At time zero, ${{\rm Y}}_0(z)=\mathrm{Im}z$. Compactness of $K\subset \mathbb{H}$ givesAlso $M_0(z)$ is uniformly bounded above on $K$. Onone has $M_{{\tau }_{ϵ}}\ge {ϵ}^{-p}{2}^{-q}$. Optional stopping, Fatou lemma, and the assumed conditional angular estimate giveThusThe exponent $(8-\kappa )/\kappa$ requested in the question does not follow and is false as written. The SLE Green-function estimate gives probability comparable to ${ϵ}^{1-\kappa /8}$, confirming that the denominator in the requested exponent should be $8$.

Putso $M_t={{\rm Y}}_t^{-p}{S}_t^q$. Before ${\tau }_{ϵ}$, one has ${{\rm Y}}_t\ge ϵ$ and $0<S_t\le 1$. ThereforeThe supplied continuous local martingale is thus bounded after stopping, and a bounded local martingale is a true martingale. Hence

It has the Loewner local growth property when, after mapping out the old hull, each short new increment is small: for every $T,ϵ>0$ there is $\delta >0$ such thatEquivalent formulations use a crosscut of diameter below $ϵ$ separating the new increment from infinity.

It has the half-plane-capacity parameterization under the standard chordal convention when

The family is a nondecreasing family of sets when

The statement is true. In the notation of part (ii), equality of the capacities forces $\mathrm{hcap}(D)=0$. Every nonempty compact H-hull has strictly positive half-plane capacity: by the Brownian representation of half-plane capacity, Brownian motion started sufficiently high has positive harmonic measure of a boundary portion of positive height. Hence $D=\varnothing$, so $C\setminus A=\varnothing$ and

Map out $A$ first. The imagewith its bounded filling is a compact H-hull, and uniqueness of hydrodynamic normalization givesComparing the coefficients of $1/z$ at infinity yields the half-plane-capacity composition ruleThus half-plane capacity is monotone under inclusion.

Let $g=g_A$ and define $u(z)=\mathrm{Im}(z-g(z))$. This is harmonic on $\mathbb{H}\setminus A$, has boundary values $\mathrm{Im}z$ on the hull boundary and zero on the real boundary, and tends to zero at infinity. The representation by harmonic measure and optional sampling theorem therefore giveAt $z=iy$, the hydrodynamic expansion givesConsequently the Brownian representation of half-plane capacity is

Apply the Feynman-Kac formula with $V(x)=\sigma x$ and terminal function $1$. The ansatz $u(t,x)=e^{A(t)x+C(t)}$ givesThus $A(t)=-\sigma t$ and $C(t)={\sigma }^2{t}^3/6$, soEquivalently, the Integral of Brownian motion is Gaussian with mean $xt$ and variance $t^3/3$.

Let ${\tau }_n$ be the first exit from a compact interval on which $H<n$, chosen so that ${\tau }_n\uparrow T$, and let $m=\inf H>-\infty$. By Itô formula,The assumption gives the drift bound $\frac{1}{2}{H}^{\prime \prime }-(H^{\prime }{)}^2\le C$. After stopping,If $h_n=\inf \{H(y):y\text{ lies outside the }n\text{th compact interval}\}$, then $h_n\to \infty$ because $H$ is coercive, andThus $\mathbb{P}(T\le t)=0$ for every $t$, and $T=\infty$ almost surely.

Itô formula givesso $M$ is a positive local martingale. On every finite interval $[0,T]$, the hypotheses implyThe Gaussian tail bound of the Brownian maximum has finite exponential moments of every subquadratic power. Therefore Novikov condition holds on $[0,T]$, and the stochastic exponential $M$ is a true martingale there. Since $T$ was arbitrary, $M$ is a true martingale.

For $\theta >0$, the Exponential martingale for Brownian motion and the Doob maximal inequality for a nonnegative submartingale giveTaking $\theta =a/s$ gives $e^{-a^2/(2s)}$. Apply the same argument to $-B$ and use the union bound:

Put $A_t={\int }_0^t{H}_s^{2}ds$. It is continuous and tends to infinity almost surely, so $A_{T_{\sigma }}={\sigma }^2$. Define the right-continuous inverse $T(u)=\inf \{t:A_t>u\}$ andThe time-change theorem for local martingales shows that $W$ is a continuous local martingale in the time-changed filtration, andFor completeness, this proves the required case of the Dambis-Dubins-Schwarz theorem: for every $\theta \in \mathbb{R}$, Itô formula makes $\exp (i\theta W_u+{\theta }^{2}u/2)$ a local martingale; stopping and conditioning show that $W_v-W_u$ has conditional characteristic function $e^{-{\theta }^2(v-u)/2}$. Hence the increments are independent centered normal variables with the Brownian variances, and continuity makes $W$ a Brownian motion. ConsequentlyThus $X_{\sigma }$ is centered Gaussian with variance ${\sigma }^2$.

