Nonsymmetric extension of the Green-Osher inequality

TL;DR

This paper extends the Green-Osher inequality to a nonsymmetric setting for smooth, convex planar bodies, establishing conditions for equality and broadening its applicability.

Contribution

It introduces a nonsymmetric extension of the Green-Osher inequality with necessary and sufficient conditions for equality in the planar convex case.

Findings

Extended Green-Osher inequality to nonsymmetric convex bodies

Established necessary and sufficient conditions for equality

Applied to smooth, planar strictly convex bodies

Abstract

In this paper we obtain the extended Green-Osher inequality when two smooth, planar strictly convex bodies are at a dilation position and show the necessary and sufficient condition for the case of equality.

Equations

$$K+L=\{x+y\mid x\in K,y\in L\}$$

$$tK=\{tx\mid x\in K\}.$$

$$V(K+tL)=V(K)+2V(K,L)t+V(L)t^2,$$

$$\frac{1}{V(E)}{\int }_0^{2\pi }F(\rho (\theta ))h_E(\theta )(h_E(\theta )+h_E^{\prime \prime }(\theta ))\mathrm{d}\theta \ge F(-t_1)+F(-t_2),$$

$$r(K,L)L\subseteq K\subseteq R(K,L)L.$$

$$r(K,L)=\max \{t>0\mid x+tL\subseteq K\text{and}x\in {\mathbb{R}}^n\},$$

$$R(K,L)=\max \{t>0\mid x+tL\supseteq K\text{and}x\in {\mathbb{R}}^n\}$$

$$\frac{1}{V(L)}{\int }_0^{2\pi }F(\rho (\theta ))h_L(\theta )(h_L(\theta )+h_L^{\prime \prime }(\theta ))\mathrm{d}\theta \ge F(-t_1)+F(-t_2),$$

$$h_K(\theta )=\underset{x\in K}{\sup }⟨x,\mathbf{u}(\theta )⟩,\mathbf{u}(\theta )\in S^1.$$

$$V(K,L)=\frac{1}{2}{\int }_0^{2\pi }(h_K(\theta )h_L(\theta )-h_K^{\prime }(\theta )h_L^{\prime }(\theta ))\mathrm{d}\theta .$$

$$V(K,L)=\frac{1}{2}{\int }_0^{2\pi }{h}_K(\theta )(h_L(\theta )+h_L^{\prime \prime }(\theta ))\mathrm{d}\theta =\frac{1}{2}{\int }_0^{2\pi }{h}_L(\theta )(h_K(\theta )+h_K^{\prime \prime }(\theta ))\mathrm{d}\theta .$$

$$t_1=-\frac{V(K,L)}{V(L)}+\frac{\delta }{V(L)},t_2=-\frac{V(K,L)}{V(L)}-\frac{\delta }{V(L)},\delta =\sqrt{V(K,L{)}^2-V(K)V(L)}.$$

$$\sup \left\{{\int }_I\rho (\theta )h_L(\theta )(h_L(\theta )+h_L^{\prime \prime }(\theta ))\mathrm{d}\theta \mid I\subseteq S^1,{\int }_I{h}_L(\theta )(h_L(\theta )+h_L^{\prime \prime }(\theta ))\mathrm{d}\theta =V(L)\right\}.$$

$$I_1\subseteq \{\theta \mid \rho (\theta )\ge a\},I_2\subseteq \{\theta \mid \rho (\theta )\le a\}.$$

$${\rho }_i=\frac{1}{V(L)}{\int }_{I_i}\rho (\theta )h_L(\theta )(h_L(\theta )+h_L^{\prime \prime }(\theta ))\mathrm{d}\theta ,i=1,2,$$

$${\rho }_1+{\rho }_2=\frac{2V(K,L)}{V(L)}\text{and}{\rho }_1\ge {\rho }_2,$$

$${\rho }_1=\frac{V(K,L)}{V(L)}+b\text{and}{\rho }_2=\frac{V(K,L)}{V(L)}-b.$$

$${\rho }_1\ge -t_2.$$

$$-t_1\le r(K,L)\le R(K,L)\le -t_2.$$

$$-\frac{\delta }{V(L)}{h}_L(\theta )\le h_K(\theta )-\frac{V(K,L)}{V(L)}{h}_L(\theta )\le \frac{\delta }{V(L)}{h}_L(\theta ),\delta =\sqrt{V(K,L{)}^2-V(K)V(L)}\ge 0.$$

