Abstract Generated abstract
This paper proves a quantitative stability result for Liebmann’s rigidity theorem for the sphere. For a convex n-dimensional surface whose Gaussian curvature satisfies 1 minus epsilon less than or equal to K less than or equal to 1 plus epsilon, the author establishes linear estimates for the radius of the largest inscribed ball and the radius of the smallest circumscribed ball, showing that both differ from 1 by at most a constant times epsilon, with the constant depending only on n. The proof develops comparison lemmas for convex surfaces of revolution, uses curvature and volume bounds, and applies Steiner symmetrization to reduce the general convex case to a rotationally symmetric one. The order of the stability estimate is shown to be sharp by comparison with spheres of constant curvature 1 minus epsilon and 1 plus epsilon.
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V. I. DISKANT
STABILITY IN LIEBMANN’S THEOREM
(Presented by Academician S. L. Sobolev, 4 V 1964)
1. A stability theorem corresponding to Liebmann’s theorem on the rigidity of the sphere was proved by A. I. Fet in (⁴). In the present work an exact estimate is given for the order of the stability function \(\varphi(\varepsilon)\) (see (⁴)) with respect to \(\varepsilon\). Namely, we prove
Theorem 1. If the Gaussian curvature \(K\) of a convex \(n\)-dimensional surface \(\Gamma\) satisfies the inequalities \(1-\varepsilon \leqslant K \leqslant 1+\varepsilon\), then for the radius \(r\) of the largest inscribed ball and for the radius \(R\) of the smallest circumscribed ball the estimates
\[ r \geqslant 1-C\varepsilon,\qquad R \leqslant 1+C\varepsilon, \]
hold, where \(C\) is a number depending only on \(n\), \(0 \leqslant \varepsilon \leqslant \tfrac12\).
The sharpness of the estimate follows from consideration of spheres of curvature \(1-\varepsilon\) and \(1+\varepsilon\).
2. We first note two lemmas for a surface of revolution.
Lemma 1. If the Gaussian curvature \(K\) of a convex surface \(\Sigma\), formed by the rotation of a curve \(y=y(x)\) \((y \geqslant 0)\) about the \(x\)-axis, satisfies the inequality \(K \geqslant a^n\), where \(a>0\), then \(\max y(x) \leqslant 1/a\).
Proof. Let \(\max y(x)=y(0)\). Suppose that \(y(0)>1/a\). Consider the curve
\[ x=x_a(y)= \begin{cases} \displaystyle \int_y^{y(0)} \frac{\bigl[a^n\bigl(y^n-y^n(0)\bigr)+1\bigr]^{1/n}} {\sqrt{1-\bigl[a^n\bigl(y^n-y^n(0)\bigr)+1\bigr]^{2/n}}}\,dy, & \text{if } y\ne y(0),\\[2ex] 0, & \text{if } y=y(0). \end{cases} \]
The surface of revolution of the curve \(x=x_a(y)\) about the \(x\)-axis has curvature \(K(x)\equiv a^n\). Applying Lemma 1 (³) to the curves \(x=x_a(y)\) and \(x=x(y)\), we obtain \(x'(\bar y)=0\), where \(\bar y=\sqrt[n]{\,y^n(0)-1/a^n\,}>0\). The latter is impossible.
Lemma 2. If the curve \(y=y(x)\) is defined on \([0,\eta]\) and satisfies the conditions
\[
y(x)\geqslant 0,\qquad y'(x)\leqslant 0,\qquad y''(x)<0,\qquad y'(\eta)=0,\qquad y(0)\leqslant 1/a,
\]
\[
K(x)\geqslant a^n,
\]
then all points of this curve belong to the figure \(G\), whose boundary consists of the curve \(x=x_a(y)\) \((0\leqslant y\leqslant y(0))\), the segments \([0,x_a(0)]\) of the \(x\)-axis, and \([0,y(0)]\) of the \(y\)-axis.
Lemma 2 is a direct consequence of Lemma 1 (³).
