathrm{\mid }b(s,X_0){\mathrm{\mid }}^{2}ds\le 4K^2{T}^2(1+E\mathrm{\mid }{X}_0{\mathrm{\mid }}^2)$; step 1.2 and [F1] bound the diffusion term by $8m{\int }_0^{T}E\mathrm{\parallel }\sigma (s,X_0){\mathrm{\parallel }}_F^{2}ds\le 16m{K}^{2}T(1+E\mathrm{\mid }{X}_0{\mathrm{\mid }}^2)$. Their sum is the stated base-iterate bound.

If $T=0$, both integrals vanish, $\mathrm{\Phi }(Z{)}_0=X_0$, and all three estimates hold with zero integral terms; for $T>0$, the preceding steps establish every assertion. The coordinate sums are finite and no nondegeneracy of $\sigma$ is used. The Axiom of Choice enters only through the interfaces recorded in [F6].

If $T=0$, $Y_0=X_0$ and the claim is $E\mathrm{\mid }{X}_0{\mathrm{\mid }}^2<\mathrm{\infty }$. Suppose $T>0$. Use the everywhere-continuous representative permitted by [F1]; changing it on the ${\mathcal{F}}_0$-measurable continuity-null complement preserves the equation up to indistinguishability. Since $Y$ is continuous and adapted, [F2] gives the required progressive/predictable representatives. For $R>0$ define ${\tau }_R:=\inf \{s\in [0,T]:\mathrm{\mid }{Y}_s\mathrm{\mid }\ge R\}\wedge T$, with $\inf \mathrm{\varnothing }=+\mathrm{\infty }$. For $t<T$, $\{{\tau }_R\le t\}=\{{\sup }_{s\le t}\mathrm{\mid }{Y}_s\mathrm{\mid }\ge R\}$, which is ${\mathcal{F}}_t$-measurable because the supremum equals the supremum over rational times in $[0,t]$ and $t$ by [F4]; for $t\ge T$ the event is all of $\mathrm{\Omega }$. Thus ${\tau }_R$ is a stopping time. On $[0,{\tau }_R]$ the process is bounded by $R$, unless $\mathrm{\mid }{X}_0\mathrm{\mid }\ge R$, when ${\tau }_R=0$.

The multidimensional Itô formula [F2] for $U(x)=1+\mathrm{\mid }x{\mathrm{\mid }}^2$, stopped at ${\tau }_R$, has drift $2Y_s\cdot b(s,Y_s)+\mathrm{\parallel }\sigma (s,Y_s){\mathrm{\parallel }}_F^2$ and martingale integrand $2Y_s^{\mathsf{T}}\sigma (s,Y_s)$. By step 1.1 and the linear-growth bound, the stopped martingale integrand has finite energy on $[0,T]$, so [F3] and the stopping identity make its expectation zero. Since $2K\mathrm{\mid }y\mathrm{\mid }(1+\mathrm{\mid }y\mathrm{\mid })\le 3K(1+\mathrm{\mid }y{\mathrm{\mid }}^2)$ and $K^2(1+\mathrm{\mid }y\mathrm{\mid }{)}^2\le 2K^2(1+\mathrm{\mid }y{\mathrm{\mid }}^2)$, the generator is at most $C_0(1+\mathrm{\mid }y{\mathrm{\mid }}^2)$, where $C_0=3K+2K^2$. Hence, for $g_R(t):=E(1+\mathrm{\mid }{Y}_{t\wedge {\tau }_R}{\mathrm{\mid }}^2)$, $g_R(t)\le E(1+\mathrm{\mid }{X}_0{\mathrm{\mid }}^2)+C_0{\int }_0^t{g}_R(s)ds\mathrm{.}$ The stopped variables are integrable because $Y$ is bounded before the stop and $X_0\in L^2$.

Set $H_R(t)=E(1+\mathrm{\mid }{X}_0{\mathrm{\mid }}^2)+C_0{\int }_0^t{g}_R(s)ds$. Then $g_R\le H_R$ and $H_R^{\mathrm{\prime }}\le C_0{H}_R$ almost everywhere, so $(e^{-C_{0}t}{H}_R(t){)}^{\mathrm{\prime }}\le 0$. Therefore $g_R(t)\le E(1+\mathrm{\mid }{X}_0{\mathrm{\mid }}^2)e^{C_{0}t}$.

