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Work on a filtered probability space with the usual conditions for a filtration. A continuous martingale belongs to the class of L2-bounded continuous martingales whenThis means a uniform bound over the entire time interval, not merely being square-integrable at each individual time. The L2 martingale convergence theorem gives a terminal variable $M_{\infty }$ and almost sure convergence and convergence in the Lebesgue space $L^2$; moreover $M_t=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_t]$ by conditional expectation.

The predictable sigma-algebra $\mathcal{P}$ on $\Omega \times [0,\infty )$ is the smallest sigma-algebra making every left-continuous adapted process measurable. Equivalently, it is generated bywhere the sample coordinate comes first. A previsible process is precisely a process measurable for this predictable sigma-algebra. Values at time zero matter for its definition, although they make no contribution to integration against a continuous quadratic variation starting at zero.

A simple predictable process has the form $H_s={\sum }_{j=0}^{m-1}{h}_j{1}_{(t_j,t_{j+1}]}(s)$, where $0=t_0<\cdots <t_m<\infty$ are deterministic and $h_j$ is bounded and ${\mathcal{F}}_{t_j}$-measurable. A possible ${\mathcal{F}}_0$-measurable value at zero is irrelevant to this Itô integral. DefineEach coefficient is known before the corresponding increment. The conditional expectation of each subsequent increment is zero, so the result is a continuous martingale starting at zero. It is an L2-bounded continuous martingale because it is a finite sum of bounded coefficients times stopped increments of an L2-bounded continuous martingale. This definition is independent of the chosen subdivision: splitting an interval simply splits its increment into a telescoping sum. General admissible predictable processes are then integrated by completion using the Itô isometry; unbounded step coefficients are allowed when the weighted $L^2$ condition in the quadratic-variation measure in the next part holds.

Fix a L2-bounded continuous martingale $M$. Its quadratic-variation measure on the predictable sigma-algebra isIt is finite, since ${\nu }_M(\Omega \times [0,\infty ))=\mathbb{E}[M{]}_{\infty }=\mathbb{E}{M}_{\infty }^2-\mathbb{E}{M}_0^2$. The source Hilbert space is $L^2(M)=L^2(\Omega \times [0,\infty ),\mathcal{P},{\nu }_M)$: predictable processes with finite $\mathbb{E}\int H^{2}d[M]$, identified when equal ${\nu }_M$-almost everywhere. Its norm is $\Vert H{\Vert }_{L^2(M)}^2=\mathbb{E}\int H^{2}d[M]$.

The target ${\mathcal{M}}_0^2$ consists of L2-bounded continuous martingales starting at zero, identified up to indistinguishability of stochastic processes, with norm $\Vert K{\Vert }_{{\mathcal{M}}_0^2}=\Vert K_{\infty }{\Vert }_2$. It is a Hilbert space: terminal values belong to the closed linear subspace of $L^2({\mathcal{F}}_{\infty })$ whose conditional expectation given ${\mathcal{F}}_0$ is zero and whose associated martingales have continuous versions. Closure of that continuous-version subspace follows from the Doob L2 maximal inequality and an almost surely uniformly convergent subsequence.

The Itô isometry says that integration extends uniquely from simple predictable processes to a linear isometry $L^2(M)⟶{\mathcal{M}}_0^2$, withFor every finite $t$, the same identity holds with the integral restricted to $[0,t]$ and the left side $\mathbb{E}|(H\cdot M{)}_t{|}^2$. No claim of surjectivity onto all of ${\mathcal{M}}_0^2$ is needed: that would require an additional Martingale representation theorem.

Put $S=H\cdot M$ and $A_t={\int }_0^t{H}_s^{2}d[M{]}_s$. The candidate $A$ is continuous, adapted, nondecreasing, starts at zero and has integrable terminal value. We prove $S^2-A$ is a martingale, rather than assuming the desired quadratic variation of a stochastic integral.