$$-\left(h_K(\theta )-\frac{V(K,L)}{V(L)}{h}_L(\theta )\right)(\rho (\theta )-a)\le \frac{\delta }{V(L)}{h}_L(\theta )(\rho (\theta )-a).$$

$$-\frac{1}{V(L)}{\int }_{I_1}\left(h_K(\theta )-\frac{V(K,L)}{V(L)}{h}_L(\theta )\right)(\rho (\theta )-a)(h_L(\theta )+h_L^{\prime \prime }(\theta ))\mathrm{d}\theta \le \frac{\delta }{V(L)}({\rho }_1-a).$$

$$-\frac{1}{V(L)}{\int }_{I_2}\left(h_K(\theta )-\frac{V(K,L)}{V(L)}{h}_L(\theta )\right)(\rho (\theta )-a)(h_L(\theta )+h_L^{\prime \prime }(\theta ))\mathrm{d}\theta \le -\frac{\delta }{V(L)}({\rho }_2-a).$$

$$-\frac{1}{V(L)}{\int }_0^{2\pi }\left(h_K(\theta )-\frac{V(K,L)}{V(L)}{h}_L(\theta )\right)(\rho (\theta )-a)(h_L(\theta )+h_L^{\prime \prime }(\theta ))\mathrm{d}\theta \le \frac{2b\delta }{V(L)}$$

$${\rho }_1>-t_2.$$

$$-t_1<r(K,L)<R(K,L)<-t_2.$$

$$-\frac{\delta }{V(L)}{h}_L(\theta )<h_K(\theta )-\frac{V(K,L)}{V(L)}{h}_L(\theta )<\frac{\delta }{V(L)}{h}_L(\theta ),\delta =\sqrt{V(K,L{)}^2-V(K)V(L)}>0.$$

$$-\left(h_K(\theta )-\frac{V(K,L)}{V(L)}{h}_L(\theta )\right)(\rho (\theta )-a)<\frac{\delta }{V(L)}{h}_L(\theta )(\rho (\theta )-a).$$

$$-\frac{1}{V(L)}{\int }_{I_1}\left(h_K(\theta )-\frac{V(K,L)}{V(L)}{h}_L(\theta )\right)(\rho (\theta )-a)(h_L(\theta )+h_L^{\prime \prime }(\theta ))\mathrm{d}\theta <\frac{\delta }{V(L)}({\rho }_1-a),$$

$$-\frac{1}{V(L)}{\int }_0^{2\pi }\left(h_K(\theta )-\frac{V(K,L)}{V(L)}{h}_L(\theta )\right)(\rho (\theta )-a)(h_L(\theta )+h_L^{\prime \prime }(\theta ))\mathrm{d}\theta <\frac{2b\delta }{V(L)}.$$

$$-\frac{\delta }{V(L)}{h}_L(\theta )<h_K(\theta )-\frac{V(K,L)}{V(L)}{h}_L(\theta )<\frac{\delta }{V(L)}{h}_L(\theta )$$

Full text

Nonsymmetric extension of the Green-Osher inequality

111The author is supported by the Doctoral Scientific Research Foundation of Liaoning Province (No.20170520382) and the Fundamental Research Funds for the Central Universities (No.3132017046).

**Yunlong Yang

College of Science,  Dalian Maritime University,

Dalian, 116026, P. R. China

email: ylyang@dlmu.edu.cn

**

Abstract  In this paper we obtain the extended Green-Osher inequality when two smooth, planar strictly convex bodies are at a dilation position and show the necessary and sufficient condition for the case of equality.

