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First, the absolute values in the printed limit are an error: the correct long-run variance of a stationary process is the signed sum of autocovariances. Directly,soEach coefficient tends to $1$ and has absolute value at most $1$. Absolute summability and the dominated convergence theorem therefore giveThe sum of absolute values is an upper bound, not the general limit. For a concrete counterexample to the printed assertion, take $X_t={\eta }_t-{\eta }_{t-1}$ with unit-variance iid noise. Then $\gamma (0)=2$, $\gamma (\pm 1)=-1$, and other covariances vanish. The absolute sum is $4$, but ${\overline{X}}_T=({\eta }_T-{\eta }_0)/T$, so $T\mathrm{Var}({\overline{X}}_T)=2/T\to 0$.

For correlated data the central limit theorem can still hold under suitable strong mixing of a stationary process and moment conditions, but the limiting variance is the long-run variance of a stationary process, rather than the one-observation variance. When it is positive,Positive serial dependence usually increases the standard error, while negative dependence can reduce it. The approximate effective sample size of a stationary sample is $T{\gamma }_X(0)/{\sum }_h{\gamma }_X(h)$ when that denominator is positive. A long-memory time series can require a different normalization or a different limit law. Absolute covariance summability alone does not prove a CLT: if $X_t=V{\eta }_t$ with a common independent random scale $V$ taking values $1$ and $2$ with equal probabilities and iid ${\eta }_t$ with the standard normal distribution, the off-diagonal covariances are zero, yet $\sqrt{T}{\overline{X}}_T$ has the nonnormal scale-mixture law $VN(0,1)$. The example is not an ergodic stationary process. If the long-run variance of a stationary process is zero, the usual nondegenerate square-root-$T$ CLT is unavailable.

For the causal AR(1), $|\varphi |<1$ andThe conditional distribution follows because the current innovation is independent of the past.

For a fixed known $\varphi$, condition on the observed $X_1$ and use the $n=T-1$ transitions $t=2,\dots ,T$. Their conditional maximum likelihood criterion is, up to constants,Differentiating in $\mu$ yieldsSince $X_t-\varphi X_{t-1}=(1-\varphi )\mu +{\epsilon }_t$,Thus it is an unbiased estimator, even conditionally on $X_1$, andIt has statistical consistency with mean-square convergence, and the iid-noise strong law of large numbers also gives almost-sure statistical consistency. Its conditional distribution is exactly normal with the displayed mean and variance. If $X_0$ is also observed and all $T$ transitions are used, replace $T-1$ by $T$.

The fixed-$\varphi$ qualification is necessary for the exact finite-sample claims. If $\varphi$ is jointly estimated, conditional likelihood function is linear regression with an intercept: writing ${\bar{X}}_{-}=n^{-1}{\sum }_{t=2}^T{X}_{t-1}$ and ${\bar{X}}_{+}=n^{-1}{\sum }_{t=2}^T{X}_t$, the unconstrained estimators areThis ratio is not generally an unbiased estimator and does not have the preceding finite-sample variance. Under the usual stationary regression conditions it has statistical consistency; its asymptotic variance is ${\sigma }_{\epsilon }^2/(1-\varphi {)}^2$. Indeed, it differs from ${\bar{X}}_{+}$ by $\hat{\varphi }(X_T-X_1)/(n(1-\hat{\varphi }))$, an asymptotically negligible endpoint term. Profiling an unknown innovation variance does not change the fixed-$\varphi$ estimate of $\mu$.

A linear regularization is a family of bounded linear operators $R_{\alpha }:V\to U$, $\alpha >0$, for which $R_{\alpha }f\to K^{†}f$ for every $f\in \mathcal{D}(K^{†})$ as $\alpha \downarrow 0$. To obtain a convergent regularization of an inverse problem for noisy data, choose a regularization parameter so that the exact-data approximation error and the amplified noise both vanish.

