UDC 513

MATHEMATICS

P. S. SOLTAN

ILLUMINATION FROM WITHIN FOR UNBOUNDED CONVEX BODIES

(Presented by Academician P. S. Aleksandrov on 2 III 1970)

Let (K) be a convex body of the (n)-dimensional Euclidean space (E^n); let (\operatorname{bd} K) be its boundary and (\operatorname{int} K) its interior. A point (y \in \operatorname{bd} K) will be called illuminated from within by a point (x \in \operatorname{bd} K) if (x \ne y) and the interval ((x,y)) is contained in (\operatorname{int} K). A set (N \subset \operatorname{bd} K) will be called illuminated from within by a set (M \subset \operatorname{bd} K) if every point (y \in N) is illuminated from within by some point (x \in M). We shall assume throughout that (\operatorname{bd} K \ne \varnothing).

Theorem 1. The boundary (\operatorname{bd} K) of a convex unbounded body (K \subset E^n) is illuminated from within if and only if (K) is not a cone.

Proof. If (K) is a cone with vertex (a), then, evidently, the point (a) is not illuminated from within by any point (x \in \operatorname{bd} K).

Conversely, if the boundary (\operatorname{bd} K) of the body (K) is not illuminated from within, then there exists a point (z \in \operatorname{bd} K) that is not illuminated from within by any point (x \in \operatorname{bd} K). Let (y \in \operatorname{int} K). If the ray (zy) did not belong entirely to the body (K), then on this ray there would be found such a point (x \in \operatorname{bd} K) that (y \in (x,z)). But this would mean that the point (z) is illuminated from within by the point (x \in \operatorname{bd} K), contradicting the assumption. Thus the ray (zy) is contained in the body (K). Consequently, ({z} \cup \operatorname{int} K) is a cone, and therefore, passing to the closure, we find that (K) is a cone (with vertex at the point (z)).

Theorem 1 resolves the question under what condition the entire boundary (\operatorname{bd} K) of the body (K) is illuminated from within by some set (M \subset \operatorname{bd} K). The least cardinality of an illuminating set (M) will be denoted by (p(K)). It is known ((^3,^5)) that for every bounded convex body (K \subset E^n) the inequalities (2 \le p(K) \le n+1) hold. Furthermore ((^5)), for every natural (p) satisfying the inequalities (2 \le p \le n+1), there exists such a bounded convex body (K \subset E^n) that (p(K)=p), and the equality (p(K)=n+1) holds only for the (n)-dimensional simplex. In this note the indicated results are extended to the case of unbounded convex bodies.

A point (x \in \operatorname{bd} K) will be called regular if, first, through (x) there passes a unique supporting hyperplane (\Gamma_x) of the body (K), and, second, the point (x) is an interior point of the convex set (\Gamma_x \cap \operatorname{bd} K) (relative to the carrier plane of this set). If (x) is a regular point, then the closed convex set (F_x=\Gamma_x \cap \operatorname{bd} K) will be called a regular face.

Lemma 1. If (x) is a regular point and (y) is an interior point of the corresponding regular face (F_x) (relative to its carrier plane), then (y) is also a regular point and (F_y=F_x).

Lemma 2. The set (M) of all regular points is dense in the boundary of the body (K), i.e. (\overline{M}=\operatorname{bd} K).

For the proof of these lemmas see ((^2)), p. 110, and ((^1)), p. 24.

Lemma 3. An unbounded convex body (K \subset E^n) is a cone if and only if, for any (n) regular points (x_1,x_2,\ldots,x_n), the corresponding regular faces (F_{x_1},F_{x_2},\ldots,F_{x_n}) have a nonempty intersection.

In the proof only the sufficiency of this condition is needed.

First of all it is established that if the intersection of any (n) regular faces is nonempty, then the intersection of all regular faces is nonempty (for this one considers all possible intersections of regular faces with one of the planes (\Gamma_x), and applies Helly’s theorem ((^3)) to these intersections). Next it is shown (using Lemma 2) that if (a) is a common point of all regular faces, then (\operatorname{bd} K) is a cone with vertex (a). From this it follows without difficulty that (K) is a convex cone with vertex (a).

