Does the surface area of the unit Lp ball go to zero for all $p < \infty$? 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.)
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. Tools from this mathoverflow Q&A and linked paper might be useful.
 A: 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 to 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 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.
A: This is another proof which rests on the explicit formula for the volume $c_p(n)$ of the n-dimensional $l^p$ ball $B_p(n)=\{x \in \mathbb R^n: \sum_{i=1}^n |x_i|^p \leq 1\}$, namely $$c_p(n)=\frac{\left (\frac 2p \Gamma (1/p) \right )^n}{\Gamma (1+n/p)}.$$ If $\phi(x)=\sum_{i=1}^n |x_i|^p$, then $|\nabla \phi|^2 =p^2 \sum_{i=1}^n |x_i|^{2(p-1)}$ and using the coarea formula
$$
r^{n/p} c_p(n)=m_n\{\phi \leq r\}=\int_0^r dt \int_{\{\phi=t\}}|\nabla \phi|^{-1}\, d\sigma_t.
$$
Differentiating with respect to $r$ and setting $r=1$ we get
$$\frac np c_p(n)= \int_{\{\phi=1\}}|\nabla \phi|^{-1}\, d\sigma_1 \geq \frac{1}{p \sqrt n} m_{n-1}\{\phi=1\}
$$
so that $m_{n-1}\{\phi=1\} \leq n^{\frac 32}c_p(n) \to 0$, as $n \to \infty$, by Stirling's formula for the $\Gamma$ function.
Note that when $p \geq 2$ then $2p-2 \geq p$ and $|\nabla \phi|^2 \leq p^2$ on $\{\phi=1\}$ so that the exponent $3/2$ improves to 1. When $p <2$ a better exponent follows from Holder.
