Volume ratio of general $\ell_p$ balls and surfaces

Question

This question is a generalization of the question Volume ratio of $\ell_1$ balls and $\ell_1$ surfaces

For any $p\in[1,\infty]$ define $\|x\|_p := (|x_1|^p+\cdots+|x_d|^p)^{1/p}$ for $p\in[1,\infty)$ and $\|x\|_\infty := \max_{1\leq i\leq d}|x_i|$ for $p=\infty$. Denote $B_p^d := \{x\in\mathbb R^d: \|x\|_p\leq 1\}$ as the unit $\ell_p$-ball in $d$ dimension, and let $\partial B_p^d := \{x\in\mathbb R^d: \|x\|_p = 1\}$. The quantity of interest is the following "ratio" $$ \mathfrak d_{p,d} := \frac{\mathrm{vol}_{d-1}(\partial B_p^d)}{\mathrm{vol}_{d}(B_p^d)}. $$

My question is the following:

For any fixed $p\in[1,\infty]$, how does $\mathfrak d_{p,d}$ asymptotically scale with dimension $d$, as $d\to\infty$?

For some special cases, both $\mathrm{vol}_d(B_p^d)$ and $\mathrm{vol}_{d-1}(\partial B_p^d)$ have closed form solutions and the ratio $\mathfrak d_{p,d}$ can be calculated explicitly. Below are three examples:

1. $p=2$: in this case $\mathrm{vol}_d(B_2^d) = \frac{\pi^{d/2}}{\Gamma(d/2+1)}$, $\mathrm{vol}_{d-1}(\partial B_2^d) = \frac{2\pi^{d/2}}{\Gamma(d/2)}$ and therefore $\mathfrak d_{2,d} = d$;

2. $p=1$: in this case $\mathrm{vol}_d(B_1^d) = \frac{2^d}{d!}$ and $\mathrm{vol}_{d-1}(\partial B_1^d) = \frac{2^d \sqrt{d}}{(d-1)!}$, and therefore $\mathfrak d_{1,d} = d\sqrt{d}$

3. $p=\infty$: in this case $\mathrm{vol}_d(B_\infty^d) = 2^d$ and $\mathrm{vol}_{d-1}(\partial B_\infty^d) = 2^d n$. Therefore $\mathfrak d_{\infty,d} = d$.

From the above examples, my (very wild) guess is that $\mathfrak d_{p,d} \asymp d^{1+1/p-1/2}$ for $1\leq p\leq 2$ and $\mathfrak d_{p,d} \asymp d$ for $p\geq 2$. But of course I could be very wrong. The challenge for general $p$ is the apparent difficulty in evaluating the volume of $L_p$ sphere areas (e.g., Surface area of an $\ell_p$ unit ball?), but I'm hoping that the (asymptotic) ratio $\mathfrak d_{p,d}$ for general $p$ is potentially easier to evaluate.

Edit: I realized that, because the volume of a unit $\ell_p$ ball does have closed forms for general $p\in[1,\infty]$, the question can be answered if we know the asymptotic dependency of dimension $d$ of the (unit) $\ell_p$-surface area. This question was explicitly mentioned in the comment under the accepted answer of the following question Surface area of superellipsoid (dice) Unfortunately, the OP in that question did not pursue this direction.

Answer 1

We have managed to obtain a complete solution to this problem, by applying the divergence theorem and the classical results of Naor and Romik on the affinity between cone and surface measures in $\ell_p$ balls.

The results are different than what I have conjectured in the problem statement. In particular, it is proved that $\mathfrak d_{p,d} \asymp d^{1/2+1/p}$ for all fixed $p<\infty$ and $d\to\infty$. Therefore, there is a phase transition between $p\to\infty$ and $p=\infty$. Such phase transition is because of the asymptotic regime we're considering ($p$ fixed, and $d$ goes to infinity afterwards). In cases where both $p$ and $d$ go to infinity simultaneously, the question remains unsolved.