Mathematics Subject Classification 2010: 52A40,  52A10

Key words:  dilation position,  Green-Osher’s inequality,  nonsymmetric,  relative Steiner polynomial

1 Introduction

We denote by $\mathbb{R}^{n}$ the usual $n$-dimensional Euclidean space with the canonical inner product $\langle\cdot,\cdot\rangle$. A compact convex set $K$ in $\mathbb{R}^{n}$ is called a convex body if it contains the origin and has nonempty interior. When $n=2$, it is called a planar convex body. The volume of a set $S\subseteq\mathbb{R}^{n}$ is denoted by $V(S)$. The Minkowski sum of convex bodies $K$ and $L$, and the Minkowski scalar product of $K$ for positive real number $t$ are, respectively, defined by

[TABLE]

and

[TABLE]

For two planar convex bodies $K$ and $L$, the volume of the Minkowski sum $K+tL$ gives the relative Steiner polynomial of $K$ with respect to $L$:

[TABLE]

where $V(K,L)$ is the mixed area of $K$ and $L$. Formula (1.1) is closely related to the classical isoperimetric inequality, the Brunn-Minkowski inequality and the log-Brunn-Minkowski inequality. Many proofs, sharpened forms and generalization of the isoperimetric inequality can be found in Chavel [2], Dergiades [3], Osserman [9] and Schneider [10].

Using remarkable symmetrization, Gage [4] successfully obtained an inequality for the total squared curvature for convex curves. Following his work, for a planar strictly convex body $K$ and a symmetric, planar strictly convex body $E$, Green and Osher [8] (see also [12]) obtained a generalized formula:

[TABLE]

where $\rho(\theta)$ is the relative curvature radius of $K$ with respect to $E$, $F(x)$ is a strictly convex function on $(0,+\infty)$, $t_{1}$ and $t_{2}$ are the two roots of the relative Steiner polynomial of $K$ with respect to $E$. Inequality (1.2) plays a significant role in studying the curve shortening flow (see Gage [5, 6] and Gage-Hamilton [7]).

A natural question is whether the Green-Osher inequality holds without symmetric condition. Similar question is asked by the log-Brunn-Minkowski inequality (see Böröczky-Lutwak-Yang-Zhang [1], Xi-Leng [11] and Yang-Zhang [13]). Xi and Leng [11] gave the definition of dilation position for the first time to prove the log-Brunn-Minkowski inequality and solve the planar Dar’s conjecture.

Let $K$ and $L$ be two convex bodies. Convex bodies $K$ and $L$ are at a dilation position, if the origin $o\in K\cap L$ and

[TABLE]

Here $r(K,L)$ and $R(K,L)$ are the inradius and outradius of $K$ with respect to $L$, i.e.,

[TABLE]

Noticing that there is a common center when $K$ and $L$ are at a dilation position, then the ratio of the support functions of $K$ and $L$ belongs to the range from $r(K,L)$ to $R(K,L)$, which leads to the Green-Osher inequality holds without symmetric condition. Properties of convex bodies are at a dilation position can be found in Lemma 3.1 (see also Xi-Leng [11]).

In this paper, inspired by the impressive work in [11], we obtain the main result.

Theorem 1.1**.**

Let $K,L$ be two smooth, planar strictly convex bodies and $\rho(\theta)$ the relative curvature radius of $K$ with respect to $L$. If $K$ and $L$ are at a dilation position and $F(x)$ is a strictly convex function on $(0,+\infty)$, then

[TABLE]

where $t_{1}$ and $t_{2}$ are the two roots of the relative Steiner polynomial of $K$ with respect to $L$, and the equality in (1.4) holds if and only if $K$ and $L$ are homothetic.

This paper is organized as follows. In Section 2, we give some basic facts about planar convex bodies. In Section 3, we get the extended Green-Osher inequality when two smooth, planar strictly convex bodies are at a dilation position.

2 Preliminaries

Let $K$ be a planar convex body. A line $l$ is called a support line of $K$ if it passes through at least one boundary point of $K$ and if the entire planar convex body $K$ lies on one side of $l$. Let $l(\theta)$ be the support line of $K$ in the direction $\mathbf{u}(\theta)=(\cos\theta,\sin\theta)$, where $\theta$ is the oriented angle from the positive $x$-axis to the perpendicular line of $l(\theta)$. The support function of $K$ is defined by

[TABLE]

It is easy to see that $h_{K}(\theta)$ is the signed distance of the support line $l(\theta)$ of $K$ with exterior normal vector $\mathbf{u}(\theta)$ from the origin. Clearly, $h_{K}$, as a function of $\theta$, is single-valued and $2\pi$-periodic.