An example is Tikhonov regularization:The positive quadratic term makes $K^{\ast }K+\alpha I$ invertible. The Tikhonov filter norm bound follows from the filter $s/(s^2+\alpha )$, whose maximum for $s\ge 0$ is $1/(2\sqrt{\alpha })$. Therefore $\Vert R_{\alpha }\Vert \le 1/(2\sqrt{\alpha })$. The spectral factors $s^2/(s^2+\alpha )$ tend to one on the positive spectrum; dominated convergence theorem gives consistency on $\mathcal{D}(K^{†})$. The noise-bias decomposition for linear regularization then proves convergence whenever $\alpha (\delta )\to 0$ and $\delta /\sqrt{\alpha (\delta )}\to 0$, for example $\alpha (\delta )=\delta$ for sufficiently small positive noise levels.

Use the Fourier transform convention $\hat{\phi }(\xi )=\int e^{-ix\cdot \xi }\phi (x)dx$, with inverse factor $(2\pi {)}^{-n}$. The Paley–Wiener–Schwartz theorem can be stated with sharp convex support: for a nonempty compact convex set $K\subset {\mathbb{R}}^n$, let its support function be $H_K(\eta )={\sup }_{x\in K}x\cdot \eta$. Then $F$ is the Fourier transform of a unique distribution supported in $K$ if and only if it is entire andfor some $C$ and nonnegative integer $M$. For $K={\overline{B}}_R$, this is the familiar bound $C(1+|z|{)}^M{e}^{R|\mathrm{Im}z|}$. Allowing some $R$ characterizes all compactly supported distributions. Here the extension of the Fourier transform is $F(z)=⟨u,e^{-ix\cdot z}⟩$, interpreted with a cutoff equal to one near the support of a distribution.

First suppose $\mathrm{supp}u\subset K$. Fix one smooth cutoff function equal to one near $K$. Pairing the resulting compactly supported exponential with $u$ shows that $F$ is an entire function: differentiation with respect to $z_j$ inserts $-i{x}_j$, and the power series converges in the test-function seminorms uniformly on compact subsets of ${\mathbb{C}}^n$.

To keep the exponential type exactly $H_K$, rather than that of a fixed larger neighborhood, use a shrinking-cutoff exponential-type estimate. There are cutoffs ${\chi }_{\epsilon }$ equal to one on $K+\epsilon B_1$, supported in $K+3\epsilon {\overline{B}}_1$, and satisfying $|{\partial }^{\alpha }{\chi }_{\epsilon }|\le C_{\alpha }{\epsilon }^{-|\alpha |}$ for $0<\epsilon \le 1$. One construction convolves the indicator of $K+2\epsilon B_1$ with a unit-mass mollifier supported in $\epsilon B_1$. Continuity of $u$ on a fixed compact neighborhood gives a finite order of a distribution $q$ there. By the product rule,Taking $\epsilon =(1+|z|{)}^{-1}$ proves the required bound with $M=2q$, since the extra exponential factor is at most $e^3$. The value of the pairing is independent of the chosen cutoff because all cutoffs agree near $\mathrm{supp}u$. Thus the forward direction has the exact asserted support function, without an unproved estimate on derivatives restricted only to $K$.

Conversely, suppose the entire $F$ has the stated bound. Its restriction to real frequency has polynomial growth, so define a tempered distribution $u$ byThe Schwartz space decay makes this integral absolutely convergent and continuous; by Fourier inversion, $\hat{u}=F$ on real frequency. It remains to prove the support assertion by mollifier regularization for contour recovery of support.

Choose a nonnegative unit-mass mollifier $\rho$ supported in $B_1$, and setOn real frequency, $F_{\epsilon }$ decays faster than any polynomial after enough applications of integration by parts to $\hat{\rho }$; hence $h_{\epsilon }$ is smooth, by differentiation under the integral sign. More generally, for every integer $L$ there is $C_{\epsilon ,L}$ such thatTo obtain this estimate, write the transform of ${\rho }_{\epsilon }$ as the real-frequency transform of $e^{x\cdot \eta }{\rho }_{\epsilon }(x)$ and integrate by parts; each derivative introduces at most one factor of $|\eta |$.