The proof idea is to first use the divergence theorem to obtain $$ d\times \mathrm{vol}_d(\mathbb B_p^d) = \mathrm{vol}_{d-1}(\partial\mathbb B_p^d)\times \int_{\partial\mathbb B_p^d}\frac{\mathrm{d}\sigma_p^{d-1}(v)}{\sqrt{\sum_{i=1}^d|v_i|^{2p-1}}}. $$ Here $\sigma_p^{d-1}(v)$ is the surface measure on $\partial\mathbb B_p^{d-1}$, meaning that $\sigma_p^{d-1}(A)=\mathrm{vol}_{d-1}(A)/\mathrm{vol}_{d-1}(\partial\mathbb B_p^d)$ for all measurable $A\subseteq\mathbb B_p^{d-1}$. The work of Naor and Romik showed that this surface measure is very close (in TV distance) to a cone measure $\gamma_p^{d-1}$, which is equivalently the distribution of $X=Z/\|Z\|_p$ where $Z=(Z_1,\cdots,Z_d)$ consists of i.i.d. random variables distributed according to a generalized Normal distribution $p(\cdot)\propto \exp\{-|\cdot|^p\}$. It is then possible to use Chebyshev's inequality to get an adequate estimate of the integral term above, which yields an asymptotic formula for $\mathfrak d_{p,d}$.

Answer 2

$\newcommand{\R}{\mathbb R} \newcommand{\dd}{\operatorname{d}\!}$

This is not a complete answer, but it may lead to one. Since the volume of the unit $\ell_p$ ball $B_p^d$ is known in closed form, we "only" need to find the asymptotics of the $(d-1)$-dimensional surface area of this ball. It is easy to see that for $p\in(1,\infty)\setminus\{2\}$ this area is given by the formula \begin{align*} \mathrm{vol}_{d-1}(\partial B_p^d) &=2^d\int_{S_p^{d-1}}\frac{\dd x_1\cdots\dd x_{d-1}}{\cos\theta} \\ &=2^d\int_{S_p^{d-1}} \sqrt{1+\Big(1-\sum_1^{d-1}x_i^p\Big)^{2/p-2}\,\sum_1^{d-1}x_i^{2p-2}}\, \dd x_1\cdots\dd x_{d-1} \\ &=2^d\iint\limits_{a<1} \sqrt{1+(1-a)^{2/p-2}\,b}\,\,f_{d-1}(a,b)\dd a\dd b, \end{align*} where $S_p^{d-1}:=\{(x_1,\dots,x_{d-1})\in(0,\infty)^{d-1}\colon\sum_1^{d-1}x_i^p<1\}$, $\theta$ is the angle between the vectors $(0,\dots,0,1)\in\R^d$ and $\vec\nabla G=\big(-\frac{\partial F}{\partial x_1},\dots,-\frac{\partial F}{\partial x_{d-1}},1\big)$, $G:=G(x_1,\dots,x_d):=x_d-F(x_1,\dots,x_{d-1})$, $F:=F(x_1,\dots,x_{d-1}):=\big(1-\sum_1^{d-1}x_i^p\big)^{1/p}$ (so that $(0,\infty)^d\cap\partial B_p^d=\{(x_1,\dots,x_d)\in(0,\infty)^d\colon G(x_1,\dots,x_d)=0\}$), $f_{d-1}$ is the joint probability density function (pdf) of the random variables (r.v.'s) $A:=\sum_1^{d-1}X_i^p$ and $B:=\sum_1^{d-1}X_i^{2p-2}$, and $X_1,\dots,X_{d-1}$ are independent r.v.'s uniformly distributed on the interval $(0,1)$.

Thus, it appears to mainly remain to find an appropriate asymptotics of $f_{d-1}(a,b)$ for $a<1$ and $d\to\infty$. This task may not seem exceedingly difficult, but it has eluded my efforts so far; perhaps someone else will be able to succeed here.