The assumed boundedness of $M$ also makes $M^2-[M]$ a true martingale on every finite horizon. Indeed, apply its local martingale identity at a localizing sequence; bounded stopped squares and the Fatou lemma show that $[M{]}_t$ is integrable. The stopped expressions are then dominated by ${\sup }_s|M_s{|}^2+[M{]}_t$, so the dominated convergence theorem gives the unrestricted conditional identity on that horizon.

First let $H$ be a bounded simple predictable process. On each interval with coefficient $h$, an increment of $S$ is $h(M_t-M_a)$. The defining quadratic variation identity for $M$, together with the martingale property of $M$, makesa martingale for $t\ge a$. Expanding $S_t^2-S_a^2$ gives a term $2S_{a}h(M_t-M_a)$ and the square term above; their conditional expectations yield the desired identity on that interval. Integrability follows because the variables are square-integrable; concatenating the finitely many intervals gives $S^2-A$ a martingale.

For general bounded predictable $H$, choose bounded simple predictable processes $H^n\to H$ in $L^2(M)$, which is possible since they generate the predictable sigma-algebra. Set $S^n=H^n\cdot M$ and $A_t^n={\int }_0^t(H_s^n{)}^{2}d[M{]}_s$. The Itô isometry and Doob L2 maximal inequality imply $S_t^n\to S_t$ in $L^2$; the Cauchy-Schwarz inequality gives $(S_t^n{)}^2\to S_t^2$ in $L^1$ andPass the conditional expectation identity for $(S^n{)}^2-A^n$ to the $L^1$ limit. Thus $S^2-A$ is a martingale. Uniqueness in the defining quadratic variation property now gives

The Exponential martingale for Brownian motion is a true martingale: independent increments and the normal distribution identity $\mathbb{E}{e}^{aN}=e^{a^{2}v/2}$ for $N\sim N(0,v)$ give $\mathbb{E}[Z_t\mid {\mathcal{F}}_s]=Z_s$. Its second moment isThe uniform integrability from bounded second moments criterion now applies: for $K>0$,Thus $\{Z_t:0\le t\le T\}$ has uniform integrability. In fact, all values stopped at stopping times bounded by $T$ have uniform integrability, since $Z_{\tau }=\mathbb{E}[Z_T\mid {\mathcal{F}}_{\tau }]$ and uniform integrability of conditional expectations applies. This is a finite-horizon assertion, not an assertion of uniform integrability over all $t\ge 0$.

Define a probability measure on the same full measurable space byThe Radon-Nikodym derivative is positive and has expectation one, so this is a probability measure and even equivalent to $\mathbb{P}$, in particular ${\mathbb{P}}^T\ll \mathbb{P}$. For $t\le T$ its density on ${\mathcal{F}}_t$ is $\mathbb{E}[Z_T\mid {\mathcal{F}}_t]=Z_t$.

The finite-horizon Girsanov theorem says that if $W$ is a Brownian motion and $\mathcal{E}(\int hdW)$ is a true martingale through $T$ defining the density, then $W_t-{\int }_0^t{h}_{s}ds$ is a Brownian motion under that density for $t\le T$. Here $h_s=\mu$, and the stochastic exponential is exactly $Z$. The previous part verifies the necessary martingale property; alternatively the Novikov condition is $e^{{\mu }^{2}T/2}<\infty$. Hence $B_t-\mu t$ is a standard Brownian motion on $[0,T]$ relative to the same filtration under ${\mathbb{P}}^T$.

Set $V(x)=(1+x^2{)}^{-1}$ and $A_t={\int }_0^{t}V(B_s)ds$. On a fixed horizon $T$, the Itô formula and Itô product rule giveThe drift vanishes by the partial differential equation. This makes $M$ a continuous local martingale. Since $0\le A_t\le T$,The right side is a finite deterministic bound under the stated boundedness assumption, so the dominated convergence theorem removes a localizing sequence and $M$ is a true martingale. If the derivatives are also bounded, the Itô isometry alternatively proves the stochastic integral is square-integrable because $\mathbb{E}{\int }_0^T{e}^{2A_s}|u_x(T-s,B_s){|}^{2}ds\le T{e}^{2T}\Vert u_x{\Vert }_{\infty }^2$.

