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The surface area of the unit $L^p$ ball (of radius $1$) goes to $0$ as $n$ goes to infinity, if $p<\infty$.

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: this works in general for two convex sets one contained in the other, which can be checked using the same proof as in this MSE answer to Surface area of a convex set less than that of its enclosing sphere?.

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.

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.

The surface area of the unit $L^p$ ball (of radius $1$) goes to $0$ as $n$ goes to infinity, if $p<\infty$.

Consider the orthoplex circumscribed to an $L^p$-ball, with boundary contained in the $2^n$ hyperplanes $\{\varepsilon_1x_1+\dots+\varepsilon_nx_n=\frac{n}{n^{1/p}}\}$, for $\varepsilon_i\in\{0,1\}$. The $2^n$ points of tangency are $(\pm\frac{1}{n^{1/p}},\dots,\pm\frac{1}{n^{1/p}})$, and the $2n$ vertices of the orthoplex are the points $\pm n^\frac{p-1}{p}$ of all coordinates axes.

Then the $n-1$-volume of the surface of the orthoplex is bigger than that of the $L^p$-ball: this works in general for two convex sets one contained in the other, which can be checked using the same proof as in this MSE answer to Surface area of a convex set less than that of its enclosing sphere?.

So we just have to check that for fixed $p$, the surface area of this orthoplex goes to $0$ when $n$ goes to $\infty$. Note that the orthoplex has edges of length $\sqrt{2}n^\frac{p-1}{p}$, and its surface is formed by $2^n$ simplices of dimension $n-1$. The volume of an $n-1$-dimensional simplex with unit edges is $\frac{\sqrt{n-1}}{(n-1)!\sqrt{2^{n-1}}}$, so the surface volume of the orthoplex is

$$A=2^n\cdot\left(\sqrt{2}n^\frac{p-1}{p}\right)^{n-1}\cdot\frac{\sqrt{n-1}}{(n-1)!\sqrt{2^{n-1}}}=2^nn^\frac{(n-1)(p-1)}{p}\frac{\sqrt{n-1}}{(n-1)!}$$

This converges to $0$, as can be seen by taking logarithms:

$$\ln(A)=n\ln(2)+\frac{(n-1)(p-1)}{p}\ln(n)+\frac{\ln(n-1)}{2}-\ln((n-1)!),$$

which by Stirling's approximation is

$$n\ln(2)+\frac{(n-1)(p-1)}{p}\ln(n)-(n-1)\ln(n-1)+(n-1)+O(ln(n))$$ $$=\frac{p-1}{p}n\ln(n)-(n-1)\ln(n-1)+O(n)$$ $$=\frac{p-1}{p}(n\ln(n)-(n-1)\ln(n-1))-\frac{1}{p}(n-1)\ln(n-1)+O(n).$$

As $p$ is constant, the term $\frac{1}{p}(n-1)\ln(n-1)$ dominates the others, so the expression goes to $-\infty$ when $n\to\infty$, as we wanted.

The surface area of the unit $L^p$ ball (of radius $1$) goes to $0$ as $n$ goes to infinity, if $p<\infty$.

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: this works in general for two convex sets one contained in the other, which can be checked using the same proof as in this MSE answer to Surface area of a convex set less than that of its enclosing sphere?.

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.

The surface area of the unit $L^p$ ball (of radius $1$) goes to $0$ as $n$ goes to infinity, if $p<\infty$.

Consider the orthoplex circumscribed to an $L^p$-ball, with boundary contained in the $2^n$ hyperplanes $\{\varepsilon_1x_1+\dots+\varepsilon_nx_n=\frac{n}{n^{1/p}}\}$, for $\varepsilon_i\in\{0,1\}$. The $2^n$ points of tangency are $(\pm\frac{1}{n^{1/p}},\dots,\pm\frac{1}{n^{1/p}})$, and the $2n$ vertices of the orthoplex are the points $\pm n^\frac{p-1}{p}$ of all coordinates axes.

Then the $n-1$-volume of the surface of the orthoplex is bigger than that of the $L^p$-ball: this works in general for two convex sets one contained in the other, which can be checked using the same proof as in this MSE answer to Surface area of a convex set less than that of its enclosing sphere?.

