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For standard one-dimensional Brownian motion, the Itô formula for $B^k$ gives this recursion. Stop at exits from $[-n,n]$ to make the stochastic integral a true martingale. Gaussian tail integrability and the Doob Lp maximal inequality dominate all stopped powers on a fixed horizon, justifying the limit. Odd moments vanish and $\mathbb{E}{B}_t^{2k}=(2k)!t^k/(2^{k}k!)$.

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For a nonnegative submartingale $X$, the maximal tail bound is $\lambda \mathbb{P}({\max }_{k\le n}{X}_k\ge \lambda )\le \mathbb{E}[X_n;{\max }_{k\le n}{X}_k\ge \lambda ]$. Applied to the absolute value of a martingale and integrated over $\lambda$, it gives the Doob Lp maximal inequality; at $p=2$ this is the Doob L2 maximal inequality.

Let $F,G\ge 0$, $1<p<\infty$, and $G\in L^p$. Assume the displayed tail inequality and finite positive-threshold superlevel measures. First split $G$ at $a/2$ to obtain $a\mu \{F>2a\}\le {\int }_{\{G>a\}}G$. Layer cake representation and Tonelli theorem imply $F\in L^p$. Integrating the original inequality against $p{a}^{p-2}$ then gives $\Vert F{\Vert }_p^p\le q\int G{F}^{p-1}\le q\Vert G{\Vert }_p\Vert F{\Vert }_p^{p-1}$, with $q=p/(p-1)$. This is the analytic estimate behind the Doob Lp maximal inequality.

A martingale satisfying ${\sup }_n\mathbb{E}|M_n{|}^p<\infty$, for $1<p<\infty$, has almost sure convergence and convergence in Lp to a limit $M_{\infty }\in L^p$. Also $M_n=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_n]$. The Doob Lp maximal inequality makes ${\sup }_n|M_n|$ integrable to the power $p$; the martingale convergence theorem gives the almost-sure limit, and the dominated convergence theorem gives convergence in the Lp norm. Boundedness in $L^1$ alone does not imply convergence in L1.

Let $(X_k{)}_{k=0}^n$ be a nonnegative submartingale and assume $X_n\in L^p$. Set $X^{\ast }={\max }_{0\le k\le n}{X}_k$. For $a>0$, let $A_k$ be the event that $k$ is the first index with $X_k>a$. These events are disjoint, $A_k\in {\mathcal{F}}_k$, and their union is $\{X^{\ast }>a\}$.

The submartingale property gives $\mathbb{E}[X_n\mid {\mathcal{F}}_k]\ge X_k$. HenceThus part (ii) applies on the probability space with $F=X^{\ast }$ and $g=X_n$, giving the Doob Lp maximal inequalityThe same split at $a/2$ makes $X^{\ast }/2$ satisfy part (i), recovering the preliminary constant $2^{p}q$. Probability spaces automatically have finite superlevel measures. For a right-continuous nonnegative submartingale on $[0,T]$, apply these finite-time bounds to increasing finite grids containing $T$ and a dense set of times; right continuity and monotone convergence theorem extend the estimate to ${\sup }_{t\le T}{X}_t$.

Let $(Z_k,{\mathcal{F}}_k{)}_{0\le k\le n}$ be an integrable real submartingale, and put $Z_n^{\ast }={\max }_{0\le k\le n}{Z}_k$. The refined Doob maximal inequality isIn particular, for a nonnegative submartingale the right-hand side is at most $\mathbb{E}{Z}_n$, giving the usual weak maximal bound.

For the proof, partition $A=\{Z_n^{\ast }\ge \lambda \}$ by its first crossing time: $A_k=\{Z_j<\lambda (j<k),Z_k\ge \lambda \}\in {\mathcal{F}}_k$. The submartingale property gives $\mathbb{E}[Z_n\mid {\mathcal{F}}_k]\ge Z_k$, soSumming over the disjoint crossing events proves the refined inequality, including a possible crossing at time zero. No independence assumption is needed.

The strong form is the Doob Lp maximal inequality. For $p>1$ and a nonnegative submartingale with $Z_n\in L^p$,Indeed, integrate the refined inequality against $p{\lambda }^{p-2}$ up to $K$. The tail integral formula for moments and the Tonelli theorem giveApply the Holder inequality, divide by $\Vert Z_n^{\ast }\wedge K{\Vert }_p^{p-1}$ when it is nonzero, and let $K\to \infty$ by the monotone convergence theorem. For a martingale $M$, the process $|M_k|$ is a submartingale by the conditional Jensen inequality, giving the corresponding bound for $\max |M_k|$. On a bounded continuous-time interval the nonnegative, right-continuous version follows by applying these inequalities on refining finite time grids and taking limits.