Theorem 2. For every unbounded convex body (K \subset E^n) distinct from a cone, the inequalities (2 \leq p(K) \leq n) hold.

Proof. Since the body (K) is distinct from a cone, by Lemma 3 there exist (n) regular points (x_1, x_2, \ldots, x_n \in \operatorname{bd} K) such that the corresponding regular faces (F_{x_1}, F_{x_2}, \ldots, F_{x_n}) have empty intersection. If a point (x \in \operatorname{bd} K) is not illuminated from within by some point (x_i) ((i=1,2,\ldots,n)), then the whole segment ([x_i,x]) lies on the boundary of the body (K) and, hence, in the hyperplane (\Gamma_{x_i}). In particular, (x \in \Gamma_{x_i}). Since, moreover, (x \in \operatorname{bd} K), it follows that (x \in \Gamma_{x_i}\cap \operatorname{bd} K = F_{x_i}). Thus, if a point (x \in \operatorname{bd} K) is not illuminated from within by the point (x_i), then (x \in F_{x_i}), and consequently (x) belongs to the intersection of the faces (F_{x_1}, F_{x_2}, \ldots, F_{x_n}). But since the sets (F_{x_1}, F_{x_2}, \ldots, F_{x_n}) have empty intersection, it follows that every point (x \in \operatorname{bd} K) is illuminated by at least one of the points (x_1, x_2, \ldots, x_n). Consequently, (p(K) \leq n).

The inequality (2 \leq p(K)) is obvious, since a single point (x \in \operatorname{bd} K) cannot illuminate from within the whole boundary (\operatorname{bd} K) of the body (K) (the point (x) itself will remain unilluminated from within). This completes the proof of Theorem 2.

Theorem 3. For every integer (p) satisfying the inequalities (2 \leq p \leq n), there exists an unbounded convex body (K \subset E^n) such that (p(K)=p).

Proof of Theorem 3 is carried out by the same methods as for bounded convex bodies (see ((^5))), but cylinders are considered instead of balls. Namely, if (L) is a convex set with supporting hyperplane (\Gamma \subset E^n) and if (a \notin \Gamma), then, as is not hard to prove, for the cone (aL) with base (L) and vertex (a) the relation
[
p(aL)=p(L)+1.
]
Therefore, if (C) is such a convex body of dimension (n-p+2) for which (p(C)=2), then, forming (p-2) times a cone over this body, we arrive at an (n)-dimensional body (K) satisfying the relation (p(K)=p) (here (p) may take the values (2,3,\ldots,n)). In note ((^5)), an ((n-p+2))-dimensional ball was used as (C). If, however, one takes as (C) a cylinder, i.e., the (r)-neighborhood of a straight line in ((n-p+2))-dimensional space, then this construction leads us to the required unbounded body. Thus Theorem 3 is proved.

Finally, let us note that an unbounded body (K \subset E^n) for which (p(K)) assumes its maximal possible value is no longer unique, as it was in the case of a bounded body. Indeed, if instead of (C) in the proof of Theorem 3 one considers some two-dimensional unbounded convex polygon having two unbounded parallel sides, then the body (K) to which we arrive, as is not hard to trace from the proof of Theorem 3, will satisfy the relation (p(K)=n). But there are infinitely many such polygons (C).

The author takes this opportunity to express his gratitude to Prof. V. G. Boltyanskii for his attention to the present work.

Kishinev State
University

Received
25 II 1970

REFERENCES

(^1) G. Busemann, Convex Surfaces, 1964.
(^2) N. Bourbaki, Topological Vector Spaces, 1959.
(^3) B. Grünbaum, Acta Math. Acad. Sci. Hung., 15 (1964).
(^4) L. Danzer, B. Grünbaum, V. Klee, Helly’s Theorem, Moscow, 1968.
(^5) P. S. Soltan, Mat. sbornik, new ser., 57 (99), 4 (1962).