Chebyshev center and inscribed balls: properties and calculations

Abstract

The relationship between the Chebyshev center of a closed convex bounded subset and a center of an inscribed ball for some “dual” set (a ball of maximal radius which is contained in the set) is considered in a real Hilbert space. The inscribed ball is unique in the discussed situation. We present an approximate algorithm for calculation of the Chebyshev center (or the center of the inscribed ball for the “dual” set) for a convex compact subset from ${{\mathbb {R}}}^{n}$ which is given via its supporting function. We reduce the problem to the solution of a linear programming problem and estimate the error between an approximate and the exact solutions in terms of the step of a grid. Few numerical examples are considered. The considered algorithm is efficient in the space of small dimension n.

Appendix

Proposition 4.3, sketch of the proof

Without loss of generality, let $i(B)=0$. Then, by Lemma 4.3, $v\in i_M({\hat{B}})$ implies $v+r_B B_1(0)\subset B+B_{\varepsilon }(0)$. Put $D_B = \partial \,B\cap \partial \,B_{r_B}(0)$.

Let $b\in D_B$ and p (depending on b) be the unit normal vector to the set B (and $B_{r_B}(0)$) at the point b. Consider the dual R-strong convex set C with $C+B=B_R(0)$ and $c=\arg \max \limits _{x\in C}(p,x)$. Then $Rp = b+c = r_B p+c$, hence $c=(R-r_B)p = R_C p$. Consider $D_C = C\cap \partial \,B_{R_C}(0)$. By the Jung theorem ([20], Theorem 11.5.8), there exist unit vectors $\{p_i\}_{i=1}^{k}$, $k\le n+1$, and positive numbers $\lambda _{i}^{k}$ with $\sum \nolimits _{i=1}^{k}\lambda _i R_{C}p_i=0$ and $R_{C}p_{i}\in \partial \,C$. Consequently, for particular $b_i\in D_B$ with $(p_i,b_i)=s(p_i,B)$ (i.e. $p_i=b_i/r_B$), we have $\sum \limits _{i=1}^{k}\lambda _i b_i=0$. For any $v\in i_M({\hat{B}})$, by the equality $\sum \limits _{i=1}^{k}\lambda _i (v,b_i)=0$, we obtain that there exists $1\le i\le k$ with $(v,b_i)\ge 0$ or equivalently $(v,p_{i})\ge 0$. Using the supporting principle for R-strongly convex sets ([12], Corollary 2.1), we get $B\subset B_R(b_i-Rp_i)$ and

$$\begin{aligned} \begin{array}{l} b_i+v\in (B+B_{\varepsilon }(0))\cap \{ x\ :\ (p_i,x)\ge r_B\}\qquad \qquad \qquad \qquad \qquad \qquad \\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \subset (B_R(b_i-Rp_i)+B_{\varepsilon }(0))\cap \{ x\ :\ (p_i,x)\ge r_B\}. \end{array} \end{aligned}$$

Notice that $b_{i}=r_{B}p_{i}$, $(p_{i},b_{i})=r_{B}$ and hence $\{ x\ :\ (p_i,x)\ge r_B\}=\{ x\ :\ (p_i,x)\ge (p_{i},b_{i})\}$. So, $b_i+v$ belongs to the spherical segment of radius $R+\varepsilon$ and height $\varepsilon =\Vert b_{i} - (b_{i}+\varepsilon p_{i}) \Vert$. Thus, by the Pythagoras theorem, $\Vert v\Vert \le \sqrt{(R+\varepsilon )^2-R^2}=\sqrt{2R\varepsilon +\varepsilon ^2}$.

Now consider the case $B_{\delta }(i(B))\subset \mathrm{co\,}D_{B}$. Fix $\tau \in (0,1)$. Let $D_{\tau }=\{ d_{i}\}_{i=1}^{N}\subset D_{B}$ be such a finite set that

$$\begin{aligned} h\left( \mathrm{co\,}D_{\tau }, \mathrm{co\,}D_{B}\right) <(1-\tau )\delta . \end{aligned}$$

Then $B_{\tau \delta }(0)\subset \mathrm{co\,}D_{\tau }$. Note also that $\Vert d_{i}\Vert =r_{B}$ for all i.

Let $v\in i_M({\hat{B}})$ be a shift, by Lemma 4.3, $v+D_{\tau }\subset v+B_{r_B}(0)\subset B+B_{\varepsilon }(0)$. Suppose that $v\in \mathrm{cone\,}\{ d_{i_{1}},\dots ,d_{i_{k}}\}$ and moreover, $\mathrm{co\,}\{ d_{i_{1}},\dots ,d_{i_{k}}\}$ is an edge of $\mathrm{co\,}D_{\tau }$. Define the affine hull H of the points $\{ d_{i_{j}}\}_{j=1}^{k}$. We have $h=\varrho _{H}(0)\ge \tau \delta$. Define $e=P_{H}0$. The angle $\varphi _{0}$ between $d_{i_{j}}$ (for any $1\le j\le k$) and e

$$\begin{aligned} \cos \varphi _{0}=\frac{h}{r_{B}}\ge \frac{\tau \delta }{r_{B}}. \end{aligned}$$

Similar to the proof of Theorem 4.2 [17] one can check that there exists j, $1\le j\le k$, with the angle $\angle (d_{i_{j}},v)\le \varphi _{0}$. We denote for brevity $d_{0}=d_{i_{j}}$.

Consider a cone K with the vertex $d_{0}$, the axis $\{ td_{0}\ :\ t\ge 1\}$ and the angle between the axis and a generatrix equals $\varphi _{0}$. Let $H^{+}=\{ x\in {{\mathbb {R}}}^{n}\ :\ (d_{0},x-d_{0})\ge 0\}$. The hyperplane $\partial \,H^{+}$ is supporting for B at the point $d_{0}$, the hyperplane $\partial \,H^{+}+\varepsilon \frac{d_{0}}{r_{B}}$ is supporting for $B+B_{\varepsilon }(0)$ at the point $d_{0}+\varepsilon \frac{d_{0}}{r_{B}}$.

Considering the intersection of the cone K with the $\varepsilon$-layer

$$\begin{aligned} \left( (B+B_{\varepsilon }(0))\backslash int B\right) \bigcap K \subset \left( H^{+}\cap ({{\mathbb {R}}}^{n}\backslash int H^{+}+\varepsilon \frac{d_{0}}{r_{B}}) \right) \bigcap K, \end{aligned}$$

we obtain that the point $d_{0}+v$ lies between hyperplanes $\partial \,H^{+}$ and $\partial \,H^{+}+\varepsilon \frac{d_{0}}{r_{B}}$, the angle between v and $d_{0}$ is less than $\varphi _{0}$. Herefrom

$$\begin{aligned} \frac{\varepsilon }{\Vert v\Vert }\ge \cos \varphi _{0}\ge \frac{\tau \delta }{r_{B}}. \end{aligned}$$

Hence for any vector v with $v+B_{r_B}(0)\subset B+B_{\varepsilon }(0)$ we obtain that $\Vert v\Vert \le \varepsilon \frac{r_{B}}{\tau \delta }$. Thus $h(i_{M}({\hat{B}}),i(B))\le \varepsilon \frac{r_B}{\tau \delta }$ for all $\tau \in (0,1)$. Passing $\tau \rightarrow 1-0$, we complete the proof. $\square$