3. The assertion of Theorem 1 concerning the radius \(r\) follows from the following estimate.
Proposition A. If the Gaussian curvature \(K\) of a convex surface \(\Gamma\) satisfies the inequalities \(a^n \leqslant K \leqslant b^n\), then for the radius \(r\) of the largest inscribed ball the estimate
\[ r \geqslant A+\bigl(B^{n+1}-A^{n+1}\bigr)\, \frac{(n-1)\chi_{n+1}}{\chi_n} \left[ \frac{1}{A^n}+\frac{1}{B^n}\left(\frac{A}{B}\right)^{n^2+n+1} \right], \tag{1} \]
holds, where \(\chi_n\) is the volume of the \(n\)-dimensional unit ball, \(A=1/a\), \(B=1/b\).
The proof of Proposition A is based on the following lemma.
Lemma 3. Let the convex body \(T\) be bounded by a surface of revolution \(\Sigma\) about the \(x\)-axis. Suppose that the Gaussian curvature \(K\) of the surface \(\Sigma\) satisfies the inequality \(K \geq a^n\) and that the volume of the body \(V_T \geq \varkappa_{n+1}B^{n+1}\). Then, for the distance \(\rho(Z,X_0)\) between the center of gravity \(Z=(z,0,\ldots,0)\) of the body \(T\) and the point of intersection \(X_0=(x(0),0,\ldots,0)\) of the \(x\)-axis with the surface \(\Sigma\), the estimate
\[
\rho(Z,X_0)\geq A+(B^{n+1}-A^{n+1})\frac{\varkappa_{n+1}(n+1)}{\varkappa_n}
\left[\frac{1}{A^n}+\frac{1}{B^n}\left(\frac{A}{B}\right)^{n^2+n+1}\right].
\tag{2}
\]
Proof of Lemma 3. Let \(\Sigma\) be obtained by rotating the curve \(y=y(x)\) (\(y\geq 0\)). Suppose that \(y(0)=\max y(x)\). A suitable approximation allows us to assume that \(y=y(x)\) satisfies all the conditions of Lemma 1 and, for \(x\geq 0\), the conditions of Lemma 2, for, by Lemma 1, \(y(0)\leq A\). Consider in the \(xy\)-plane the figure \(M\), composed of the figure \(G\) and the figure symmetric to \(G\) with respect to the \(y\)-axis.
Introduce the following notation: \(T_M\) is the body obtained by rotating the figure \(M\) about the \(x\)-axis; \(V_M\) is the volume of \(T_M\); \(T_{\mathrm{r}}\) (\(T_{\mathrm{l}}\)) is the part of the body \(T\) lying in the half-space \(x\geq 0\) (\(x\leq 0\)); \(V_{\mathrm{r}}\) (\(V_{\mathrm{l}}\)) is the volume of \(T_{\mathrm{r}}\) (\(T_{\mathrm{l}}\)); \(T_{\mathrm{r}}^M\) (\(T_{\mathrm{l}}^M\)) is the part of the body \(T_M\) lying in the half-space \(x\geq 0\) (\(x\leq 0\)); \(T_{\mathrm{r}}^0=T_{\mathrm{r}}^M\setminus T_{\mathrm{r}}\), \(T_{\mathrm{l}}^0=T_{\mathrm{l}}^M\setminus T_{\mathrm{l}}\); \(V_{\mathrm{r}}^0\) (\(V_{\mathrm{l}}^0\)) is the volume of \(T_{\mathrm{r}}^0\) (\(T_{\mathrm{l}}^0\)); \(z_{\mathrm{r}}\) (\(z_{\mathrm{l}}\)) is the abscissa of the center of gravity of \(T_{\mathrm{r}}\) (\(T_{\mathrm{l}}\)); \(z_{\mathrm{r}}^M\) (\(z_{\mathrm{l}}^M\)) is the abscissa of the center of gravity of \(T_{\mathrm{r}}^M\) (\(T_{\mathrm{l}}^M\)); \(z_{\mathrm{r}}^0\) (\(z_{\mathrm{l}}^0\)) is the abscissa of the center of gravity of \(T_{\mathrm{r}}^0\) (\(T_{\mathrm{l}}^0\)).
