CONVEX CURVES OF BOUNDED TYPE

Let C be a simple closed convex curve in the plane for which the radius of curvature ρ is a continuous function of the arc length. Such a curve is called a convex curve of bounded type, if ρ lies between two fixed positive bounds. Here we give a new and simpler proof of Blaschke's Rolling Theorem. We prove one new theorem and suggest a number of open problems.

at every point of C. We let CV(RI,R2) denote the set of all such curves that satisfy (1.1) for fixed R and R 2.
Theorems about the class CV(RI,R2) appear in the literature (see for example Theorem 3), but as far as I am aware, this class has not been given a specific name and symbol until now.In this work we are concerned with one type of question, namely how close can C come to its "center" and how far away from its "center" can C go.
The center can be defined in various ways.For example the center of mass of the region bounded by C when the region has a uniform mass distribution.Or the center could be the center of mass of the curve C when the mass is distributed either uniformly or as some other function of s the arc length on C. In any case we can take the origin as the center of mass without loss of generality.CV(R I, R 2) and the center is the center of mass of the curve C. If the mass distribution on C is uniform, then R =< I) _<_ D 2 =< R 2 (1. 3) The two circles of radius R and R 2 show that the inequality (1.3) is sharp.
Bose and Roy [6] call this center the perimeter centroid.
in section we review some facts about parallel curves and we give a new proof of Blaschke's Roiling Theorem [3, pp. 114-116].In section 3 we prove Theorem 1.In section 4 we suggest some topics for further research on the set GV(R1,R2).

PARALLEL CURVES.
Let C e CV(R1,R2).We select the parameter s (arc length) so that s increases as the point P P(s) traverses C in the counterclockwise direction.Let 0 denote, as usual, the angle that the unit tangent T makes with the positive x-axis, and let N be the unit inward normal to C at the point P P(s).J-(cos 0)_i + (sin 0)j (2.2) and N (-sin 0)i + (cos 0)j. ( V(s) is the vector equation of C we introduce a second curve C* defined by and dy*= dy A< sin 0 (l-A<)sin 0. ( ds ds We let s*, *, and 0* denote, arc length, curvature, and radius of curvature at the corresponding point on C*.Then (2.5) and (2.6) give ds ] \ds ] the vector equation V* V(s) + AN, where A is a constant.The curve C* is said to be parallel to C, see [13   (2.4)If I/0 is the curvature of C at P, then < d0/ds and from (2.4)  CV(R I, R 2) and A < R I, then C a is locally convex and at corres- ponding points 0 n 0 A.
By locally convex we mean *' (s*) 0 at each point of Ca.
Proof.From Lemma we have * at corresponding points.(2.9) , Of course 0 n 0 A is geometrically obvious from the definition of Ca.Q.E.D.
If A R 2, the factor 1/(I-A/o) in (2.9) is replaced by 0/(A-0).Again C a is locally convex, but in this case 0* A 0.
It is geometrically obvious that if A R or A R2, then C* is a simple closed curve.
It seems that a direct proof is rather elusive.The difficulty may lie in the following example.Let C a be the image of Izl under the complex function f(z) z + z 2. Then C a is convex in the sense that * 0 at every point, so C* is locally convex.But this curve fails to be simple.Nevertheless we have  (2.12) Proof.We have already seen that C'is locally convex, but the example shows this is not sufficient to prove that C a is simple.On C a let L* A* *(e*) *(0) f ds*, Since C* is locally convex and A* 2, we see that C* is a simple curve.
If A R2, then * + .Hence (2.14) is still true and the proof remains valid.
Theorem 3. Let C CV(R1,R2) and let K be a circle tangent internally to C at any point PO of C. If K has radius R I, then K is contained in C. If K has radius R 2, then This theorem is often called Blaschke's Rolling Theorem, because it states that (a) a circle of radius R can roll around the inside of C, and (b) a circle of radius R 2 can roll around the outside of C. Blaschke has extended his theorem to 3-dimensional space [3, p. 118].For further work on this theorem, and various extensions see [II, 17,  20, and 22].
To be precise the phrase "internally tangent" means that K is tangent to C at PO and the center of K lies on the inward normal to C at P0" Thus the location of the center is given by equation set (2.4) with A replaced by R= the radius of the tangent circle (a 1,2).We say that K is contained in C if K is contained in the closure of the region bounded by C. Further K contains C, if C is in the closed disk bounded by K.
Proof of Theorem 3. We first show that the curve C cannot cross the circle K in a neighborhood of P0' the point of contact.Without loss of generality let PO be the origin and let K and C be tangent to the x-axis at the origin.Further suppose that both the circle and the curve lie above the x-axis, except at the origin.In this position the lower half of the circle will have equation (2.16) If y f(x) is the equation of C in a suitable neighborhood, I -e < x < e, then we have f'(x) sgn x 0 and f''(x) > 0 in I.
Lemma 3. Suppose that 0 R in I, where 0 is the radius of curvature on C.Then, under the conditions described above x g I.
Thus in I, the curve C cannot cross from outside to inside K, but of course C may coincide with K. We omit the proof of Lemma 3, but it follows directly from two inte- grations, starting with the inequality y''(x) By reversing the inequality signs we have Lemma 4. Suppose that 0 R in I. Then under the conditions on K and C des- cribed above y(x) J Y(x) R-/_X x e 1 Thus in I, the curve C cannot cross from inside to outside K, but of course C may coincide with K.
From these two lemmas we see that if R R or R R 2, then C cannot cross into or out of K in a neighborhood of a point of tangency.To complete the proof of Theorem 3, we must obtain this same result in the large.
First suppose that K has radius R and is tangent internally to C at PO" If K is not contained in C, then K crosses C at a point P2 distinct from PO" Then we may find a smaller circle K 0 with radius R 0 < R I, and such that K 0 is tangent internally to C at PO' and is tangent to C at another point PI' see Fig. 2.