If $h_{K}(\theta)$ and $h_{L}(\theta)$ are continuously differentiable, then

[TABLE]

Furthermore, if $h_{K}(\theta)$ and $h_{L}(\theta)$ are smooth, then

[TABLE]

From the Minkowski inequality, it follows that the expression $V(K+tL)=0$ has two negative real roots. Denote by $t_{1}$ and $t_{2}$ ($t_{1}\geq t_{2}$) the two roots of the relative Steiner polynomial of $K$ with respect to $L$, that is,

[TABLE]

In order to prove the extended Green-Osher inequality, we have the following definition that is similar to the Definition 3.3 of [8].

Definition 2.1** ([8]).**

Let $K,\,L$ be two smooth, planar strictly convex bodies. Consider

[TABLE]

Let $I_{1}$ denote the smallest subset of $S^{1}$ with measure $V(L)$ and realizing the above supremum, and let $I_{2}$ be its complement. Then, there exists an $a\in\mathbb{R}^{+}$ such that

[TABLE]

Set

[TABLE]

which yield that

[TABLE]

and there is a real number $b\geq 0$ such that

[TABLE]

3 Nonsymmetric extension of the Green-Osher inequality

In order to prove the main result, we first give four lemmas, in which Lemma 3.1 shows that convex bodies are at a dilation position by appropriate translations and the location of “dilation position” (detailed proof can be found in [11, Lemma 2.1]), Lemmas 3.2 and 3.3 are used to prove inequality (1.4), and Lemma 3.4 is used to deal with its equality case.

Lemma 3.1** ([11]).**

Let $K,L$ be two convex bodies in $\mathbb{R}^{n}$.

• (i)

There are a translate of $L$, say $\bar{L}$, and a translate of $K$, say $\bar{K}$, so that $\bar{K}$ and $\bar{L}$ are at a dilation position.

• (ii)

If $K$ and $L$ are at a dilation position, then the origin $o\in\mathrm{int}(K\cap L)\cup(\partial K\cap\partial L)$.

Lemma 3.2**.**

Let $K,L$ be two smooth, planar strictly convex bodies. If $K$ and $L$ are at a dilation position, then the origin $o\in\mathrm{int}(K\cap L)$ or $o$ is the point of tangency of $\partial K$ and $\partial L$ such that $K\subseteq L$ (or $L\subseteq K$).

Proof.

By Lemma 3.1(ii), the origin $o\in\mathrm{int}(K\cap L)\cup(\partial K\cap\partial L)$. If the origin $o\in\mathrm{int}(K\cap L)$, we are done. If the origin $o\in\partial K\cap\partial L$, then $o$ must be the point of tangency of $\partial K$ and $\partial L$ such that $K\subseteq L$ (or $L\subseteq K$). Otherwise, $o$ is the point of intersection of $\partial K$ and $\partial L$, which contradicts to (1.3). ∎

Lemma 3.3**.**

Let $K,L$ be two smooth, planar strictly convex bodies. If $K$ and $L$ are at a dilation position, then

[TABLE]

Proof.

From [1, Lemma 4.1] and the Minkowski inequality, it follows that

[TABLE]

By Lemma 3.2, the origin $o\in\mathrm{int}(K\cap L)$ or $o$ is the point of tangency of $\partial K$ and $\partial L$ such that $K\subseteq L$ (or $L\subseteq K$).

If the origin $o\in\mathrm{int}(K\cap L)$, then $r(K,L)\leq\frac{h_{K}(\theta)}{h_{L}(\theta)}\leq R(K,L)$, which implies

[TABLE]

On $I_{1}$, $\rho(\theta)-a\geq 0$, combining with the above inequality, it yields

[TABLE]

By integrating this on the interval $I_{1}$,

[TABLE]

Similarly, on $I_{2}$, we have

[TABLE]

It can be seen from (3.2) and (3.3) that

[TABLE]

and its left-hand side can be simplified to $\frac{2\delta^{2}}{V(L)^{2}}$, thus we have, $b\geq\frac{\delta}{V(L)}\geq 0$, that is, $\rho_{1}\geq-t_{2}$.

