A Lipschitz SDE solution has a finite path-sup second moment

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].