Po
Figure 2 C* Now consider the parallel curve with A R 0 < R I. By Theorem 2, this curve is a simple close curve.On the other hand, the center D of the circle K 0 is at least a double point of C* because it is the corresponding point for both P0 and PI" Hence we have a contradiction.
For the second part of Theorem 3 let K be a circle with radius R 2 and tangent in- ternally to C at P0" If K does not contain C, then K crosses C at a point P2 distinct from PO" Then we may find a larger circle K 0 with radius R 0 > R 2 and such that K 0 is tangent internally to C at P0 and is tangent to C at another point PI" Again consider the parallel curve C* with A R 0 > R 2. By Theorem 2 this curve C* is a simple closed curve.Just as before we obtain a contradiction because D the center of K 0 is at least a double point on C*.Q.E.D.
Corollary I. Let L(C) denote the length of C and let A(C) denote the area of the region enclosed by C. If C E CV(RI,R2), then

19)
The circles of radius R and R 2 show that both of these inequalities are sharp.

PROOF OF THEOREM I.
Let C g CV(RI,R2) and let (s) be a mass distribution of C. We exclude the trivial case in which all of the mass is concentrated at one point.Then the center of mass will be an interior point of the region bounded by C. Without loss of generality we select the center of mass to be the origin.If L is the length of C, then  then this mass distribution will give Mx* * 0. By Theorem 2, the curve C* is a simple closed convex curve and the origin which is also the center of mass lles inside the region bounded by C*.
No let P be a point on C furthest from the origin.Then OP is normal to C at P.
If P* is the point on C* corresponding to P, then PP* is also normal to C at P. Hence the points P, O, and P* are collinear.
Finally we observe that by Lemma I, the directed tangents to C and C* at the points P and P* have opposite directions.Hence the origin is an interior point of the line segment PP*.Therefore, IOP < IPP*I A. Since A may be taken arbitrarily close to R 2, we have D 2 R 2. Q.E.D. 4. FURTHER QUESTION FOR STUDY.
We observe that the inequality R D D 2 R 2 is sharp for the circles of radius R and R 2. But in each extreme case does not vary throughout the interval [RI,R2] but instead is a constant at one end of the interval.The question naturally arises, can we find better bounds for D and D 2 p is a continuous function of s whose values fill out the interval [RI,R2].A first candidate for consideration is the ellipse x a cos t, y b sin t, 0 _< t _< 27, with 0 < b < a.If we set b (RI2R2)3, and a (RIR22) 3, a, so the expressions in (4.1) may appear as the proper lower and upper bounds for D. If true, this would improve the bounds R and R 2 given in Theorem I.However, by piecing together arcs of circles, we can show that no better bounds than R < D < D 2 < R 2 can be obtained.To see this, we give C only in the first quadrant and complete the curve by reflecting C respectively.The endpoints of the two arcs will meet when t T if we select a (R2-RI) cos T and b (R2-R I) sin T where 0 < R < R 2. If we compute the first deriva- tive for the two arcs at t T, they will not be equal, but the tangent vectors will be parallel, so that for this choice of a and b, the curve C C u C 2 is a smooth curve.
Further p R on C and 0 R 2 on C 2. Finally D R sin T + R2(l-sin T)and DI/R as T/ 7/2.Similarly D 2 R 2 cos T + R (l-cos T) and D2R2 as T/0.Thus no better bounds than D 2 R 2 and D R can be proved under the hypotheses stated.Of course p is no___t continuous in a neighborhood of P(T), where C and C 2 meet, but it is merely a matter of labor to alter the curve slightly at P(T) to make 0 continuous.
