3 Grammar is fixed.

I'd like to add a few words on what happens in higher dimensions. First, a convexity assumption becomes essential (as in the second proof of Hurwitz which works only for convex domains). The isoperimetric inequalities in $\mathbb R^n$, $n>2$, are much easier to deal with in case of convex bodies, and the whole problem in some sense looks most natural under the convexity assumption. Second, there are many different isoperimetric inequalities in higher dimensions. And Fourier analysis (or rather harmonic analysis on a sphere) can be successfully applied to prove at least some of them.

There is a classical approach to isoperimetric problems based on Steiner's theorem. Let $K$ be a convex body in $\mathbb R^n$ and let $K_r$ denote the "parallel" body $$K_r=\{x\in\mathbb R^n|\ dist(x, K)\leq r \},\quad r>0.$$ Then, by Steiner's theorem, there exist $n+1$ numbers $W_0^n(K),W_1^n(K),\dots,W_n^n(K)$, such that $$V(K_r)=Vol(K_r)=\sum\limits_{i=0}^{n}{n \choose i}W^n_i(K)r^{i}.$$ It can be shown that $$W^n_0(K)=V(K),\quad W^n_1(K)=\frac{S(K)}{n},\qquad(1)$$ where $S(K)$ is the surface area of $\partial K$. Moreover, $W^n_n(K)$ is equal to the volume $\pi_n$ of the unit ball in $\mathbb R^n$ and $$W^n_{n-1}(K)=\frac{\pi_n}{2}w(K),\qquad\qquad\qquad\quad (2)$$ where $w(K)$ is the mean width of $K$. Note that for $n=2$ the perimeter $P(K)$ equals $\pi w(K).$ The numbers $W_i^n(K)$ give some information on how the convex body $K$ is different from a ball (for the unit ball in $\mathbb R^n$, obviously $W^n_i(K)=\pi_n$ for all $i$.)

A convex body is completely determined by its support function $$h(x)=\sup\{x\cdot y|\ y\in K\}$$ which measures the directed distance of the origin to the tangent plane of $K$ at direction $x\in S^{n-1}.$ Now, the second proof of Hurwitz deals with the Fourier decomposition of the support function of a convex 2D domain. The problem is that in dimension $n>2$ the formulas for volume and surface area in terms of the support function cannot be expressed nicely by means of spherical harmonics. However, it is still possible to derive an isoperimetric inequality for the numbers $W^n_{n-2}$ and $W^n_{n-1}$ via harmonic analysis, namely $$W^n_{n-1}\geq\sqrt{\pi_n W^n_{n-2}}. \qquad\qquad\qquad(3)$$

When $n=2$ this is the standard isoperimetric inequality $P^2\geq 4\pi A$.

If $n=3$ $(3)$ gives the isoperimetric inequality between the mean width and surface area of a convex body $$\pi [w(K)]^2\geq S(K).\qquad\qquad\qquad (4)$$

The proof is a straightforward extension of the second Hurwitz proof (using a decomposition of the support function into a series of spherical harmonics) and can be found here.

Update (concerning the question in Victor's comment below). If we assume as known the inequality $$W_{1}^3\geq \sqrt{W_{0}^3W_{2}^3},$$ then together with (1),(2) and (4) it implies that $S^3\geq 36\pi V^2$. ("Known" means that I don't know how to obtain the inequality using only harmonic analysis. It follows from the Alexandrov-Fenchel inequality for mixed volumes.)

2 added 549 characters in body; added 3 characters in body

I'd like to add a few words on what happens in higher dimensions. First, a convexity assumption becomes essential (as in the second proof of Hurwitz which works only for convex domains). The isoperimetric inequalities in $\mathbb R^n$, $n>2$, are much easier to deal with in case of convex bodies, and the whole problem in some sense looks most natural under the convexity assumption. Second, there are many different isoperimetric inequalities in higher dimensions. And Fourier analysis (or rather harmonic analysis on a sphere) can be successfully applied to prove at least some of them.

There is a classical approach to isoperimetric problems based on Steiner's theorem. Let $K$ be a convex body in $\mathbb R^n$ and let $K_r$ denote the "parallel" body $$K_r=\{x\in\mathbb R^n|\ dist(x, K)\leq r \},\quad r>0.$$ Then, by Steiner's theorem, there exist $n+1$ numbers $W_0^n(K),W_1^n(K),\dots,W_n^n(K)$, such that $$V(K_r)=Vol(K_r)=\sum\limits_{i=0}^{n}{n \choose i}W^n_i(K)r^{i}.$$ It can be shown that $$W^n_0(K)=V(K),\quad W^n_1(K)=\frac{S(K)}{n},$$ W^n_1(K)=\frac{S(K)}{n},\qquad(1)$$where S(K) is the surface area of \partial K. Moreover, W^n_n(K) is equal to the volume \pi_n of the unit ball in \mathbb R^n and$$W^n_{n-1}(K)=\frac{\pi_n}{2}w(K),$$W^n_{n-1}(K)=\frac{\pi_n}{2}w(K),\qquad\qquad\qquad\quad (2)$$ where $w(K)$ is the mean width of $K$. Note that for $n=2$ the perimeter $P(K)$ equals $\pi w(K).$ The numbers $W_i^n(K)$ give some information on how the convex body $K$ is different from a ball (for the unit ball in $\mathbb R^n$, obviously $W^n_i(K)=\pi_n$ for all $i$.)