Put ${\mathcal{F}}_n=\sigma (B_{k{2}^{-n}}:0\le k\le 2^n)$. These sigma-fields increase, and their union generates the continuous Brownian path. Let ${\mathbf{B}}^{(n)}=S_2(B^{(n)})$ be the polygonal enhancement. Its coordinates are determined by its displacement $b_{s,t}^{(n)}$ and antisymmetric area $a_{s,t}^{(n)}$; the symmetric second level is $\frac{1}{2}{b}_{s,t}^{(n)}\otimes b_{s,t}^{(n)}$.

The given conditional-expectation identities implyFor the second assertion it is important to use area coordinates rather than asserting that the whole tensor signature is a conditional expectation. Write $I_t={\int }_0^t\beta d\tilde{\beta }$. The local off-diagonal second integral isConditional on their separate dyadic values, the two independent Brownian components remain independent. Thus the conditional expectation of the product in the last term is ${\beta }_s^{(n)}({\tilde{\beta }}_t^{(n)}-{\tilde{\beta }}_s^{(n)})$. The assumed identity for $I$ gives the polygonal $12$ entry. The reverse entry follows by integration by parts, since the product of the two component increments also has the product of their conditional means. Taking the antisymmetric part proves the area identity. Diagonal second levels are recovered from the geometric identity, rather than from a conditional-expectation claim about squares.

For each fixed dyadic time $t$, the first-level and area conditional expectations are uniformly integrable martingales in $n$. Their targets are measurable with respect to ${\mathcal{F}}_{\infty }=\sigma ({\bigcup }_n{\mathcal{F}}_n)$: Brownian area is a limit of stochastic integral sums determined by the path. The martingale convergence theorem gives convergence to $B_t$ and $A_{0,t}$, respectively. The continuous group coordinate formulas then giveat all dyadic times on a common event of probability one.

For the uniform Hölder estimate, fix $\alpha <1/2$ and put $K=\Vert \mathbf{B}{\Vert }_{\alpha \text{-Ho¨lder};[0,1]}$. The given all-order integrability and the group norm comparison implyUse the conditional-expectation formulas, the conditional Jensen inequality, and $\mathbb{E}[K\mid {\mathcal{F}}_n]\le \sqrt{\mathbb{E}[K^2\mid {\mathcal{F}}_n]}$. The group norm comparison yieldsFirst take all dyadic pairs, a countable collection, and then extend the inequality to all $s,t$ by continuity. Hence the Brownian rough path martingale Hölder bound isFor any $q>1$, the Doob Lp maximal inequality givesTherefore

To upgrade convergence from dyadic times to every time simultaneously, apply this bound at one positive exponent $\beta <1/2$. The polygonal enhancements are then equicontinuous in the group metric, with a common finite random constant. For a finite dyadic time net, convergence holds at every net point. The triangle inequality bounds the errors at the remaining times by the two Hölder errors to the nearest net point and the error at that point. First let $n\to \infty$, and then let the net mesh tend to zero. This proves uniform convergence, and in particular convergence for every $t\in [0,1]$ on one event of probability one.

There is also convergence in each weaker Hölder rough path metric. Uniform convergence of the group coordinates gives uniform convergence of their levelwise increments through the Chen formulas. If $\alpha <\beta <1/2$, the uniform $\beta$ bounds bound the level-$k$ increment differences by a random constant times $|t-s{|}^{k\beta }$, while their unweighted suprema tend to zero. Interpolating these two bounds proves convergence with denominators $|t-s{|}^{k\alpha }$, for $k=1,2$. This establishes the nested polygonal approximation of Brownian rough paths in the topology needed for the support argument.

Fix $T<\infty$ and a standard $d$-dimensional Brownian motion $B$ starting at zero. Let $D_n$ be deterministic nested finite partitions of $[0,T]$ with mesh tending to zero; dyadic partitions are the basic example. Let $B^n$ be the polygonal interpolation through the partition values and put ${\mathbf{B}}^n=S_2(B^n)$. The limit is enhanced Brownian motion, with incrementsHere the integral is a Stratonovich integral. In terms of Itô integrals,Its symmetric part is $\frac{1}{2}{B}_{s,t}^{\otimes 2}$, while its antisymmetric part is Lévy area,These identities and splitting the integral at an intermediate time give the Chen identity, so $\mathbf{B}$ has values in $G^2({\mathbb{R}}^d)$.

