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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.

Letting $t\to \infty$ in the Burkholder-Davis-Gundy inequalities with exponent two gives absolute constants $c,C>0$ such thatSince $[X]$ is nondecreasing, $\mathbb{E}[X{]}_{\infty }={\sup }_t\mathbb{E}[X{]}_t$. Thus one of the two quantities in the question is finite exactly when the other is.

Let ${\tau }_n$ localize $M$. The Itô isometry givesThus the stopped variables are bounded in $L^2$, and localization plus weak compactness shows that $M$ is a true martingale. The Itô formula applied to $f(x)=x^2$ givesAfter localization this is a martingale; the Burkholder-Davis-Gundy inequalities and $\mathbb{E}[M{]}_t<\infty$ supply the required local integrability, so it is a true martingale.

For a continuous local martingale $M$ starting at zero, Burkholder-Davis-Gundy inequalities compare its maximal size with its quadratic-variation clock:They remain valid with infinite values, by stopping and passage to the limit. Their significance here is that the horizontal and vertical martingales share the same stopping-time clock, so one coordinate's maximum can determine moments of the other and of the clock. For a process started away from zero, first subtract its initial value.

Starting at $i$, the vertical coordinate starts at one and is stopped on first hitting zero. For $\lambda >1$, it reaches level $\lambda$ before zero with probability $1/\lambda$, by optional stopping theorem of one-dimensional Brownian motion on $[0,\lambda ]$. ThusThe tail integral formula givesand it diverges for $p\ge 1$.

The stopped martingale $Y_{t\wedge T_{\mathbb{H}}}-1$ has quadratic variation $t\wedge T_{\mathbb{H}}$. Its maximum in absolute value is $\max (1,Y^{\ast }-1)$, between $Y^{\ast }/2$ and $Y^{\ast }$, so it has precisely the same finite positive moments as $Y^{\ast }$. BDG therefore gives $\mathbb{E}{T}_{\mathbb{H}}^{p/2}<\infty$ exactly for $p<1$. Applying BDG to $X_{t\wedge T_{\mathbb{H}}}$, which starts at zero and has the same clock, yieldsTaking the displayed $1/p$ or $1/q$ powers does not change finiteness. In particular the exit time is almost surely finite while its mean is infinite.

For $0<\alpha <2$, write $z=r{e}^{i\theta }$ with $0<\theta <\pi$. The chosen power map sends this to $r^{\alpha }{e}^{i\alpha \theta }$, soAt $\alpha =2$ this is the plane slit along the nonnegative real axis. For every $0<\alpha \le 2$, the map is conformal on the upper half-plane.

To check finiteness of the physical exit time, start a Brownian motion $B$ at $i$ and let $T$ be its real-axis exit. The time change of a continuous process in conformal invariance of planar Brownian motion isWe know $T<\infty$ almost surely. The Cauchy exit law from a Brownian half-plane assigns probability zero to $B_T=0$. On such a path, $B_s\ne 0$ throughout the compact interval $[0,T]$, because it is in the open upper half-plane before exit and has nonzero endpoint. Its modulus therefore has a positive minimum and finite maximum. The displayed integrand is bounded on that interval, proving $A_T<\infty$. The image Brownian motion exits $D$ at this finite changed time.

The endpoint $\alpha =0$ printed in the source is degenerate: $z^0=1$, so its image is the singleton $\{1\}$, not an open domain, and the map is not conformal. If exit from that set is defined for Brownian motion started at one, its exit infimum is zero. It cannot be included in a nondegenerate conformal-domain argument.

Here is the martingale argument for maximal comparisons, including the starting-value issue. For $P=f+ig$, the Cauchy-Riemann equations and Itô's formula show thatare orthogonal continuous local martingales with the same quadratic variation. Localize before large disks and before exit. The Burkholder-Davis-Gundy inequalities then givewith a constant depending only on $p$; Fatou's lemma removes the localization. These are centered martingales. For the uncentered processes this impliessince $|f(i^{\alpha })|\le \sup |f(Z_t)|$.

The source normalizes $P(0)=0$ at a boundary point, rather than the value of $g$ at the starting point. Consequently its displayed constant cannot be uniform in $\alpha$ with only $p$ dependence. To see this, take $P(z)=iz$ and $0<\alpha <\min (1/2,1/p)$. Then $f=-\mathrm{Im}z$ and $g=\mathrm{Re}z$. Inside this narrow wedge,Both $p$th moments are finite and the right one is positive; finiteness follows from the wedge radial-maximum tail of order $R^{-1/\alpha }$, justified below. The source inequality would force $C_p\ge {\cot }^p(\alpha \pi )$, unbounded as $\alpha \downarrow 0$. At $\alpha =0$, the immediate-exit interpretation gives the direct contradiction $1\le C_{p}0$ for the same polynomial.