Fix a unit vector $v$ and take $L>M+n+1$. A contour-shift proof of the Paley–Wiener–Schwartz theorem moves the inverse-transform contour to ${\mathbb{R}}^n+itv$:Here is a justification of the shift. Rotate coordinates so $v$ is the first coordinate direction, apply the Cauchy integral theorem on a rectangle in the first complex variable, and integrate over the other real variables. For fixed $t$, the vertical sides at real part $\pm A$ have an integrated bound proportional to $A^{M-L+n-1}$, and hence vanish as $A\to \infty$. The same decay bounds give absolute convergence on the horizontal sides. No contour shift of an unregularized polynomially growing integral is needed.

Positive homogeneity of the support function now givesIf $x\notin K+\epsilon {\overline{B}}_1$, the Hahn-Banach separation theorem supplies a unit vector $v$ with $x\cdot v>H_K(v)+\epsilon$. Letting $t\to \infty$ proves $h_{\epsilon }(x)=0$. Therefore $h_{\epsilon }\in C_c^{\infty }$ and $\mathrm{supp}{h}_{\epsilon }\subset K+\epsilon {\overline{B}}_1$.

Since $\hat{\rho }(\epsilon \xi )\to 1$ and its modulus is at most one on real frequency, the dominated convergence theorem in the formula for $⟨u,\phi ⟩$ gives $h_{\epsilon }\to u$ as tempered distributions, and thus as distributions. A test function supported outside $K$ has positive distance from $K$, so its pairing with $h_{\epsilon }$ is zero for all sufficiently small $\epsilon$. This proves $\mathrm{supp}u\subset K$. The Fourier transform of a compactly supported distribution constructed in the forward direction equals $F$ on ${\mathbb{R}}^n$; applying the one-variable identity theorem successively in each coordinate extends the equality to ${\mathbb{C}}^n$. Injectivity of the Fourier transform of a tempered distribution proves uniqueness and completes both directions.

For the independence application, put $g_m(z)=e^{iz\cdot y_m}{f}_m(z)$ and $r_m=1/(m+1)$. The ball version of the Paley–Wiener–Schwartz theorem supplies a nonzero distribution $v_m$ supported in ${\overline{B}}_{r_m}(0)$ with ${\hat{v}}_m=g_m$. By the Translation property of the Fourier transform,where $⟨{\tau }_{y}v,\phi ⟩=⟨v,\phi (\cdot +y)⟩$. Distinct integer points are at least distance one apart. For distinct positive indices $j,k$, the largest possible radius sum is $1/2+1/3=5/6<1$. Thus these closed balls, and hence the distribution supports, are pairwise disjoint.

If ${\sum }_{m=1}^N{c}_m{f}_m=0$, injectivity of the Fourier transform gives ${\sum }_m{c}_m{w}_m=0$. For each $j$, choose a smooth cutoff function equal to one near its ball and zero near all other balls. Multiplying the distributional identity by this cutoff isolates $c_j{w}_j=0$. Since $f_j$ is not identically zero, $w_j\ne 0$, so $c_j=0$. Consequently $f_1,\dots ,f_N$ are linearly independent over $\mathbb{C}$. This Fourier independence from disjoint distribution supports uses the quantitative exponential types to establish support separation.

With $⟨\theta ⟩=(1+|\theta {|}^2{)}^{1/2}$, the symbol class $\mathrm{Sym}(X,{\mathbb{R}}^k;N)=S_{1,0}^N$ consists of complex-valued $a\in C^{\infty }(X\times {\mathbb{R}}^k)$ such that, for every compact $K⋐X$ and every pair of multi-indices $\alpha ,\beta$,An $x$ derivative preserves symbol order, while a $\theta$ derivative lowers it by one. The definition is local in $x$ and imposes estimates of every derivative order.