A precise nested polygonal approximation of Brownian rough paths statement is: for every $0<\alpha <1/2$,where the rough path metric used here isIn particular, for each $2<p<3$, choosing $1/p<\alpha <1/2$ gives convergence in the inhomogeneous $p$-variation rough path metric. The limiting signed area is the Lévy area, and the limiting full second level is Stratonovich, not Itô. This distinction is necessary: a polygonal lift has the prescribed symmetric second level, whereas an Itô lift would lack half the quadratic-variation correction.

Here is a proof sketch emphasizing the nested martingale structure. SetA Brownian bridge between two prescribed endpoints has the affine interpolation as conditional mean. ConsequentlyFor $i\ne j$ the coordinate processes remain independent when their respective partition values are conditioned on. Approximating the Itô integral by its left sums, conditioning each product, and then taking its $L^2$ limit givesIndeed each conditioned factor is its polygonal interpolation, and their sums converge to the ordinary integral of those piecewise linear functions. For a fixed pair $s,t$, the right side is therefore a martingale in the refinement index $n$. It is not being asserted to be a martingale in physical time; polygonal interpolation uses the right endpoint of its current cell.

The union of the partitions is dense. Continuity of $B$ implies that ${\mathcal{G}}_{\infty }=\sigma ({\bigcup }_n{\mathcal{G}}_n)$ contains the full Brownian path and its stochastic integrals, up to completion. The martingale convergence theorem for conditional expectations therefore yields convergence of $B_t^n$ and all off-diagonal ${\mathbb{B}}_{s,t}^{n,ij}$ at every fixed pair, almost surely and in $L^q$ for finite $q$. Simultaneously it holds for a countable dense collection of pairs. The diagonal terms must be handled separately:They are not conditional expectations of the diagonal Brownian second level: the latter would also contain a conditional variance.

Pointwise convergence alone is insufficient for the rough path metric. The Burkholder-Davis-Gundy inequalities, Brownian scaling and Gaussian increment moments give, for arbitrarily large $q$,Equivalently, using the homogeneous norm on $G^2$, the $q$-th moment of the Carnot-Carathéodory distance of an increment is bounded by $C_q|t-s{|}^{q/2}$. The metric form of the Kolmogorov continuity theorem, with $q$ chosen large, gives for every $\beta <1/2$ a random $K_{\beta }$ having all finite moments such thatConditioning these bounds givesThe diagonal bounds follow by squaring the first bound. The two conditional-expectation processes are nonnegative martingales; the Doob Lp maximal inequality makes their suprema over $n$ finite almost surely, with all needed moments. Thus all the polygonal lifts have a common random bound at exponent $\beta$.

These bounds and the Chen identity give joint equicontinuity of the increments as functions of $(s,t)$. Convergence on the dense collection of pairs consequently gives uniform convergence of both levels on the time triangle. Take $\alpha <\beta <1/2$. For either level $k=1,2$, with $M$ a common bound and ${\epsilon }_n$ its uniform difference, interpolation givesThis proves the stated Hölder rough path metric convergence. Finally, for a partition $D$, the $k$-th level difference satisfiesbecause $\alpha p>1$. Taking the appropriate powers and suprema proves $p$-variation convergence too. Nesting is what permits the conditional-expectation martingale proof; without nesting one instead needs direct approximation estimates.

Fix a finite horizon $T$ and define ${\tau }_n=\inf \{s\ge 0:|B_s|\ge n\}$. These stopping times increase to infinity almost surely because continuous paths are bounded on each finite interval. Applying the Itô formula to $B_{t\wedge {\tau }_n}^k$, for $k\ge 2$, givesThe first integrand is bounded by $k{n}^{k-1}$, so its integral is square-integrable and has expectation zero. ThusTo justify both limits explicitly, let $S_T={\sup }_{s\le T}|B_s|$ and choose an integer $q>\max \{k,1\}$. The Gaussian density gives $\mathbb{E}|B_T{|}^q<\infty$ by its exponential tail. The Doob Lp maximal inequality statesTherefore $S_T^k$ and $1+S_T^{k-2}$ are integrable. The stopped terminal powers converge almost surely to $B_t^k$ and are dominated in absolute value by $S_T^k$. The integrands in the time integral converge almost surely and are bounded by $1+S_T^{k-2}$, so their integrals are dominated by $T(1+S_T^{k-2})$. Dominated convergence theorem on both sides, and then Fubini justified by the same bound, give the Brownian moment recursionThis uses only finiteness of Gaussian moments, not the closed moment formula requested in the next part.

