S_{\alpha,n}}v_{\Omega_{\alpha,n}}+\int_{A_n\setminus S_{\alpha,n}}v_{\Omega_{\alpha,n}}\nonumber \\ &
\le\frac{c^3n^{-3\alpha}}{12}+|A_n|\max\{v_{\Omega_{\alpha,n}}:x\in \Omega_{\alpha,n}\setminus S_{\alpha,n}\}\nonumber \\ &
=\frac{c^3n^{-3\alpha}}{12}+\frac{|A_n|}{8n^2}.\end{aligned}$$ By , $\|v_{\Omega_{\alpha,n}}\|_1\ge \frac{1}{45n^{2}}$ so that $$\label{f4}
\frac{\int_{A_n}v_{\Omega_{\alpha,n}}}{\|v_{\Omega_{\alpha,n}}\|_1}\le\frac{15c^3n^{2-3\alpha}}{4}+\frac{45|A_n|}{8}.$$ Since $\alpha>\frac23$ we have that the right-hand side of tends to $0$ as $n\rightarrow\infty$ by the hypothesis on $|A_n|$. That hypothesis also shows that $$\lim_{n\rightarrow\infty}\frac{|A_n|}{|\Omega_{\alpha,n}|} \le \lim_{n\rightarrow\infty}|A_n|=0.$$ This proves that $\kappa=0$.

\(iii) We will show that for $n\rightarrow\infty$ all of the torsion mass localises at $S_{\alpha,n}$. This is equivalent to showing that $$\lim_{n\rightarrow\infty}\frac{\int_{\Omega_{\alpha,n}\setminus S_{\alpha,n}}v_{\Omega_{\alpha,n}}}{\int_{\Omega_{\alpha,n}}v_{\Omega_{\alpha,n}}}=0.$$ This follows from $$\int_{\Omega_{\alpha,n}\setminus S_{\alpha,n}}v_{\Omega_{\alpha,n}}\le \frac{1}{12n^2},$$ the estimate , $$\int_{\Omega_{\alpha,n}}v_{\Omega_{\alpha,n}}\ge \frac{c^3n^{-3\alpha}}{45},$$ and the hypotheses on $\alpha$ and on $n$. This proves assertion (iii).

\(iv) To obtain a lower bound for $\kappa$ in with $V=0,\, \Omega_n=\Omega_{\frac23,n}$ we choose $A_n=S_{\frac23,n}$. Then $ \lim_{n\rightarrow\infty} | S_{\frac23,n} | = 0$, and it follows by that $$\lim_{n\rightarrow\infty}n^2\int_{S_{\frac23,n}}v_{S_{\frac23,n}}=\frac{c^3}{12},$$ and $$\lim_{n\rightarrow\infty}n^2\int_{\Omega_{\frac23,n}\setminus S_{\frac23,n}}v_{\Omega_{\frac23,n}}=\lim_{n\rightarrow\infty}n^2\int_{\Omega_{\frac23,n}\setminus S_{\frac23,n}}v_{\Omega_{\frac23,n}\setminus S_{\frac23,n}}=\frac{1}{12}.$$ This implies $$\label{h8}
 \kappa_c\ge \frac{c^3}{1+c^3}.$$ To prove the reverse inequality we let $(A_n)_n$ be an arbitrary sequence of measurable sets with $A_n\subset \Omega_{\frac23,n},\,n\in{\mathbb{N}}$ and $\lim_{n\rightarrow\infty}|A_n|=0$. Then, by , $$\label{h6}
\int_{A_n}v_{\Omega_{\frac23,n}}\le
\frac{1}{n^2}\bigg(\frac{c^3}{12}+\frac{|A_n|}{8}\bigg).$$ By the first inequality in , $$\label{h7}
\|v_{\Omega_{\frac23,n}}\|_1\ge \frac{n^{-2}}{12}+\frac{c^3n^{-2}}{12}-\frac{11n^{-3}}{180}-\frac{11c^4n^{-8/3}}{180}.$$ By and , $$\label{h9}
\limsup_{n\rightarrow\infty}\frac{\int_{A_n}v_{\Omega_{\frac23,n}}}{\|v_{\Omega_{\frac23,n}}\|_1}\le \frac{c^3}{1+c^3}.$$ Taking the supremum in the left-hand side of over all sequences $(A_n)$ in yields, $$ \kappa_c\le \frac{c^3}{1+c^3}.$$ This, together with gives . Furthermore, the supremum 