In the article The sound of Symmetry in Proposition $6$, there is a problem that bothers me and I cannot solve. Could anyone is able to explain to me why if $f_{\infty} \geq f(R_n)$, then no subsequence of $P_k$ can collapse to a segment?
Definitions :
- $f$ is the isoperimetric quotient $f(\Omega) = \frac{|\Omega|}{|\partial \Omega|}$.
- $R_n$ is a regular $n$-gon.
- $f_{\infty}$ is the limit of $f(P_k)$ as $k \to \infty$ for a sequence of convex $n$-gons $\{P_k\}$
Rowlett is hosting a preprint of the paper here, where it is not behind a paywall.
Setup: Notice that $f$ is positive and bounded on $\mathfrak{M}_n$. We let $f_\infty = \sup f(\mathfrak{M}_n) \in [0,f(\mathbb{D})]$, which exists because $f$ is bounded above by $f(\mathbb{D})$. As $f_\infty$ is the supremum of the image of $f$, there exists a sequence $P_k\in \mathfrak{M}_n$ such that $f(P_k)\to f_\infty$.
Crucial observation The fact that $\lim_{k\to\infty} f(P_k) = f_\infty > 0$ precludes a subsequence of the $P_k$ from degenerating to a line segment, as that subsequence $P_{k_i}$ would have $f(P_{k_i})\to_{i\to\infty} 0$. (This is key in the base case $n=3$, as it forces the $P_k$ to converge to $R_3$, the equilateral triangle.)