For the bounded parabolic differentiability class, distinguish global bounds from bounds on each finite time slab. Only the latter are needed here; a single global bound on the classical solution $u$ would conflict with its requested long-time growth in part (c).

For the true martingale of part (a), $M_0=u(T,x)$ and $M_T=f(B_T)e^{A_T}$. Taking expectations gives the Feynman-Kac formula with a bounded potential:The right side is finite since its absolute value is at most $e^t\Vert f{\Vert }_{\infty }$. This also proves uniqueness among classical solutions bounded on every finite time slab: applying the formula to their difference with zero initial data gives zero. The positive sign of the potential in the partial differential equation explains the positive exponent; changing it to a negative exponent would solve a different equation.

Put $u(z)=\mathrm{Im}(z-g_A(z))$ on $D_A$. This is a harmonic function, tends to zero at infinity, and is bounded. To justify the boundary behavior without requiring a locally connected space as its boundary, let $f_A=g_A^{-1}$. Its expansion at infinity implies that bounded $z$ cannot have $|g_A(z)|\to \infty$. If $z_n$ approaches a finite point of $\partial D_A$ and $g_A(z_n)$ had a subsequential limit inside $\mathbb{H}$, continuity of $f_A$ would force that point to lie inside $D_A$, a contradiction. Thus boundary degeneration under a mapping-out function gives $\mathrm{Im}{g}_A(z_n)\to 0$. Consequently $u$ extends continuously to the finite boundary with value $\mathrm{Im}z$.

Let $B$ be planar Brownian motion started at $z\in D_A$, and $\tau$ its Brownian exit time from $D_A$. This time is finite almost surely: it is at most the first time the imaginary coordinate, a one-dimensional Brownian motion, hits zero. By the Itô formula, $u(B_{t\wedge \tau })$ is a bounded martingale. The optional stopping theorem and dominated convergence theorem therefore giveBoundedness is important here: directly stopping the unbounded imaginary-coordinate martingale would require an unjustified uniform integrability assertion.

At $z=iy$, the Laurent series gives $u(iy)=a/y+O(y^{-2})$. HenceThe sign follows because the exit point lies on $\partial D_A\subseteq \overline{\mathbb{H}}$. This proves the Brownian representation of half-plane capacity.

Before the Loewner swallowing time $T_z$, put $Z_t=g_t(z)-U_t=X_t+i{Y}_t$ and ${\theta }_t=\arg Z_t\in (0,\pi )$. The Itô formula applied to the holomorphic logarithm in the complex upper half-plane givesTaking imaginary parts yields the SLE angle processFor $\kappa =4$ the drift vanishes, sois a continuous local martingale. Since $0<{\theta }_t<\pi$, every localized stopped version is a bounded true martingale. After resolving the lifetime issue below, the dominated convergence theorem applied to conditional expectations removes localization and gives a true bounded martingale.

For completeness, there is no finite lifetime ambiguity for a fixed interior $z$ in this parameter range. The preceding Bessel process and Conformal Markov property of SLE argument gives a simple trace staying in the interior, so no point is swallowed in a disconnected pocket. If $T_z$ were finite, the trace would have to reach $z$, forcing the Loewner conformal radius to tend to zero by the Koebe quarter theorem. ButThe Dambis-Dubins-Schwarz theorem says that a bounded continuous local martingale cannot have infinite quadratic variation before a finite terminal time: that would require a Brownian motion to stay in a bounded interval for all clock times. Therefore ${{\rm Y}}_t$ cannot tend to zero at a finite $T_z$, a contradiction. Thus $T_z=\infty$ almost surely for each fixed $z$. This proves the SLE4 angle martingale assertion: the angle is a bounded continuous martingale. In particular $\mathbb{E}[{\theta }_t]=\arg z$. The upper-half-plane branch of the argument is used throughout.