Isoperimetric inequality, but $L_p$ norm

Question

I would like to consider the isoperimetric problem of $L_p$ norm:

Given a region in $\mathbb R^2$ such that the boundary is a curve $C(x,y)$, where $\int_{C}(|\mathrm dx|^p+|\mathrm dy|^p)^{1/p}$ is a constant. Find the curve that makes it to have a largest area.

Some example are here (in 2D we can WLOG the region is convex)

• If $p=1$, this means that $\int_C|\mathrm dx|+|\mathrm dy|$ is a constant. So, this means that the span of $y$-coordinate plus the span of $x$-coordinate is a constant. The area is no more than $(y_{\max}-y_{\min})\cdot(x_{\max}-x_{\min})$ and be maximized when these two are equal. Hence, the curve is $\max(|x|,|y|)=C$, or the infinity norm of $(|x|,|y|)=C$.

• If $p=\infty$, it is analogous (just skewed version) of $p=1$. So the corresponding curve is a rotated square, which is $|x|+|y|=C$, the $L_1$ norm is constant.

• If $p=2$ it is the regular isoperimetric problem. The optimal curve is a circle, which means the $L_2$ norm of $(x,y)$ is a constant.

So, I have two questions:

1. For $\mathbb R^2$ case, if $p>1$, is it correct that the best curve for maximizing the area given the $L_p$ norm is a curve like $|x|^q+|y|^q=C$ where $1/p+1/q=1$?

2. For general $\mathbb R^3,\mathbb R^4,\dots$ case, is it still true (especially for $p=1,2,\infty$ which is true for $\mathbb R^2$?)

Note: the thing fixed in the second question correspond to the $L_p$ norm of each normal. For example, for $\mathbb R^3$, let $n(x,y,z)=(n_x,n_y,n_z)$ be the normal vector. For $p=2$ case, $\iint_S \sqrt{n_x^2+n_y^2+n_z^2} \, \mathrm dS=\iint_S 1 \, \mathrm dS$ is its surface area, and for $p=1$ case, $\iint_S |n_x|+|n_y|+|n_z| \, \mathrm dS$ is the sum of the projection of the surface element.

The second question, even if $\mathbb R^3$ with $p=1,\infty$ turns out to be not as easy as in 2-D case, since we cannot assume easily that the body is convex (even if we only consider the case the body is star-shaped, i.e. the segment from some interior point (say, the origin) to all the points on its boundary is inside the body.)

Thanks a lot for the help!

• Related questions:

A combinatorial one (as it is close to $\infty$ norm case) https://math.stackexchange.com/questions/4436190/combinatorial-isoperimetric-inequality

Another one: https://math.stackexchange.com/questions/4907036/what-is-the-minimal-number-of-pieces-to-surround-n-pieces-in-a-go-game

The discrete version of the second question with $p=1$ case is an math olympiad problem (actually stronger, by AM-GM), 1992 IMO P5, see https://artofproblemsolving.com/community/c6h60719p366415

Answer 1

The answer to both questions is affirmative. I do not discuss below the smoothness conditions on curves/surfaces under which this all is true.

We start with dimension 2.

Let $\|\cdot\|$ be a norm on $\mathbb{R}^2$. Our goal is to find the centrally symmetric (with respect to the origin) convex set $\Phi\subset \mathbb{R}^2$ such that for any convex compact set $K\subset \mathbb{R}^2$, $\|\cdot\|$-perimeter $p_{\|\cdot\|}(K)$ of $K$ is given by the formula $$p_{\|\cdot\|}(K)=\lim_{t\to +0}\frac{|K+t\Phi|-|K|}{t},\tag{1}$$ where $|A|$ denotes the area of $A\subset \mathbb{R}^2$, and $A+B$ denores Minkowski sum of $A, B\subset \mathbb{R}^2$. Moreover, (1) should hold for non-convex set $K$ bounded by a smooth closed curve.

If we find such $\Phi$, then the solution of the $\|\cdot\|$-isoperimetric problem is given by homothetic images of $\Phi$ due to Brunn–Minkowski inequality: for fixed $|K|$ and fixed $t>0$, $|K+t\Phi|$ is minimal if $K$ is homothetic to $\Phi$ (with a positive coefficient).

It suffices to verify (1) for polygons (by some standard approximation mumbo-jumbo), and more specifically, what we really need is that for every line segment $[PQ]$ the half-area of $[PQ]+t\Phi$ equals $t\|Q-P\|+o(t)$ for small $t$. that means that we need $$\sup_{x\in \Phi} |x\times (Q-P)|=1,$$ where $\times$ stands for the cross product. If $\Phi_{\pi/2}$ denotes the set $\Phi$ rotated by $\pi/2$, we have $$\sup_{x\in \Phi} |x\times (Q-P)|=\sup_{y\in \Phi_{\pi/2}} |y\cdot (Q-P)|,$$ where $\cdot$ stands for the inner product. Thus, we need $\Phi_{\pi/2}$ to be the unit ball of the dual norm.

If the original norm is $\ell^p$-norm, we see that $\Phi$ is indeed $\ell^q$ unit ball for $1/p+1/q=1$, since $\ell^q$ is both the dual norm and $\pi/2$-rotation invariant.

Now about higher dimension.

Let $\|\cdot \|$ be a norm in $\mathbb{R}^d$, $\Phi$ the unit ball of the dual norm, $K\subset \mathbb{R}^d$ a body bounded by a closed surface $S$. Let $n(x)$ demote the outer (Euclidean)-unit normal vector at a point $x\in S$. I claim that $$\int_S \|n(x)\|dS(x)=\lim_{t\to +0}\frac{|K+t\Phi|-|K|}{t}. \tag{2}$$ For polyhedral $S$, (2) reduces to the following claim: for a hyperplane $\alpha\ni 0$ with unit normal $n$, $$\sup_{x\in \Phi} d(x, \alpha)=\|n\|,$$ where $d(x,\alpha)$ denotes the distance from $x$ to $\alpha$. This immediately follows from $d(x, \alpha)=|x\cdot n|$ and the very definition of the dual norm.