Fix $\epsilon >0$ and set $d=2/\epsilon$. Suppose, towards a contradiction, thatwhere $\lambda$ will absorb constants depending only on $\epsilon$. Put $n=\lfloor |G{|}^{\epsilon }\rfloor$. Once $\lambda$ is large enough, $n\ge N(d)$, $n<D$, andPart (a) yields $H⊴C<G$ with $[G:C]=O_{\epsilon }(1)$, $H\subseteq S^{\lfloor n/2\rfloor }$, and $C/H$ nilpotent of class $O_{\epsilon }(1)$.

The subgroup core ${\bigcap }_{g\in G}gC{g}^{-1}$ is normal in $G$ and has index at most $[G:C]!$. Since $G$ is a simple group, the core is either $\{1\}$ or $G$. In the first case $|G|\le [G:C]!=O_{\epsilon }(1)$, which is excluded by increasing $\lambda$. Hence the core is $G$, so $C=G$.

Now $H⊴G$. Since $n/2<D$, the ball $S^{\lfloor n/2\rfloor }$ is not all of $G$, so $H\ne G$. Simplicity gives $H=\{1\}$, and therefore $G=C/H$ is a nilpotent group. A nontrivial finite nilpotent group has nontrivial center of a group; simplicity would force that center to be all of $G$, making $G$ Abelian. This contradicts the assumption that $G$ is non-abelian. Consequently

The Cayley graph ${\mathrm{Cay}}_S(G)$ is a connected graph with $|G|$ vertices. A shortest path never repeats a vertex, so it has at most $|G|-1$ edges. Therefore

Use the product from part (ii). Move each selected element of each small set $X_i$ to the left, conjugating every approximate-group factor that it crosses. For each tuple $\mathbf{x}\in X_1\times \cdots \times X_m$, the corresponding part of the product is therefore contained inwhere every $D_{j,\mathbf{x}}$ is a conjugate subset of some $C_j$. Conjugation preserves cardinality, the approximation parameter, and the property that the generated subgroup is abelian. The conjugating elements belong to $A^{O_s(1)}$, so $D_{j,\mathbf{x}}\subseteq A^{O_s(1)}$ after enlarging the implicit constant.

The number of tuples is at mostThese translated products cover $A$, so one has size at least the reciprocal fraction of $|A|$. For that tuple, put $D_j=D_{j,\mathbf{x}}$. Thenwhere $q=n\le O_s({\log }^{O_s(1)}(2K))$, and each $D_j$ is a $K^{O_s(1)}$-approximate group generating an abelian subgroup.

We induct on the nilpotency class $s$. For $s=1$, the group is Abelian, so take $C_1=A$ and no $X_i$. For $s>1$, part (i) writeswith $X$ small and every $⟨B_i⟩$ of class at most $s-1$. Apply the induction hypothesis to each $B_i$. Since $B_i\subseteq A^{O(1)}$ and its approximation parameter is $K^{O(1)}$, all resulting small sets lie in $A^{O_s(1)}$, all their sizes are at mostand all resulting approximate groups lie in $A^{O_s(1)}$ and have approximation parameter $K^{O_s(1)}$. Their generated subgroups are abelian groups.

There are $O({\log }^{O(1)}(2K))$ factors at each of at most $s$ induction levels. Absorbing the resulting products of the bounds into the $O_s$ notation givesKeeping the factors in the order supplied by the induction yields the required product of the $X_i$ and $C_j$ containing $A$.

WriteUnder the improved bounds allowed in the question, $r=O({\log }^{O(1)}(2K))$ andMoreover $AC\subseteq A^{O(1)}$, whose size is at most $K^{O(1)}|A|$ by the higher product bound for an approximate group. ThusThe Ruzsa covering lemma supplies $X\subseteq A$ of size at most $\exp (O({\log }^{O(1)}(2K)))$ such thatTaking the $B_i$ to be the factors in this last product giveswhere $k\le O({\log }^{O(1)}(2K))$ and every $B_i$ is a $K^{O(1)}$-approximate group in $A^{O(1)}$ generating a subgroup of step less than $s$.

Assume $\dim (N)=|N|=\kappa$, with $\kappa \ge |L|+{\aleph }_0$. Let $A\subseteq N$ have size below $\kappa$ and let $p(x)\in S_1(A)$ be a complete type. If $p$ is algebraic, its realizations lie in $\mathrm{acl}(A)\subseteq \mathrm{acl}(N)=N$. If $p$ is nonalgebraic, strong minimality makes it the unique generic type over $\mathrm{acl}(A)$. The pregeometry exchange property givesso some basis element of $N$ lies outside $\mathrm{acl}(A)$ and realizes $p$.

Thus $N$ realizes every one-type over every parameter set of size below $\kappa$. The standard one-type characterization, applied successively to tuple coordinates, makes $N$ $\kappa$-saturated. Consequently

Let $\kappa =|N|$. If the saturated model $N$ had a basis $B$ of cardinality below $\kappa$, then $N=\mathrm{acl}(B)$ would omit the consistent generic type in a strongly minimal theory over $B$. This contradicts $\kappa$-saturation. A basis is a subset of $N$, so its cardinality is at most $\kappa$. Therefore