Take $R=n\in \mathbb{N}$ and let $n\to \mathrm{\infty }$. Every continuous path on $[0,T]$ is bounded, so $Y_{t\wedge {\tau }_n}\to Y_t$ almost surely for each fixed $t$. Fatou's lemma [F5] gives ${\sup }_{t\le T}E\mathrm{\mid }{Y}_t{\mathrm{\mid }}^2\le E(1+\mathrm{\mid }{X}_0{\mathrm{\mid }}^2)e^{C_{0}T}=:M_T<\mathrm{\infty }$.

The growth bound and step 4.1 imply $E{\int }_0^T\mathrm{\mid }b(s,Y_s){\mathrm{\mid }}^{2}ds\le 2K^{2}T(1+M_T)$ and $E{\int }_0^T\mathrm{\parallel }\sigma (s,Y_s){\mathrm{\parallel }}_F^{2}ds\le 2K^{2}T(1+M_T)$. Pathwise Cauchy–Schwarz for the drift and finite-energy Itô theory therefore put $Y$ in the continuous Brownian Itô-process class of [F1]. The vector martingale part $N_t=({\sum }_k{\int }_0^t{\sigma }^{ik}(s,Y_s)d{W}_s^k{)}_{i=1}^d$ satisfies $E{\sup }_{t\le T}\mathrm{\mid }{N}_t{\mathrm{\mid }}^2\le 4mE{\int }_0^T\mathrm{\parallel }\sigma (s,Y_s){\mathrm{\parallel }}_F^{2}ds,$ by the finite-coordinate inequality $\mathrm{\mid }{\sum }_k{z}_k{\mathrm{\mid }}^2\le m{\sum }_k\mathrm{\mid }{z}_k{\mathrm{\mid }}^2$ and the scalar bound in [F3]. Likewise, pathwise Cauchy–Schwarz gives $E{\sup }_{t\le T}\mathrm{\mid }{\int }_0^{t}b(s,Y_s)ds{\mathrm{\mid }}^2\le TE{\int }_0^T\mathrm{\mid }b(s,Y_s){\mathrm{\mid }}^{2}ds$. Thus the SDE equation and $\mathrm{\mid }{X}_0+D+N{\mathrm{\mid }}^2\le 3(\mathrm{\mid }{X}_0{\mathrm{\mid }}^2+\mathrm{\mid }D{\mathrm{\mid }}^2+\mathrm{\mid }N{\mathrm{\mid }}^2)$ imply $E{\sup }_{t\le T}\mathrm{\mid }{Y}_t{\mathrm{\mid }}^2<\mathrm{\infty }$. AC is inherited exactly as recorded in [F6].

If $T=0$, set $X_0$ as the sole value of the process; both integrals vanish, the path-sup bound follows from $(1+3E\mathrm{\mid }{X}_0{\mathrm{\mid }}^2)\ge E\mathrm{\mid }{X}_0{\mathrm{\mid }}^2$, and any two solutions with that initial value agree. For $T>0$, let $\mathrm{\Phi }$ be the map of [F1], set $X_t^0=X_0$, and recursively set $X^{n+1}=\mathrm{\Phi }(X^n)$.

Put $a_n(t)=E{\sup }_{u\le t}\mathrm{\mid }{X}_u^n-X_u^{n-1}{\mathrm{\mid }}^2$ for $n\ge 1$. By [F1], $a_1(T)\le A_T:=(4K^2{T}^2+16m{K}^{2}T)(1+E\mathrm{\mid }{X}_0{\mathrm{\mid }}^2)$ and $a_{n+1}(t)\le C_d{\int }_0^t{a}_n(s)ds$. Induction on $n$ gives $a_n(t)\le A_T{C}_d^{n-1}{t}^{n-1}\mathrm{/}(n-1)!$.