For a nonnegative submartingale $(M_k{)}_{0\le k\le n}$ and $M_n^{\ast }={\max }_{0\le k\le n}{M}_k$, the weak form of the Doob maximal inequality isTo prove it, let $A_k$ be the event that $k$ is the first index with $M_k\ge u$. These events are disjoint, $A_k\in {\mathcal{F}}_k$, and their union is $A=\{M_n^{\ast }\ge u\}$. The submartingale property gives $\mathbb{E}[M_n\mid {\mathcal{F}}_k]\ge M_k$. Multiplying by $1_{A_k}$, taking expectations and summing yields $\mathbb{E}[M_n{1}_A]\ge {\sum }_k\mathbb{E}[M_k{1}_{A_k}]\ge u\mathbb{P}(A)$. Nonnegativity gives the second inequality.

The associated Doob Lp maximal inequality, for $p>1$ and $M_n\in L^p$, isIndeed $M_k\le \mathbb{E}[M_n\mid {\mathcal{F}}_k]$, so conditional Jensen inequality first ensures finite $p$th moments for all $M_k$ and hence for the finite maximum. Integrating the weak inequality against $p{u}^{p-2}du$ and using the Tonelli theorem givesThe last step is Holder inequality. Dividing, unless the maximum is zero, proves the stated norm bound. For an arbitrary martingale, its absolute value is a nonnegative submartingale by conditional Jensen inequality, so both corresponding absolute-maximum bounds follow.

Put $V_t=⟨X{⟩}_t$ and $S_t={\sup }_{s\le t}|X_s|$. Suppose first that $|X|\le K$. The Itô formula for $|x{|}^p$, which is twice continuously differentiable for $p\ge 2$, givesFor $p=2$, the second derivative is interpreted as the constant $2$. The stochastic term has mean zero: its integrand is bounded, and $\mathbb{E}{V}_t<\infty$ makes it a square-integrable martingale. ConsequentlyThis is the first required estimate.

A bounded local martingale is a true martingale. Apply the allowed Doob Lp maximal inequality, and then Hölder's inequality with conjugate exponents $p/(p-2)$ and $p/2$ for $p>2$. With $D_p=(p/(p-1){)}^p$, this yieldsIf the left side is zero there is nothing to prove; otherwise divide by its indicated power and raise to $p/2$. For $p=2$, the same conclusion follows directly from $\mathbb{E}{X}_t^2=\mathbb{E}{V}_t$ and Doob's inequality. A usable constant is thereforeIn particular $C_2=4$. This is the upper maximal moment bound for a continuous local martingale.

For an unbounded continuous local martingale, stop at ${\tau }_n=\inf \{s:|X_s|\ge n\}$. Its stopped path is bounded, and its bracket at $t$ is $V_{t\wedge {\tau }_n}\le V_t$. The proved inequality gives a bound by $C_p\mathbb{E}{V}_t^{p/2}$, independent of $n$. Continuity makes ${\tau }_n\uparrow \infty$, and the stopped maxima increase to $S_t$. Monotone convergence proves the same inequality for the original unbounded process, with the same constant. This localization step also shows that the assumed bracket moments supply all the needed maximal moments.

For $1<p<\infty$, the Lp martingale convergence theorem states that a martingale with ${\sup }_n\mathbb{E}|M_n{|}^p<\infty$ has a limit $M_{\infty }\in L^p$ such thatMoreover, $M_n=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_n]$ and $\mathbb{E}|M_{\infty }{|}^p\le {\sup }_n\mathbb{E}|M_n{|}^p$. To see why $p>1$ matters, the Doob Lp maximal inequality gives an integrable dominating variable $({\sup }_n|M_n|{)}^p$. The martingale convergence theorem first supplies the almost-sure limit, then the dominated convergence theorem supplies convergence in the Lp norm. This maximal estimate is unavailable at $p=1$ in the required form.

After replacing $X_j$ by $X_j-\mu$, the maximal inequality for independent averages givesThis follows from the Doob Lp maximal inequality by dyadically grouping the partial sums. The strong law makesalmost surely. The displayed maximal function is in $L^p$, so dominated convergence applied to its $p$th power proves convergence in $L^p$.

For $p\ge 2$ and a continuous local martingale starting at zero, the Itô formula for $|X{|}^p$, the Doob Lp maximal inequality and Hölder's inequality yield $\mathbb{E}{\sup }_{s\le t}|X_s{|}^p\le C_p\mathbb{E}⟨X{⟩}_t^{p/2}$. One possible constant is $C_p=[p(p-1)(p/(p-1){)}^p/2{]}^{p/2}$. Stop at increasing absolute-value levels to remove an initial boundedness assumption. This proves the upper half of the Burkholder-Davis-Gundy inequalities in this range.