So we just have to check that for fixed $p$, the surface area of this orthoplex goes to $0$ when $n$ goes to $\infty$. Note that the orthoplex has edges of length $\sqrt{2}n^\frac{p-1}{p}$, and the surface of the orthoplex is formed by $2^n$ simplices of dimension $n-1$. The volume of an $n-1$-dimensional simplex with unit edges is $\frac{\sqrt{n-1}}{(n-1)!\sqrt{2^{n-1}}}$, so the surface volume of the orthoplex is

$$A=2^n\cdot\left(\sqrt{2}n^\frac{p-1}{p}\right)^{n-1}\cdot\frac{\sqrt{n-1}}{(n-1)!\sqrt{2^{n-1}}}=2^nn^\frac{(n-1)(p-1)}{p}\frac{\sqrt{n-1}}{(n-1)!}$$

This converges to $0$, as can be seen by taking logarithms:

$$\ln(A)=n\ln(2)+\frac{(n-1)(p-1)}{p}\ln(n)+\frac{\ln(n-1)}{2}-\ln((n-1)!),$$

which by Stirling's approximation is

$$n\ln(2)+\frac{(n-1)(p-1)}{p}\ln(n)-(n-1)\ln(n-1)+(n-1)+O(ln(n))$$ $$=\frac{p-1}{p}n\ln(n)-(n-1)\ln(n-1)+O(n)$$ $$=\frac{p-1}{p}(n\ln(n)-(n-1)\ln(n-1))-\frac{1}{p}(n-1)\ln(n-1)+O(n).$$

As $p$ is constant, the term $\frac{1}{p}(n-1)\ln(n-1)$ dominates the others, so the expression goes to $-\infty$ when $n\to\infty$, as we wanted.

The surface area of the unit $L^p$ ball (of radius $1$) goes to $0$ as $n$ goes to infinity, if $p<\infty$.

Consider the orthoplex circumscribed to an $L^p$-ball, with boundary contained in the $2^n$ hyperplanes $\{\varepsilon_1x_1+\dots+\varepsilon_nx_n=\frac{n}{n^{1/p}}\}$, for $\varepsilon_i\in\{0,1\}$. The $2^n$ points of tangency are $(\pm\frac{1}{n^{1/p}},\dots,\pm\frac{1}{n^{1/p}})$, and the $2n$ vertices of the orthoplex are the points $\pm n^\frac{p-1}{p}$ of all coordinates axes.

Then the $n-1$-volume of the surface of the orthoplex is bigger than that of the $L^p$-ball: this works in general for two convex sets one contained in the other, which can be checked using the same proof as in this MSE answer to Surface area of a convex set less than that of its enclosing sphere?.

So we just have to check that for fixed $p$, the surface area of this orthoplex goes to $0$ when $n$ goes to $\infty$. Note that the orthoplex has edges of length $\sqrt{2}n^\frac{p-1}{p}$, and its surface is formed by $2^n$ simplices of dimension $n-1$. The volume of an $n-1$-dimensional simplex with unit edges is $\frac{\sqrt{n-1}}{(n-1)!\sqrt{2^{n-1}}}$, so the surface volume of the orthoplex is

$$A=2^n\cdot\left(\sqrt{2}n^\frac{p-1}{p}\right)^{n-1}\cdot\frac{\sqrt{n-1}}{(n-1)!\sqrt{2^{n-1}}}=2^nn^\frac{(n-1)(p-1)}{p}\frac{\sqrt{n-1}}{(n-1)!}$$

This converges to $0$, as can be seen by taking logarithms:

$$\ln(A)=n\ln(2)+\frac{(n-1)(p-1)}{p}\ln(n)+\frac{\ln(n-1)}{2}-\ln((n-1)!),$$

which by Stirling's approximation is

$$n\ln(2)+\frac{(n-1)(p-1)}{p}\ln(n)-(n-1)\ln(n-1)+(n-1)+O(ln(n))$$ $$=\frac{p-1}{p}n\ln(n)-(n-1)\ln(n-1)+O(n)$$ $$=\frac{p-1}{p}(n\ln(n)-(n-1)\ln(n-1))-\frac{1}{p}(n-1)\ln(n-1)+O(n).$$

As $p$ is constant, the term $\frac{1}{p}(n-1)\ln(n-1)$ dominates the others, so the expression goes to $-\infty$ when $n\to\infty$, as we wanted.