The majorant series ${\sum }_{n\ge 1}\sqrt{A_T{C}_d^{n-1}{T}^{n-1}\mathrm{/}(n-1)!}$ converges. If $A_T=0$, it is identically zero; if $C_{d}T=0$, every term after the first is zero; otherwise the ratio of consecutive terms is $\sqrt{C_{d}T\mathrm{/}n}\to 0$. By [F4], $E\sup \mathrm{\mid }{X}^n-X^{n-1}\mathrm{\mid }\le \sqrt{a_n(T)}$ and monotone convergence gives $E{\sum }_n\sup \mathrm{\mid }{X}^n-X^{n-1}\mathrm{\mid }={\sum }_{n}E\sup \mathrm{\mid }{X}^n-X^{n-1}\mathrm{\mid }<\mathrm{\infty }$; hence that nonnegative series is finite almost surely. On this event, for every $t$ the vector sequence $(X_t^n{)}_n$ is Cauchy, since ${\sup }_{u\le T}\mathrm{\mid }{X}_u^m-X_u^n\mathrm{\mid }\le {\sum }_{j=n+1}^m{\sup }_{u\le T}\mathrm{\mid }{X}_u^j-X_u^{j-1}\mathrm{\mid }$; [F7] gives the coordinate limits and therefore a vector limit $X_t$. Letting $m\to \mathrm{\infty }$ in the same inequality bounds ${\sup }_{t\le T}\mathrm{\mid }{X}_t^n-X_t\mathrm{\mid }$ by the tail of the summable series, which tends to zero. Thus the convergence is almost surely uniform. For finite tails the $L^2$ triangle inequality follows by expanding the square and applying Cauchy–Schwarz, so $\mathrm{\parallel }{X}^m-X^n{\mathrm{\parallel }}_{L^2(\sup )}\le {\sum }_{j=n+1}^m\sqrt{a_j(T)}$. Fatou's lemma [F4] then gives $E{\sup }_{t\le T}\mathrm{\mid }{X}_t^n-X_t{\mathrm{\mid }}^2\le ({\sum }_{j=n+1}^{\mathrm{\infty }}\sqrt{a_j(T)}{)}^2\to 0$.

Let $A$ be the measurable probability-one event on which that series of path suprema is finite; measurability follows because each supremum is a supremum over the countable dense set in [F5]. Uniform convergence makes $X$ continuous on $A$; completeness gives $A^c\in {\mathcal{F}}_0$, so set $X_t=X_0$ on $A^c$. For each fixed $t$, the limit on $A$ of the ${\mathcal{F}}_t$-measurable $X_t^n$, together with the ${\mathcal{F}}_0$-measurable value on $A^c$, is ${\mathcal{F}}_t$-measurable. Thus $X$ is adapted and continuous, and its $L^2$ path-sup convergence places it in the class of [F2].

For every $n$, $X^{n+1}=\mathrm{\Phi }(X^n)$. The difference estimate of [F1] gives $E{\sup }_{u\le T}\mathrm{\mid }\mathrm{\Phi }(X^n{)}_u-\mathrm{\Phi }(X{)}_u{\mathrm{\mid }}^2\le C_{d}TE{\sup }_{u\le T}\mathrm{\mid }{X}_u^n-X_u{\mathrm{\mid }}^2\to 0$. Since $X^{n+1}\to X$ in the same norm by step 3.1, uniqueness of an $L^2$ limit implies $\mathrm{\Phi }(X)=X$ indistinguishably. Hence $X$ satisfies the SDE.

Write $f(t)=E{\sup }_{u\le t}\mathrm{\mid }{X}_u{\mathrm{\mid }}^2$. The growth estimate of [F1] and $X=\mathrm{\Phi }(X)$ imply $f(t)\le 3E\mathrm{\mid }{X}_0{\mathrm{\mid }}^2+C_T{\int }_0^t(1+f(s))ds$. Put $g=1+f$ and $H(t)=1+3E\mathrm{\mid }{X}_0{\mathrm{\mid }}^2+C_T{\int }_0^{t}g(s)ds$. Then $g\le H$, $H(0)=1+3E\mathrm{\mid }{X}_0{\mathrm{\mid }}^2$, and $H^{\mathrm{\prime }}(t)=C_{T}g(t)\le C_{T}H(t)$ almost everywhere. Consequently $(e^{-C_{T}t}H(t){)}^{\mathrm{\prime }}\le 0$ almost everywhere, so $f(T)\le g(T)\le (1+3E\mathrm{\mid }{X}_0{\mathrm{\mid }}^2)e^{C_{T}T}$.