In the homogeneous convention, a phase function is real, smooth on $X\times ({\mathbb{R}}^k\setminus \{0\})$, positively homogeneous of degree one in $\theta$, and satisfiesOnly the large-frequency behavior is relevant to the regularization argument; any prescribed smooth modification near $\theta =0$ contributes an ordinary convergent integral. If the phase is presented on all of $X\times {\mathbb{R}}^k$, homogeneity can be required outside a bounded frequency region. Equivalently for this construction, its large-frequency symbol bounds and the following uniform estimate are sufficient:For a homogeneous phase function, this follows from compactness of $K\times S^{k-1}$. Nonvanishing of the total gradient is the condition; ${\nabla }_{\theta }\Phi$ may vanish. The displayed definition does not impose the stronger rank condition sometimes called a nondegenerate phase in Fourier-integral-operator theory.

We construct the oscillatory integral by integration by parts in both $x$ and $\theta$. For $|\theta |\ge 1$, setThen $L(e^{i\Phi })=e^{i\Phi }$. The coefficients of the $x$ derivatives have symbol order $-1$ and those of the $\theta$ derivatives have order zero. Use the bilinear formal transpose of a differential operator, not the Hermitian adjoint:The symbol order reduction by a phase integration operator follows from the product rule and the symbol class estimates: $L^t:S_{1,0}^s\to S_{1,0}^{s-1}$ locally in $x$. Differentiating the coefficients preserves precisely these orders because $q$ is elliptic of degree two on each compact $x$ set.

Let $\zeta \in C_c^{\infty }({\mathbb{R}}^k)$ equal one on $|\theta |\le 1$ and vanish on $|\theta |\ge 2$. The low-frequency part is an ordinary integral. For an integer $r>N+k$, define the high-frequency part byIf the homogeneous phase is only defined off zero, its value at the single point zero is immaterial to the low-frequency integral; its modulus remains one. In the high-frequency term all coefficients are used away from zero. For $\mathrm{supp}\phi \subset K$, repeated product rules giveSince $N-r<-k$, both terms converge absolutely. They are linear in the test function, with the continuity boundThus $I_{\Phi }(a)\in {\mathcal{D}}^{\prime }(X)$, with order of a distribution at most $r$ on each compact set.

To check that this construction is the intended oscillatory integral and does not depend on $r$, $\zeta$ or the large-frequency cutoff, let $\chi \in C_c^{\infty }({\mathbb{R}}^k)$ equal one near zero and formAfter $r$ applications of integration by parts, the term without a derivative on $\chi$ tends to the absolutely convergent high-frequency term by the dominated convergence theorem. Each extra term is supported in an annulus $|\theta |\asymp R$. A derivative of $\chi (\theta /R)$ contributes $R^{-1}$, so those terms have total absolute value at most $C_K{R}^{N-r+k}{\max }_{|\alpha |\le r}\Vert {\partial }^{\alpha }\phi {\Vert }_{\infty }\to 0$. The low-frequency part is unchanged for large $R$. Hence $J_R(\phi )$ tends to the displayed functional, independently of all cutoffs and of the permissible number of integrations. This proves the cutoff independence of an oscillatory integral as well as its continuity.

For the final distribution, the derivative convention $⟨{\delta }^{\prime },g⟩=-g^{\prime }(0)$ identifies it as $-x_2{\delta }^{\prime }(x_1)$. The delta derivatives from polynomial oscillatory amplitudes identity yieldsHere $k=1$, $a\in \mathrm{Sym}({\mathbb{R}}^2,\mathbb{R};1)$, and ${\nabla }_x\Phi =(\theta ,0)$ is nonzero for $\theta \ne 0$. Thus the phase is admissible even though ${\partial }_{\theta }\Phi =x_1$ vanishes on the distribution's support.

For a direct proof of the sign and normalization, put $g(s)={\int }_{\mathbb{R}}{x}_2\phi (s,x_2)d{x}_2\in C_c^{\infty }(\mathbb{R})$. The cutoff integral isThe Fourier transform $\hat{g}$ is rapidly decreasing, so the dominated convergence theorem and differentiated Fourier inversion giveThis is exactly the required pairing, so the oscillatory representation gives the prescribed distribution, not its negative or a multiple of it.

