 Solution

Fix $x\in (-1,1)$ and start at $(x,y)$ with $y>1$. Write the independent Brownian coordinates as $x+W_t^1$ and $y+W_t^2$. LetOn $\{H<L_y\}$ the path has stayed in the upper strip and first exits the cross-shaped domain at its right or left side. The boundary value is $a$ or $b$, respectively. Let $V=a$ if $x+W_H^1=1$ and $V=b$ if $x+W_H^1=-1$. The Brownian exit from an interval probabilities giveCouple all heights $y$ using the same two Brownian paths. By part (a), $H<\infty$ almost surely, and continuity makes ${\min }_{0\le t\le H}{W}_t^2$ finite. ThusLet $K=\max \{|a|,|b|,|c|,|d|\}$. On the complement of this exceptional event, $f(B_T)=V$; on the event itself, $|f(B_T)-V|\le 2K$. The representation in part (c) therefore yieldsThis proves the harmonic limit along an infinite strip arm:Far up the arm, the bottom is reached with vanishing probability before the horizontal coordinate exits. The limit is consequently determined by the two side values alone.