A convex body is completely determined by its support function $$h(x)=\sup\{x\cdot y|\ y\in K\}$$ which measures the directed distance of the origin to the tangent plane of $K$ at direction $x\in S^{n-1}.$ Now, the second proof of Hurwitz deals with the Fourier decomposition of the support function of a convex 2D domain. The problem is that in dimension $n>2$ the formulas for volume and surface area in terms of the support function cannot be expressed nicely by means of spherical harmonics. However, it is still possible to derive an isoperimetric inequality for the numbers $W^n_{n-2}$ and $W^n_{n-1}$ via harmonic analysis, namely $$W^n_{n-1}\geq\sqrt{\pi_n W^n_{n-2}}. \qquad\qquad(*)$$qquad\qquad\qquad(3)$$When n=2 this is the standard isoperimetric inequality P^2\geq 4\pi A. If n=3 (*) (3) gives the isoperimetric inequality between the mean width and surface area of a convex body$$\pi [w(K)]^2\geq S(K).$$S(K).\qquad\qquad\qquad (4)$$

The proof is a straightforward extension of the second Hurwitz proof (using a decomposition of the support function into a series of spherical harmonics) and can be found here.

Update (concerning the question in Victor's comment below). If we assume as known the inequality $$W_{1}^3\geq \sqrt{W_{0}^3W_{2}^3},$$ then together with (1),(2) and (4) it implies $S^3\geq 36\pi V^2$. ("Known" means that I don't know how to obtain the inequality using only harmonic analysis. It follows from the Alexandrov-Fenchel inequality for mixed volumes.)

1

I'd like to add a few words on what happens in higher dimensions. First, a convexity assumption becomes essential (as in the second proof of Hurwitz which works only for convex domains). The isoperimetric inequalities in $\mathbb R^n$, $n>2$, are much easier to deal with in case of convex bodies, and the whole problem in some sense looks most natural under the convexity assumption. Second, there are many different isoperimetric inequalities in higher dimensions. And Fourier analysis (or rather harmonic analysis on a sphere) can be successfully applied to prove at least some of them.

There is a classical approach to isoperimetric problems based on Steiner's theorem. Let $K$ be a convex body in $\mathbb R^n$ and let $K_r$ denote the "parallel" body $$K_r=\{x\in\mathbb R^n|\ dist(x, K)\leq r \},\quad r>0.$$ Then, by Steiner's theorem, there exist $n+1$ numbers $W_0^n(K),W_1^n(K),\dots,W_n^n(K)$, such that $$V(K_r)=Vol(K_r)=\sum\limits_{i=0}^{n}{n \choose i}W^n_i(K)r^{i}.$$ It can be shown that $$W^n_0(K)=V(K),\quad W^n_1(K)=\frac{S(K)}{n},$$ where $S(K)$ is the surface area of $\partial K$. Moreover, $W^n_n(K)$ is equal to the volume $\pi_n$ of the unit ball in $\mathbb R^n$ and $$W^n_{n-1}(K)=\frac{\pi_n}{2}w(K),$$ where $w(K)$ is the mean width of $K$. Note that for $n=2$ the perimeter $P(K)$ equals $\pi w(K).$ The numbers $W_i^n(K)$ give some information on how the convex body $K$ is different from a ball (for the unit ball in $\mathbb R^n$, obviously $W^n_i(K)=\pi_n$ for all $i$.)

A convex body is completely determined by its support function $$h(x)=\sup\{x\cdot y|\ y\in K\}$$ which measures the directed distance of the origin to the tangent plane of $K$ at direction $x\in S^{n-1}.$ Now, the second proof of Hurwitz deals with the Fourier decomposition of the support function of a convex 2D domain. The problem is that in dimension $n>2$ the formulas for volume and surface area in terms of the support function cannot be expressed nicely by means of spherical harmonics. However, it is still possible to derive an isoperimetric inequality for the numbers $W^n_{n-2}$ and $W^n_{n-1}$ via harmonic analysis, namely $$W^n_{n-1}\geq\sqrt{\pi_n W^n_{n-2}}. \qquad\qquad(*)$$

When $n=2$ this is the standard isoperimetric inequality $P^2\geq 4\pi A$.

If $n=3$ $(*)$ gives the isoperimetric inequality between the mean width and surface area of a convex body $$\pi [w(K)]^2\geq S(K).$$

The proof is a straightforward extension of the second Hurwitz proof (using a decomposition of the support function into a series of spherical harmonics) and can be found here.