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Are there any known formulas or approximations for the surface area of a unit ball in $d$ dimensions under the $\ell_p$ norm? As obvious examples, it is of course well-known that the surface area of a $d$-dimensional unit ball in $\ell_2$ is $\frac{2\pi^{d/2}}{\Gamma(d/2)}$ and that the surface area of a $d$-dimensional unit box (which we could also call the "unit ball under the $\ell_\infty$ norm") is $d\cdot2^d$.

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    $\begingroup$ Already for $d=2$, Mathematica can't evaluate the length of the curve ($4 \int_0^1 \sqrt{1 + x^{2 (p-1)} \left(1-x^p\right)^{\frac{2}{p}-2}} \, dx$) in closed form, so I wouldn't be too optimistic about finding a useful expression for the general case. $\endgroup$
    – Gro-Tsen
    Mar 30, 2016 at 22:07

2 Answers 2

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Note: I agree with the criticism of Sawin mentioned by Wong: the information about the volume of the unit ball of $\ell_p^n$ (which you can find, for example, on p. 47 of Brazitikos, Silouanos; Giannopoulos, Apostolos; Valettas, Petros; Vritsiou, Beatrice-Helen Geometry of isotropic convex bodies. American Mathematical Society, 2014) does not lead (at least immediately) to the answer.

What I suggest below is not an answer, it is just an information which can be useful in this context; and everyone who rated my posting is expected to reconsider the rating (I also deleted my erroneous comment).

For subsets of the unit sphere of $\ell_p^n$ one can consider the usual surface measure and the cone measure, which is the measure of the corresponding part of the ball. It turns out that if we normalize both measures, the results will be rather close to each other; the best known to me result on this matter is due to Naor (Trans. Amer. Math. Soc. 359 (2007), no. 3, 1045–1079).

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    $\begingroup$ @MikhailOstrovskii: you cannot see it, but Andreas Thom has posted basically the identical answer a week ago. He deleted it after Will Sawin made the following comment: "The surface area is only the derivative of the volume if the derivative of the ℓp norm with respect to the perpendicular direction is 1, equivalently, if the dual convex body to the unit ball is the convex hull of some set of points on the unit sphere. This holds for ℓ2 and ℓ∞ but not for any other p". $\endgroup$ Mar 30, 2016 at 17:53
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    $\begingroup$ Below is a paraphrase of Will Savin's second comment from Mar 23: "Let $n = 2$ and $p = 1$. The $\ell^1$ ball has circumference $4\sqrt{2}$. Your formula would give $2 * 4 * \Gamma(2)^2 / \Gamma(3) \neq 4 \sqrt{2}$." $\endgroup$ Mar 30, 2016 at 17:56
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    $\begingroup$ @WillieWong I agree, I decided to keep my "answer", so that the information contained in our discussion will be seen. $\endgroup$ Mar 30, 2016 at 19:35
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    $\begingroup$ Wikipedia suggests that "the surface area of an $\ell_1$-(n − 1)-sphere of radius $R$ is $\sqrt{n}$ times the derivative at $R$ of the volume of an $\ell_1$-n-ball." They refer to the coarea formula, and mention that a similar correction factor exists for other values of $p$, although it might be a "complicated integral". $\endgroup$
    – Eckhard
    Mar 30, 2016 at 20:18
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    $\begingroup$ @MikhailOstrovskii: I agree that it is good to keep some version of this discussion visible! I feel a bit embarrassed to be in this discussion since my role is nothing more than that of a fax machine. $\endgroup$ Mar 30, 2016 at 21:36
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$\newcommand{\R}{\mathbb R} \newcommand{\dd}{\operatorname{d}\!}$

This is not a complete answer, but it may lead to finding the asymptotics of the surface area; an exact closed-form expression for the area seems unlikely to exist. (This answer is almost the same as my answer to almost the same question.) 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.

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  • $\begingroup$ It seems to me that your formula is correct also for $p=2$. Is there a special reason to exclude $p=2$? $\endgroup$ Sep 12, 2022 at 18:09

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