We know about volume: The $L\_{\infty}$ ball of radius one-half, i.e. the hypercube, has volume $1$ in all dimensions. On the other hand, I believe that for every $1 \leq p < \infty$, the volume of the inscribed $L\_p$ ball $\{x : \|x\|\_p \leq \tfrac{1}{2}\}$ goes to zero as the dimension $d \to \infty$. (For example, a simple generalization of [this argument](https://mathoverflow.net/a/8270/29697).)


My question is the analogous one for surface area. The surface area of the $L\_2$ ball goes to zero as the dimension diverges. But the surface area of the $L\_{\infty}$ ball is $2d \to \infty$. So, we might wonder if there is a "happy medium" $p \in (2,\infty)$ where the surface area is constant in all dimensions. But I doubt it. **Does the surface area of the radius-$\tfrac{1}{2}$ $L\_p$ ball go to zero for all $p < \infty$**?


Asymptotics about the surface area of Lp balls are [apparently hard to pin down](https://mathoverflow.net/questions/234314/surface-area-of-an-ell-p-unit-ball). Tools from [this mathoverflow Q&A](https://mathoverflow.net/questions/293840/volume-ratio-of-general-ell-p-balls-and-surfaces) and linked paper might be useful.



Does the surface area of the unit Lp ball go to zero for all $p < \infty$?
The surface area of the unit $L^p$ ball (of radius $1$) goes to $0$ as $n$ goes to infinity, if $p<\infty$. Below I show it for $p>2$, but this is enough because if we have convex sets $A\subseteq B$, then the surface area of $B$ is bigger than that of $A$, which follows from the proof in [this MSE answer](https://math.stackexchange.com/a/58556/807670) to [Surface area of a convex set less than that of its enclosing sphere?](https://math.stackexchange.com/questions/58456/surface-area-of-a-convex-set-less-than-that-of-its-enclosing-sphere).


First note that the minimum value of $\sum\_{i=1}^n|x\_i|^p$ in the sphere $\mathbb{S}^{n-1}\subseteq\mathbb{R}^n$ is achieved when $x\_1=\dots=x\_n$. This can be deduced by setting $y\_i=x\_i^2$ and then $\sum y\_i^{p/2}$ is convex, so for any $y\_1,\dots,y\_n\geq0$, if $\sum y\_i=1$ then $\sum y\_i^{p/2}\geq n\left(\frac{1}{n}\right)^{p/2}$, with equality when $y\_1=\dots=y\_n$.


This implies that the ball $B$ of radius $n^\frac{p-2}{2p}$, which is tangent to the $L^p$ ball at the points $(\pm\frac{1}{n^{1/p}},\dots,\pm\frac{1}{n^{1/p}})$, contains the $L^p$ ball. So the $n-1$-volume of $\partial B$ is bigger than that of the $L^p$-ball, and we just have to prove that the area of $B$ goes to $0$ when $n\to\infty$.


Now, the area of $\mathbb{S}^{n-1}$ is $\frac{2\pi^{n/2}}{\Gamma(\frac{n}{2})}$, so the area of $\partial B$ is $A=\frac{2\pi^{n/2}}{\Gamma(\frac{n}{2})}\left(n^\frac{p-2}{2p}\right)^{n-1}$. So


$$\ln(A)=-\ln\left(\Gamma\left(\frac{n}{2}\right)\right)+n\frac{p-2}{2p}\ln(n)+O(n),$$


which by [Stirling's approximation](https://en.wikipedia.org/wiki/Stirling%27s_approximation#Stirling%27s_formula_for_the_gamma_function) is
$$-\frac{n}{2}\ln\left(\frac{n}{2}\right)+n\ln(n)\frac{p-2}{2p}+O(n)=-\frac{n}{2}\ln(n)+n\ln(n)\frac{p-2}{2p}+O(n),$$
which goes to $-\infty$ as $n$ tends to $\infty$, concluding the proof.