Perhaps some better bounds for D and D 2 can be obtained if we impose a further re- striction that the average values of 0 over the curve be a fixed number such as (RI+R2)/2.One can also examine the problem of finding sharp bounds for D and D 2 i the mass distribution has some fixed pattern, other than uniform.For example, Steiner [23], and [24, pp.99-159] has considered curves in which the mass distribution on C is pr6portional to the curvature at each point of C.More generally one can select the mass distribution to be some other function of I/0.One can also consider Theorem i, when the center of mass of C is replaced by the center of mass of the region enclosed by C. With this replacement, Theorem was proved earlier by Nikllborc [21] and Blaschke [2].It is reasonably clear that the center of mass of a curve C is in general different from the center of mass of the region en- closed by C, but it may be of interest to examine a particular example.
Let C be f(Izl=l) under f(z) z + az 2, where 0 < a < I/4.Then C is symmetric with respect to the x-axis and if the mass distribution is uniform on C then the center of mass will be on the x-axis.Hence it suffices to compute the x-coordinate.Let d and C denote this coordinate for the domain center of mass and the curve center of RC a(l-a It is clear that in general d # C" We may distinguish a third center of mass s, which we will call the conformal strip center.Suppose that f(z) maps E conformally onto D, with f(0) O. Set Rs(r,l) the x-coordinate of the center of mass of the strip bounded by the curves f(Izl=l) and f(IzI=r), where r I. Then by definition s lim s(r,l).In this case s # d unless a 0. Further it is clear that in general s # C" This example suggests the problem of finding max lj-kl (4.9) when C varies over the set CV(RI2) and j,k e {d,C,s}.
One can also investigate the properties of normalized univalent functions that map the unit disk conformally onto a region bounded by a curve in CV(RI,R2).Some elemen- tary results in this direction have been obtained by the author [9].
For each fixed curve in CV(RI,R2) then the curve C* may have cusps as shown in Fig. I. set ds*/ds Thus in either case s* and s increase together.In the first case, A < R I, we have dV* d* ds T* [(I-A)cos 0 i + (l-A)sin 0 j] I-A dq* ds ds* (cos )i + (sin )j T. (2.8)If A R 2, then the same type computation gives * -.Lemma I.If A R I, then the directed tangents at corresponding points of C and C* are parallel and point in the same direction.Further N_* .If A R 2, then T_* -T and N* -N.
Fig. shows a number of curves parallel to the ellipse 2/9 + y2/4 I.The curve C* is also a Bertrand mate of C, although the term Bertrand curve usually refers to twisted curves in space [4, p. 35].If P(x,y) is a point on C and P*(x*,y*) is the corresponding point on the parallel curve C

Theorem 2 .
If C E CV(RI,R2) and A R or A R 2, then C* is a simple closed convex curve.If A RI, then C a CV(RI*, R2*), where RI* R A, and R2* R 2 A.
is the length of C a. We make a change of variables from s* to s.If A < R I, consider the parallel curve C* where A R I, and let * *(s*) be a mass distrl- bution on C*.Then the moments Mx* and * are given by * /L* x* (s*) * (s*) ds* 0 and .L* MX* 0 y*(s*)*(s*)ds*.The change of variable from s* to s yelds (t (y+As) U* (s*(s))(l p--)ds.
specialize, by setting (s) on C and selecting B*(s*) so that u*(s*(s)) l-A/o(s) 0.(3.6)Then(3.4) and (3.5) give * I L (x-As)ds I L x ds A I dy 0, 1).Thus with the mass distribution(3.6), the center of mass of C* is also at the origin.Since C* is a simple closed convex curve, the origin lies inside C* and hence DA.Finally we note that A may be taken arbitrarily close to R E minlp for points on C. we consider the parallel curve C* where now A > R 2. For this curve equations (3.2)and (3.3) still hold.However, in this case we have equations(3.4)and(3.5)we must replace the factor I-A/O by A/O I.If we select(s)   on C and u* on C A so that (s*(s) out the interval [RI,R2].Further D b and D 2 easy computation shows that if f(z) z + az 2, 0 < a < I/4, and the mass distribu- "'(z)/'(z))'shows that for 0 < a < I/4, the function f(z) z + az 2 gives a convex curve for which the radius of curvature is pp. 80-84, 18 p. 67,and 19].