An L1 maximal bound for finite-energy Brownian Itô integrals

If $T=0$, both sides vanish. Henceforth $T>0$. For stopping-time notation only, extend the filtration by ${\mathcal{F}}_t={\mathcal{F}}_T$ for $t>T$; it remains complete and right-continuous, and the predictable sigma-algebra on $[0,T]$ is unchanged by [F4]. All Brownian integrals and estimates below are on $[0,T]$; no Brownian property for this extension after $T$ is asserted or used. First take $m=1$ and write $B=W^1$. On the common null exceptional set, set $M$ and $Q$ to zero; completeness keeps them adapted and makes their paths continuous everywhere. This changes no integral class. Put $Q_t={\int }_0^t{H}_s^{2}ds$ and $M^{\ast }={\sup }_{t\le T}\mathrm{\mid }{M}_t\mathrm{\mid }$. The continuous version makes $M^{\ast }$ measurable, and [F1] gives $\mathbb{E}(M^{\ast }{)}^2\le 4\mathbb{E}{Q}_T<\mathrm{\infty }$, so $\mathbb{E}{M}^{\ast }<\mathrm{\infty }$.

Fix $\lambda >0$. Let $\tau$ be the first time $\mathrm{\mid }{M}_t\mathrm{\mid }\ge \lambda$, capped at $T$, and put $A_{\lambda }=\{M^{\ast }\ge \lambda \}$. Continuity and adaptation make $\tau$ a stopping time by [F9]–[F10], and $A_{\lambda }\in {\mathcal{F}}_{\tau }$: for each $t<T$, $A_{\lambda }\cap \{\tau \le t\}=\{{\sup }_{s\le t}\mathrm{\mid }{M}_s\mathrm{\mid }\ge \lambda \}$, whose supremum is measurable by [F10]; for $t\ge T$, the intersection is $A_{\lambda }\in {\mathcal{F}}_T\subseteq {\mathcal{F}}_t$ by the same terminal supremum fact. Let $R_t=Q_t-Q_{t\wedge \tau }$; this is continuous and adapted. Let $\sigma$ be its first hitting time of ${\delta }^2{\lambda }^2$, capped at $T$; it is a stopping time by [F9]–[F10], where $0<\delta \le 1\mathrm{/}4$. The set $A_{\lambda }\cap \{\tau <s\}$ is the countable union, over rational $u>0$, of $(u,\mathrm{\infty })\times (A_{\lambda }\cap \{\tau \le u\})$. Each second factor belongs to ${\mathcal{F}}_u$ by $A_{\lambda }\in {\mathcal{F}}_{\tau }$. Thus it is predictable; intersecting with $\{s\le \sigma \}$ preserves predictability by [F4].