Since \(\rho(Z,X_0)=x(0)-z=|x(0)-z|\geq x(0)-|z|\), it suffices to find the required estimates for \(x(0)\) and \(|z|\).
Let us estimate \(|z|\). By the additivity property of the center of gravity we have
\[
\frac{V_M}{2}z_{\mathrm{r}}^M=V_{\mathrm{r}}z_{\mathrm{r}}+V_{\mathrm{r}}^0z_{\mathrm{r}}^0
\quad
\left(\frac{V_M}{2}z_{\mathrm{l}}^M=V_{\mathrm{l}}z_{\mathrm{l}}+V_{\mathrm{l}}^0z_{\mathrm{l}}^0\right);
\tag{3}
\]
\[
V_Tz=V_{\mathrm{l}}z_{\mathrm{l}}+V_{\mathrm{r}}z_{\mathrm{r}}.
\tag{4}
\]
From (3) and (4) we find
\[
z=\frac{-V_{\mathrm{l}}^0z_{\mathrm{l}}^0-V_{\mathrm{r}}^0z_{\mathrm{r}}^0}{V_T};
\tag{5}
\]
from the condition of the lemma
\[
V_T\geq \varkappa_{n+1}B^{n+1}.
\tag{6}
\]
Applying Theorem 2 (?) to the sphere of radius \(A\) and to the surface of the body \(T_M\), we obtain
\[
V_T\leq V_M\leq \varkappa_{n+1}A^{n+1}.
\tag{7}
\]
From (6) and (7) we have
\[
V_{\mathrm{r}}\leq \frac{V_M}{2}\leq \frac12\varkappa_{n+1}A^{n+1}
\quad
\left(V_{\mathrm{l}}\leq \frac12\varkappa_{n+1}A^{n+1}\right);
\]
therefore
\[
V_{\mathrm{r}}=V_T-V_{\mathrm{l}}\geq \varkappa_{n+1}B^{n+1}-\frac12\varkappa_{n+1}A^{n+1}
\quad
\left(V_{\mathrm{l}}\geq \varkappa_{n+1}B^{n+1}-\frac12\varkappa_{n+1}A^{n+1}\right),
\]
whence
\[
V_{\mathrm{r}}^0\leq \varkappa_{n+1}(A^{n+1}-B^{n+1})
\quad
\left(V_{\mathrm{l}}^0\leq \varkappa_{n+1}(A^{n+1}-B^{n+1})\right).
\tag{8}
\]
To estimate \(z_{\mathrm{r}}^0\) (\(z_{\mathrm{l}}^0\)), note that \(z_{\mathrm{r}}^0\leq x_a(0)\). For \(x_a(0)\) the inequality
\[
\frac{1}{n+1}\varkappa_n y^n(0)x_a(0)\leq \frac{\varkappa_{n+1}A^{n+1}}{2},
\tag{9}
\]
is valid, in whose left-hand side stands the volume of the cone contained in \(T_{\mathrm{r}}^M\). Obviously, \(y(0)\geq \Delta/2\), where \(\Delta\) is the width of the body \(T\).
Replace in Theorem 3 \((^3)\) the estimate for \(V_S\) in formula (7) by the inequality
\[
V_S \ll 2V_C \ll \varkappa_{n+1} A^n x(0).
\tag{10}
\]
Then in Theorem 3 \((^3)\) for \(\Delta\) we obtain the estimate
\[
\Delta \gg 2B\left(\frac{B}{A}\right)^n .
\tag{11}
\]
Let us prove (10). Note that \(V_C\) is the volume of a segment of a ball of radius \(A\), \(x(0)\) \((x(0)\ll A)\) is the height of the segment. Denote by \(S\) the area of the base of the segment. Put \(V_C=hP\) and \(\frac12\varkappa_{n+1}A^{n+1}-V_C=h'Q\), where \(h\) is the height of the segment and \(h'=A-h\). Then \(P\ll S\ll Q\), whence (10) follows.