For each $t\in [0,1]$, evaluation ${\delta }_t:f\mapsto f(t)$ is a bounded linear functional of norm one. Thus weak convergence to zero gives $f_n(t)\to 0$ at every $t$. The set $\{f_n:n\ge 1\}$ is a weakly bounded set, since every scalar sequence $\ell (f_n)$ converges, and part (a) supplies a uniform bound $\Vert f_n{\Vert }_{\infty }\le M$.

The constant $M$ is integrable for Lebesgue measure on $[0,1]$. Applying the dominated convergence theorem to $|f_n|$ gives the weakly null continuous functions converge in L1 conclusionPointwise convergence alone would not provide the needed uniform dominating function; it is the weak boundedness argument that supplies it.

For a probability measure-preserving system, the hypothesis is completely positive entropy: every finite measurable partition with positive static entropy of a finite measurable partition has positive entropy rate of a measurable partition. Fix a finite partition $\xi$ and putAll sigma-algebras are interpreted modulo null sets. We will prove that every finite partition $\alpha$ measurable with respect to this tail sigma-algebra of a measurable partition has $h_{\mu }(T,\alpha )=0$. Applying this to a binary partition will force the required triviality. This proves the needed direction of the Tail characterization of the Pinsker sigma-algebra directly, including noninvertible transformations.

Write $h=h_{\mu }(T,\xi )$. The infinite-future entropy formula and the backwards chain rule for information entropy give, for every $L\ge 1$, the block conditional entropy given the infinite future identityEach term equals $h$ by invariance of the joint probabilities under a common pullback and continuity of conditional entropy under increasing finite future blocks. Invertibility is not needed for this identity.

Fix $\epsilon >0$. Since $\alpha$ is ${\mathcal{F}}_0$-measurable and finite, the Martingale convergence theorem and continuity of finite-partition conditional entropy allow an $r\ge 0$ withTo see the continuity explicitly, for each partition atom $A$ of $\alpha$ the conditional probabilities $\mathbb{E}[1_A\mid \sigma ({\xi }_0^r)]$ tend to $1_A$ almost everywhere; apply dominated convergence theorem to the bounded continuous function $-t\log t$ on $[0,1]$ and sum over the finitely many partition atoms.

For $n\ge 1$ set $\gamma ={\alpha }_0^{n-1}$, $L=n+r$, and $\beta ={\xi }_0^{L-1}$. The conditional entropy of finite measurable partitions satisfiesThe second inequality uses conditioning reduces entropy, since $\beta$ refines each block $T^{-j}{\xi }_0^r$; the final equality uses measure preservation.

Moreover $\gamma$ is ${\mathcal{F}}_L$-measurable. For each $j\ge 0$, tail measurability gives $\alpha$ measurable with respect to ${\mathcal{F}}_L$, hence $T^{-j}\alpha$ measurable with respect to $T^{-j}{\mathcal{F}}_L={\mathcal{F}}_{L+j}\subseteq {\mathcal{F}}_L$. Thus conditioning reduces entropy and the displayed block identity implyUse the symmetric entropy identity $H(\gamma )=H(\beta )-H(\beta \mid \gamma )+H(\gamma \mid \beta )$ to obtainThe fraction on the right tends to zero: $r$ is fixed and $H({\xi }_0^{n+r-1})/(n+r)\to h$ by the definition of entropy rate of a measurable partition. Taking $n\to \infty$ and then $\epsilon \downarrow 0$ gives

Now let $A\in \mathcal{T}(\xi )$ and take $\alpha =\{A,X\setminus A\}$. If $0<\mu (A)<1$, its binary entropy is $-\mu (A)\log \mu (A)-(1-\mu (A))\log (1-\mu (A))>0$, while its entropy rate is zero, contradicting completely positive entropy. ThereforeSince $\xi$ was arbitrary, every finite-partition tail is trivial. The argument needs only that $\xi$ is finite and the measure is a probability; it does not assume a finite generator, finite total system entropy, or invertibility.