If $Y$ is any other strong solution with the same data, then $Y=\mathrm{\Phi }(Y)$ by [F2] and $Y\in {\mathcal{H}}_T$ by [F6]. The difference estimate in [F1] gives $q(t):=E{\sup }_{u\le t}\mathrm{\mid }{X}_u-Y_u{\mathrm{\mid }}^2\le C_d{\int }_0^{t}q(s)ds$. With $Q(t)=C_d{\int }_0^{t}q(s)ds$, one has $Q(0)=0$ and $Q^{\mathrm{\prime }}\le C_{d}Q$ almost everywhere, hence $Q=q=0$. Thus ${\sup }_{t\le T}\mathrm{\mid }{X}_t-Y_t\mathrm{\mid }=0$ almost surely, which is indistinguishability. AC is exactly the inherited assumption [F3].

If $T=0$, then $\mathrm{\mid }{X}_0{\mathrm{\mid }}^p\le 2^{p\mathrm{/}2}(1+\mathrm{\mid }{X}_0{\mathrm{\mid }}^p)$ and the result follows. For $T>0$, put $V(x)=(1+\mathrm{\mid }x{\mathrm{\mid }}^2{)}^{p\mathrm{/}2}$. It is $C^2$ and ${\mathrm{\partial }}_{i}V=p{x}_i(1+\mathrm{\mid }x{\mathrm{\mid }}^2{)}^{p\mathrm{/}2-1},{\mathrm{\partial }}_{ij}V=p{\delta }_{ij}(1+\mathrm{\mid }x{\mathrm{\mid }}^2{)}^{p\mathrm{/}2-1}+p(p-2)x_i{x}_j(1+\mathrm{\mid }x{\mathrm{\mid }}^2{)}^{p\mathrm{/}2-2}\mathrm{.}$ The gradient bound follows from $\mathrm{\mid }x\mathrm{\mid }\le (1+\mathrm{\mid }x{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}$. The Hessian is a nonnegative multiple of $I$ plus a nonnegative multiple of $x{x}^{\mathsf{T}}$, so it is positive semidefinite. Its eigenvalues orthogonal to $x$ are $p(1+\mathrm{\mid }x{\mathrm{\mid }}^2{)}^{p\mathrm{/}2-1}$, and the eigenvalue in the $x$ direction is $p(1+\mathrm{\mid }x{\mathrm{\mid }}^2{)}^{p\mathrm{/}2-2}(1+(p-1)\mathrm{\mid }x{\mathrm{\mid }}^2)$; each is at most $p(p-1)(1+\mathrm{\mid }x{\mathrm{\mid }}^2{)}^{p\mathrm{/}2-1}$ because $p\ge 2$. At $x=0$ all eigenvalues equal $p$, which obeys the same bound.

By [F1], $E{\sup }_{t\le T}\mathrm{\mid }{X}_t{\mathrm{\mid }}^2<\mathrm{\infty }$, so [F2] makes $b(\cdot ,X_{\cdot })$ progressive and $\sigma (\cdot ,X_{\cdot })$ predictable. The growth bound gives $E{\int }_0^T(\mathrm{\mid }b(s,X_s){\mathrm{\mid }}^2+\mathrm{\parallel }\sigma (s,X_s){\mathrm{\parallel }}_F^2)ds\le 2K^{2}T(1+E{\sup }_{s\le T}\mathrm{\mid }{X}_s{\mathrm{\mid }}^2)<\mathrm{\infty }\mathrm{.}$ Thus the drift-square integral and diffusion energy are finite almost surely; pathwise Cauchy–Schwarz gives ${\int }_0^T\mathrm{\mid }b(s,X_s)\mathrm{\mid }ds\le \sqrt{T}({\int }_0^T\mathrm{\mid }b(s,X_s){\mathrm{\mid }}^{2}ds{)}^{1\mathrm{/}2}<\mathrm{\infty }$. Hence $X$ satisfies the continuous Brownian Itô-process hypotheses in [F2].