Using (10), we find from (9)
\[
z_{\Pi}^{0}\ll
\frac{(n+1)\varkappa_{n+1}}{2\varkappa_n}\,
B\left(\frac{A}{B}\right)^{n^2+n+1}.
\tag{12}
\]
By the method which we used in the proof of Theorem 3 \((^3)\), one can show that
\[
x(0)\gg A+
\frac{\varkappa_{n+1}(n+1)}{\varkappa_n A^n}
\left(B^{n+1}-A^{n+1}\right).
\tag{13}
\]
From inequalities (6), (8), (12), (13) the estimate (2) follows.
Proof of Proposition A. Denote by \(\Pi\) the center of gravity of the body bounded by \(\Gamma\). Let \(m=\min_L\rho(\Pi,L)\), and let \(L_m\) be a supporting plane for which \(\rho(\Pi,L_m)=m\). Suppose that the \(x\)-axis is the outer normal to \(L_m\). With the aid of Steiner symmetrizations and a limiting passage one can obtain from \(\Gamma\) a surface of revolution \(\Sigma\) about the \(x\)-axis. For the surface \(\Sigma\) the conditions of Lemma 3 are fulfilled. Indeed, \(V_\Sigma=V_\Gamma\), and by Theorem 3 \((^2)\) \(V_\Gamma\gg\varkappa_{n+1}B^{n+1}\). Moreover, by Theorem 4 \((^2)\) the Gaussian curvature of \(\Sigma\) is not less than \(a^n\). Denote by \(Z=(z,0,\ldots,0)\) the center of gravity of the body bounded by \(\Sigma\), and by \(L_\Sigma\) the supporting plane to \(\Sigma\) for which the \(x\)-axis is the outer normal. Then \(\rho(Z,L_\Sigma)=x(0)-z\). On the other hand, one can show that \(\rho(Z,L_\Sigma)=\rho(\Pi,L_m)\). Consequently, \(m=x(0)-z\). But \(r\gg m_1\), since otherwise a ball of radius \(\bar r>r\) could be inscribed in \(T\) \((^1,\) p. 24). Thus,
\[
r\gg x(0)-z.
\]
- The assertion of Theorem 1 concerning the radius \(R\) follows from the following estimate.
Proposition B. If the Gaussian curvature \(K\) of the convex surface \(\Gamma\) satisfies the inequalities \(a^n\ll K\ll b^n\), then for the radius \(R\) of the minimal circumscribed ball the estimate
\[
R \ll 2B-A+\frac{B\varkappa_{n+1}}{\varkappa_n}
\left[\left(\frac{A}{B}\right)^{(n+1)^2}-1\right]+
\]
\[
+\left(A^{n+1}-B^{n+1}\right)
\frac{(\chi+1)\varkappa_{n+1}}{\varkappa_n}
\left[
\frac{1}{A^n}+\frac{1}{B^n}
\left(\frac{A}{B}\right)^{n^2+n+1}
\right].
\]
Proof. If \(\bar R\) is the radius of the minimal circumscribed ball whose center coincides with the center of the maximal inscribed ball, then \(D\gg r+\bar R\) and \(R\ll\bar R\). Using estimate (1) for \(r\) and the estimate for \(D\) in Theorem 2 \((^3)\), we obtain the estimate in Proposition B.
The author expresses gratitude to A. I. Fet for posing the problem and for valuable guidance in the course of work on it.
Received
9 IX 1963
CITED LITERATURE
\(^{1}\) T. Bonnesen, W. Fenchel, Theorie der konvexen Körper, Berlin, 1934.
\(^{2}\) A. I. Fet, DAN, 153, No. 2 (1963).
\(^{3}\) V. I. Diskant, DAN, 153, No. 3 (1963).
\(^{4}\) A. I. Fet, DAN, 153, No. 3 (1963).