Define $N_t={\int }_0^t{1}_{A_{\lambda }}{1}_{(\tau ,\sigma ]}(s)H_{s}d{B}_s\mathrm{.}$ For each $n$, let $G_n=\{jT{2}^{-n}:0\le j\le 2^n\}$ and round $\tau ,\sigma$ upward to their least grid points ${\tau }_n,{\sigma }_n\in G_n$. They are stopping times because $\{{\tau }_n\le t\}={\bigcup }_{q\in G_n,q\le t}\{\tau \le q\}\in {\mathcal{F}}_t$, and likewise for ${\sigma }_n$; monotonicity of the ceiling map gives ${\tau }_n\le {\sigma }_n$ and ${\tau }_n\downarrow \tau$, ${\sigma }_n\downarrow \sigma$. Since $A_{\lambda }\in {\mathcal{F}}_{\tau }$ and $\tau \le {\tau }_n$, one has $A_{\lambda }\in {\mathcal{F}}_{{\tau }_n}$: for every $t$, $A_{\lambda }\cap \{{\tau }_n\le t\}=(A_{\lambda }\cap \{\tau \le t\})\cap \{{\tau }_n\le t\}\in {\mathcal{F}}_t$. For an elementary predictable integrand on a refinement of this grid, the coefficient on $(t_k,t_{k+1}]$ after multiplying by $1_{A_{\lambda }}{1}_{({\tau }_n,{\sigma }_n]}$ is measurable at $t_k$: the active event $\{{\tau }_n\le t_k<{\sigma }_n\}$ is in ${\mathcal{F}}_{t_k}$, and $A_{\lambda }\cap \{{\tau }_n\le t_k\}$ is in ${\mathcal{F}}_{t_k}$. Thus the finite-sum definition [F6] gives ${\int }_0^t{1}_{A_{\lambda }}{1}_{({\tau }_n,{\sigma }_n]}{H}_{s}d{B}_s=1_{A_{\lambda }}(M_{t\wedge {\sigma }_n}-M_{t\wedge {\tau }_n})$ for elementary $H$. For general predictable $H$, approximate in predictable $L^2$ by elementary $H^k$ using [F7]. The isometry [F5] passes the left side to its limit in $L^2(P)$, and [F1] applied to $H^k-H$ gives $\mathbb{E}{\sup }_{t\le T}\mathrm{\mid }{M}_t^k-M_t{\mathrm{\mid }}^2\to 0$, passing the stopped evaluations on the right to their limits for this fixed grid. As $n\to \mathrm{\infty }$, the interval indicators converge outside the graphs of $\tau$ and $\sigma$, which are $dt\otimes P$-null by Tonelli [F8]. Dominated convergence [F11] with majorant $H^2$ and the isometry [F5] pass the identity to $(\tau ,\sigma ]$. Continuity gives convergence of the right side to $1_{A_{\lambda }}(M_{t\wedge \sigma }-M_{t\wedge \tau })$ almost surely for each fixed $t$, proving the required stopped-integral identity entirely on $[0,T]$. Both sides have continuous versions, so their equality at the countable rational times and continuity give equality up to indistinguishability. The accumulated energy of the integrand is at most ${\delta }^2{\lambda }^2{1}_{A_{\lambda }}$ on every path. Thus [F1] gives $\mathbb{E}{\sup }_{t\le T}\mathrm{\mid }{N}_t{\mathrm{\mid }}^2\le 4{\delta }^2{\lambda }^2\mathbb{P}(A_{\lambda })\mathrm{.}$ On $\{M^{\ast }>2\lambda ,Q_T^{1\mathrm{/}2}<\delta \lambda \}$ the threshold defining $\sigma$ is not reached before $T$, and after the first hit of $\lambda$ the continuous path moves by more than $\lambda$. Hence ${\sup }_{t\le T}\mathrm{\mid }{N}_t\mathrm{\mid }>\lambda$ on that event. Markov's inequality and the preceding display yield $\mathbb{P}(M^{\ast }>2\lambda ,Q_T^{1\mathrm{/}2}<\delta \lambda )\le 4{\delta }^2\mathbb{P}(M^{\ast }\ge \lambda )\mathrm{.}$

Split the event $\{M^{\ast }>2\lambda \}$ according to whether $Q_T^{1\mathrm{/}2}\ge \delta \lambda$. The bound in step 2.1 gives $\mathbb{P}(M^{\ast }>2\lambda )\le \mathbb{P}(Q_T^{1\mathrm{/}2}\ge \delta \lambda )+4{\delta }^2\mathbb{P}(M^{\ast }\ge \lambda )\mathrm{.}$ For any nonnegative $Z$, the integrals of $1_{\{\lambda <Z\}}$ and $1_{\{\lambda \le Z\}}$ over $\lambda >0$ agree pathwise, since they differ at at most one point. Integrating this inequality over $\lambda >0$ and applying Tonelli [F8] to $1_{\{(\lambda ,\omega ):\lambda <M^{\ast }(\omega )\}}$ and $1_{\{(\lambda ,\omega ):\lambda <Q_T(\omega {)}^{1\mathrm{/}2}\}}$ gives $\mathbb{E}{M}^{\ast }\le \frac{2}{\delta }\mathbb{E}{Q}_T^{1\mathrm{/}2}+8{\delta }^2\mathbb{E}{M}^{\ast }\mathrm{.}$ With $\delta =1\mathrm{/}4$, the last coefficient is $1\mathrm{/}2$, so $\mathbb{E}{M}^{\ast }\le 16\mathbb{E}{Q}_T^{1\mathrm{/}2}$.

For general finite $m$, write $M={\sum }_{j=1}^m{M}^j$ with $M_t^j={\int }_0^t{H}_s^{j}d{W}_s^j$. Apply the scalar result of step 3.1 to each coordinate and then the finite-sum Cauchy–Schwarz inequality pathwise: $\mathbb{E}{\sup }_{t\le T}\mathrm{\mid }{M}_t\mathrm{\mid }\le 16{\sum }_{j=1}^m\mathbb{E}{\left({\int }_0^T\mathrm{\mid }{H}_s^j{\mathrm{\mid }}^{2}ds\right)}^{1\mathrm{/}2}\le 16\sqrt{m}\mathbb{E}{\left({\int }_0^T{\sum }_{j=1}^m\mathrm{\mid }{H}_s^j{\mathrm{\mid }}^{2}ds\right)}^{1\mathrm{/}2}\mathrm{.}$ This is the claimed estimate. AC enters only through [A1] and the cited Brownian and Itô interfaces.