and the fact that the last equality in is the one-dimensional version of the torsion function for a ball $B_a\subset {\mathbb{R}}^m$ with radius $a$, centred at $0$. One verifies that $$\label{e49c}
v_{B_a}(x)=\frac{a^2-|x|^2}{2m},$$ satisfies for $\Omega=B_a$, [@PSZ].

[*Proof of Example* ]{} (i) We have $$\label{e49}
1\le |\Omega_{\alpha,n}|\le 1+c.$$ By , $$\|v_{\Omega_{\alpha,n}}\|_{\infty}\le \frac18\max\bigg\{\frac{1}{n^2},\frac{c^2}{n^{2\alpha}}\bigg\}.$$ Since $n\ge N_{\alpha,c}$ we obtain $$\label{e70a}
\|v_{\Omega_{\alpha,n}}\|_{\infty}\le \frac{c^2}{8n^{2\alpha}}.$$ On the other hand since $n\ge N_{\alpha,c}$, $\Omega_{\alpha,n}$ contains a ball with radius $\frac12 cn^{-\alpha}$. The maximum of the torsion function is, by domain monotonicity, bounded from below by the maximum for that ball. Since by with $m=2$, $\|v_{B_a}\|_{\infty}=\frac{a^2}{4}$ we have $$\label{e71}
\|v_{\Omega_{\alpha,n}}\|_{\infty}\ge \frac{c^2}{16n^{2\alpha}}.$$ It was shown in Proposition 3.1 in [@MvdBBV] that if $a\le b$, then $$\label{e72}
\frac{a^3b}{12}-\frac{11a^4}{180}\le \|v_{R_{a,b}}\|_1\le \frac{a^3b}{12}.$$ Since all rectangles in $\Omega_{\alpha,n}$ are disjoint $$\label{f1}
\|v_{\Omega_{\alpha,n}}\|_1\le \frac{1}{12n^2}+\frac{c^3n^{-3\alpha}}{12},$$ and $$\begin{aligned}
\label{f2}
\|v_{\Omega_{\alpha,n}}\|_1&\ge \frac{1}{12n^2}+\frac{c^3n^{-3\alpha}}{12}-\frac{11}{180n^3}-\frac{11c^4n^{-4\alpha}}{180}\nonumber \\ &
\ge \frac{1}{45}\big(n^{-2}+ c^3n^{-3\alpha}\big),\qquad n\ge N_{\alpha,c}.\end{aligned}$$ The upper bound in follows from the lower bound in , , and the upper bound in . The lower bound in follows the upper bound in , , and the lower bound in .

\(ii) Let $A_n$ be a measurable subset of $\Omega_{\alpha,n}$. Denote by $S_{\alpha,n}$ the rectangle of size $cn^{-\alpha}\times 1$. We have $$\begin{aligned}
\label{f3}
\int_{A_n}v_{\Omega_{\alpha,n}}&=\int_{A_n\cap 