For a fixed nondegenerate wedge, an uncentered inequality does hold with a constant allowed to depend on $\alpha$ as well as $p$. One rigorous outline for the polynomial class in this question is as follows. Separation of variables in a sector of radius $R$ gives, for exit through its circular arc, the expansionThus at the specified starting point the radial-maximum tail is comparable to $R^{-1/\alpha }$. Hitting any fixed interior subarc has that same power, with positive leading coefficient. If a nonzero polynomial has degree $d$, its real part has magnitude at least $c{R}^d$ on some fixed interior subarc for sufficiently large $R$. Therefore its maximal real-part $p$th moment is infinite whenever $dp\ge 1/\alpha$. In that case the desired extended-real inequality is automatic.

The remaining polynomials have $d<1/(\alpha p)$, so belong to a finite-dimensional vector space space. On that spaceis a genuine norm: if it is zero, Brownian motion's positive probability of reaching any interior neighborhood forces $\mathrm{Re}P=0$ throughout $D$, so analyticity makes $P$ a purely imaginary constant, and $P(0)=0$ makes it zero. The corresponding imaginary-part functional is bounded on this finite-dimensional vector space normed space; its finiteness follows from $dp<1/\alpha$ and the radial tail bound. equivalence of norms in finite dimensions givesThis establishes the fixed-domain interpretation and identifies exactly why the universal centered martingale inequality cannot silently be applied to the printed uncentered boundary normalization. It is the distinction in starting-value obstruction in wedge conjugate maximal inequalities.

A continuous nondecreasing function $\Phi$ with $\Phi (0)=0$ has moderate growth if for some fixed $a>1$ and finite $K$,Iteration and monotonicity then give such an inequality, with a different constant, for every fixed dilation factor. Powers $\Phi (x)=x^p$, $p>0$, are standard examples.

The Burkholder-Davis-Gundy inequalities say that for a real continuous local martingale $M$ with $M_0=0$ and any stopping time $\tau$, there are positive finite constants $c_{\Phi },C_{\Phi }$, depending only on $\Phi$, such thatHere $M_{\tau }^{\ast }={\sup }_{s\le \tau }|M_s|$. Expectations may be infinite, with the inequalities understood by localization. For $\Phi (x)=x^p$, these compare the $p$th moment of the running maximum with the $p/2$th moment of quadratic variation.

Put $Y=\sqrt{\tau }$ and $V=X_{\tau }^{\ast }$. Splitting according to whether $V>\delta \lambda$, the good-lambda bound for Brownian stopping time and maximum givesIntegrate against the positive measure $d\Phi (\lambda )$. The tail integral formula for expectation in Stieltjes form yieldsBy moderate growth, choose $K_{\beta }$ with $\Phi (\beta x)\le K_{\beta }\Phi (x)$. ThenFix $\beta >1$ and choose $\delta$ small enough that $K_{\beta }\eta <1/2$. Absorb the last term and apply moderate growth once more to the fixed dilation $1/\delta$. This provesTo avoid subtracting infinities, perform the argument first for $\tau \wedge n$, where Brownian maxima have all finite polynomial moments and moderate growth is at most polynomial, and then use monotone convergence. Applying the Dambis-Dubins-Schwarz theorem transfers this Brownian estimate to continuous local martingales, giving the lower-maximum side of the Burkholder-Davis-Gundy inequalities.

Real and imaginary parts of an analytic Brownian image, after subtracting their starting values, have equal quadratic variation. The Burkholder-Davis-Gundy inequalities compare their centered maximal moments with constants depending only on $p$. A boundary-point normalization $P(0)=0$ does not remove the starting imaginary value. For $P(z)=iz$ on a wedge of opening $\alpha \pi <\pi /2$, $\sup |\mathrm{Im}z|\le \tan (\alpha \pi )\sup |\mathrm{Re}z|$. If $\alpha p<1$ both relevant moments are finite, so an uncentered comparison requires a constant at least ${\cot }^p(\alpha \pi )$; no bound depending only on $p$ can hold over all such wedges. For fixed $\alpha >0$, polynomials with finite real-part maximal moment have degree $d<1/(\alpha p)$ by the wedge radial-maximum tail. They form a finite-dimensional vector space, and equivalence of norms in finite dimensions gives a valid constant depending on $p,\alpha$.

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.