For $R>0$ define ${\tau }_R:=\inf \{s\in [0,T]:\mathrm{\mid }{X}_s\mathrm{\mid }\ge R\}\wedge T$, with $\inf \mathrm{\varnothing }=+\mathrm{\infty }$. For $t<T$, $\{{\tau }_R\le t\}=\{{\sup }_{s\le t}\mathrm{\mid }{X}_s\mathrm{\mid }\ge R\}$; by continuity and [F4] that supremum is measurable using the countable rational times in $[0,t]$ and $t$, all of which are ${\mathcal{F}}_t$-measurable. For $t\ge T$ the event is all of $\mathrm{\Omega }$, so ${\tau }_R$ is a stopping time. On $[0,{\tau }_R]$ the process is bounded by $R$ unless $\mathrm{\mid }{X}_0\mathrm{\mid }\ge R$, in which case ${\tau }_R=0$. Hence $1_{[0,{\tau }_R]}\mathrm{\nabla }V(X_s{)}^{\mathsf{T}}\sigma (s,X_s)$ is predictable and has finite energy on $[0,T]$.

Let ${\mathcal{L}}_{s}V(x)=b(s,x)\cdot \mathrm{\nabla }V(x)+\frac{1}{2}\mathrm{tr}(\sigma {\sigma }^{\mathsf{T}}(s,x)D^{2}V(x))$. The inequality $1+\mathrm{\mid }x\mathrm{\mid }\le \sqrt{2(1+\mathrm{\mid }x{\mathrm{\mid }}^2)}$ follows by squaring and using $(\mathrm{\mid }x\mathrm{\mid }-1{)}^2\ge 0$, so the drift term has absolute value at most $\sqrt{2}pKV(x)$. For each diffusion column ${\sigma }_{\cdot k}$, positive semidefiniteness and the operator-norm bound give $0\le {\sigma }_{\cdot k}^{\mathsf{T}}{D}^{2}V{\sigma }_{\cdot k}\le \mathrm{\parallel }{D}^{2}V{\mathrm{\parallel }}_{op}\mathrm{\mid }{\sigma }_{\cdot k}{\mathrm{\mid }}^2$. Summing over the finite $m$ columns yields $0\le \frac{1}{2}\mathrm{tr}(\sigma {\sigma }^{\mathsf{T}}{D}^{2}V)\le p(p-1)K^{2}V(x)$. Thus ${\mathcal{L}}_{s}V(x)\le C_{p}V(x)$.

Apply the multidimensional Itô formula [F2] to $V(X)$ and evaluate the common continuous identity at $t\wedge {\tau }_R$. By the stopping identity and mean-zero finite-energy martingale facts in [F3], the stopped stochastic integral has expectation zero. With $g_R(t):=EV(X_{t\wedge {\tau }_R})$, nonnegativity of $V$ and step 2.1 therefore give $g_R(t)=EV(X_0)+E{\int }_0^t{1}_{[0,{\tau }_R]}(s){\mathcal{L}}_{s}V(X_s)ds\le EV(X_0)+C_p{\int }_0^t{g}_R(s)ds\mathrm{.}$ The right side is finite: before stopping $X$ is bounded, and on paths stopped at zero the value is $X_0$ with finite $p$th moment.

Set $H_R(t)=EV(X_0)+C_p{\int }_0^t{g}_R(s)ds$. By step 3.1, $g_R\le H_R$; also $H_R$ is absolutely continuous and $H_R^{\mathrm{\prime }}(t)=C_p{g}_R(t)\le C_p{H}_R(t)$ almost everywhere. Thus $(e^{-C_{p}t}{H}_R(t){)}^{\mathrm{\prime }}\le 0$ almost everywhere, giving $g_R(t)\le EV(X_0)e^{C_{p}t}$.

Take $R=n\in \mathbb{N}$ and let $n\to \mathrm{\infty }$. Every continuous path on $[0,T]$ is bounded, so for each fixed $t$ one has $X_{t\wedge {\tau }_n}\to X_t$ almost surely. Fatou's lemma [F5] and step 4.1 give $EV(X_t)\le EV(X_0)e^{C_{p}T}$. Also $V(X_0)=(1+\mathrm{\mid }{X}_0{\mathrm{\mid }}^2{)}^{p\mathrm{/}2}\le 2^{p\mathrm{/}2}(1+\mathrm{\mid }{X}_0{\mathrm{\mid }}^p)$ and $\mathrm{\mid }{X}_t{\mathrm{\mid }}^p\le V(X_t)$. Taking the supremum over $0\le t\le T$ proves the stated moment bound. AC is only the inherited interface assumption [F6].