For $m\ge 1$, the increments have exponential momentThus $Z_n=2^{S_n}$ is an exponential martingale of a random walk relative to the natural filtration. We justify its stopping limit. From any surviving state in $(-m,m)$, a block of $2m$ successive increments all equal to $-1$ forces an exit. The block has probability $q=(6/7{)}^{2m}>0$, independently of the preceding increments. The geometric tail bound from a uniform escape probability givesso $T$ is finite almost surely and has finite expectation.

At the lower exit $S_T=-m$ exactly; at the upper exit $S_T$ is either $m$ or $m+1$. Before exit the same overall bound $-m\le S_{n\wedge T}\le m+1$ holds. Hence $Z_{n\wedge T}$ is bounded by $2^{m+1}$, uniformly in $n$. The bounded optional stopping theorem gives $\mathbb{E}{Z}_{n\wedge T}=Z_0=1$, and the dominated convergence theorem now yieldsAllowing the upper overshoot by one is essential; the terminal state need not equal $m$ on upper exit.

The mean increment is $\mathbb{E}{X}_1=-6/7+2/7=-4/7$, so $M_n=S_n+(4/7)n$ is a martingale. Applying the bounded optional stopping theorem at $n\wedge T$ givesThe terminal sums are bounded by the exit-state bounds from (b), while $n\wedge T$ increases to the integrable stopping time $T$. The dominated convergence theorem on the left and the monotone convergence theorem on the right therefore proveThis is also Wald's equation here. In fact, the stopped identity already implies $\mathbb{E}(n\wedge T)\le 7m/4$ from $S_{n\wedge T}\ge -m$, giving another direct finite-expectation justification.

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.

Let ${\tau }_r=\inf \{t\ge 0:|B_t|=r\}$, ${\tau }_R=\inf \{t\ge 0:|B_t|=R\}$, and $\tau ={\tau }_r\wedge {\tau }_R$. The function $u(y)=|y{|}^{-1}$ is harmonic away from the origin: its radial Laplacian in dimension three isThe Itô formula therefore makes $u(B_{t\wedge \tau })$ a local martingale, bounded between $1/R$ and $1/r$, hence a true martingale. The annulus exit time is finite almost surely. For instance, exit from the containing open ball is finite because from every point in the open ball a sufficiently large Gaussian vector increment has a uniform positive probability of leaving it in unit time; iterate the geometric tail bound from a uniform escape probability.

The optional stopping theorem at $t\wedge \tau$, followed by the dominated convergence theorem, gives $\mathbb{E}u(B_{\tau })=u(x)$. By continuity, there is no radial overshoot and the two exit spheres cannot be hit simultaneously. Writing $p={\mathbb{P}}_x({\tau }_r<{\tau }_R)$ givesAll harmonicity and boundedness statements apply only in the annulus, which stays away from the singularity at zero.

For $a>0$, the Brownian reflection principle gives, for $t>0$,where $\Phi$ is the standard normal distribution function. Hence $H_a<\infty$ almost surely; for $a=0$, $H_0=0$.

For $u\ge 0$, use the Exponential martingale for Brownian motion $Z_t=\exp (u{B}_t-u^{2}t/2)$. At $H_a\wedge t$, it is bounded by $e^{ua}$ because $B_s<a$ before the first hit and $B_{H_a}=a$ by continuity. The bounded optional stopping theorem gives $\mathbb{E}{Z}_{H_a\wedge t}=1$. Its limit is $e^{ua-u^2{H}_a/2}$, and the dominated convergence theorem yields $\mathbb{E}{e}^{ua-u^2{H}_a/2}=1$. Equivalently, the Brownian first-passage Laplace transform isTaking $\lambda =u^2/2$ proves the displayed exponential identity. The cases $u=0$ and $a=0$ are included. The restriction $u\ge 0$ is what gives the upper bound on the stopped exponential.

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

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).

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.