If the origin $o$ is the point of tangency of $\partial K$ and $\partial L$ such that $L\subseteq K$ (the case of $K\subseteq L$ is similar), then $r(K,L)\leq\frac{h_{K}(\theta)}{h_{L}(\theta)}\leq R(K,L)$ for $\theta\in\tilde{I}$ ($\tilde{I}$ is a subset of $S^{1}$) and $h_{K}(\theta)=h_{L}(\theta)=0$ for $\theta\in S^{1}\setminus\tilde{I}$. A similar discussion implies that $\rho_{1}\geq-t_{2}$. ∎

Lemma 3.4**.**

Let $K,L$ be two smooth, planar strictly convex bodies. If $K$ and $L$ are at a dilation position but not homothetic, then

[TABLE]

Proof.

Since $K$ and $L$ are not homothetic, by [1, Lemma 4.1] and the fact that $K$ and $L$ are smooth and strictly convex,

[TABLE]

By Lemma 3.2, the origin $o\in\mathrm{int}(K\cap L)$ or $o$ is the point of tangency of $\partial K$ and $\partial L$ such that $K\subseteq L$ (or $L\subseteq K$).

If the origin $o\in\mathrm{int}(K\cap L)$, then

[TABLE]

For $I_{1}$ and $I_{2}$, $\rho(\theta)\equiv a$ holds on at most one interval, unless $K$ and $L$ are homothetic. Without loss of generality, assume that $\rho(\theta)>a$ on a subinterval $I_{1}^{\prime}$ of $I_{1}$. On $I_{1}^{\prime}$, $\rho(\theta)>a$ and

[TABLE]

Integrating this expression over the interval $I_{1}$ yields

[TABLE]

which, together with (3.3), gives

[TABLE]

By a similar argument as in Lemma 3.3, $b>\frac{\delta}{V(L)}>0$, which implies that $\rho_{1}>-t_{2}$.

If the origin $o$ is the point of tangency of $\partial K$ and $\partial L$ such that $L\subseteq K$ (the case of $K\subseteq L$ is similar), then

[TABLE]

for $\theta\in\tilde{I}$. Similar with the case that the origin $o\in\mathrm{int}(K\cap L)$, one can get $\rho_{1}>-t_{2}$. ∎

Now, we give the proof of Theorem 1.1.

Proof of Theorem 1.1 By Jensen’s inequality on $I_{i}$, $i=1,2$, one has

[TABLE]

Then

[TABLE]

where $\rho_{1}=\frac{V(K,L)}{V(L)}+b$, $\rho_{2}=\frac{V(K,L)}{V(L)}-b$ and $b\geq 0$. Again from (3.1), it follows that $b\geq\frac{\delta}{V(L)}\geq 0$ and $\delta=\sqrt{V(K,L)^{2}-V(K)V(L)}$. Since function $F(x)$ is strict convexity,

[TABLE]

which together with (3.5) yields inequality (1.4).

On one hand, if $K$ and $L$ are homothetic, then $-t_{1}=-t_{2}=\rho(\theta)$, it is clear that the equality holds in (1.4). On the other hand, in order to prove that $K$ and $L$ are homothetic when the equality holds in (1.4), it is enough to show that inequality (1.4) is strict when $K$ and $L$ are not homothetic. If $K$ and $L$ are not homothetic, then $\delta=\sqrt{V(K,L)^{2}-V(K)V(L)}>0$, and by (3.4), one has $b>\frac{\delta}{V(L)}>0$. It follows from the strict convexity of function $F(x)$ that (3.6) is strict, which together with (3.5) implies that (1.4) is strict.∎

Remark 3.5**.**

If $\mathbb{R}^{2}$ is equipped with a suitable Minkowski metric such that $\partial L$ becomes the isoperimetrix of the Minkowski plane, then (1.4) turns into an inequality in Minkowski geometry (see [12, Remark 3.6]).

Acknowledgements

I am grateful to the anonymous referee for his or her careful reading of the original manuscript of this paper and giving us many invaluable comments. I would also like to thank Professor Shengliang Pan for posing this problem to me.