Isoperimetric deficit and convex plane sets of maximum translative discrepancy

STEFANO CAMPI

I S O P E R I M E T R I C D E F I C I T A N D C O N V E X P L A N E S E T S

O F M A X I M U M T R A N S L A T I V E D I S C R E P A N C Y *

ABSTRACT. The paper deals with the following question: Among the convex plane sets of fixed isoperimetric deficit, which are the sets of maximum translative deviation from the circular shape? The answer is given for the cases in which the deviation is measured either by the translative Hausdorff metric or by the translative symmetric difference metric.

AMS SUBJECT CLASSIFICATION: 52 A40

It is well known that, for a simple closed plane curve of length I enclosing an area a, the quantity l z - 4ha is a nonnegative number that vanishes if and only if the curve is a circle.

A natural question arises: What is the largest possible deviation from the circular shape for a curve whose isoperimetric deficit l 2 - 4 h a does not exceed a fixed positive number? Of course, the answer depends on how we choose to measure such a deviation.

Meaningful definitions of the discrepancy measure from a circle are suggested by the so-called Bonnesen-style inequalities involving the isoperi-

metric deficit; for a complete description of such classical results one can refer

to the textbook of Bonnesen and Fenchel I-4] and to the surveys of Osserman [7] and Groemer I-6].

We choose to measure this deviation with a translative distance, that is the minimal distance (in some standard metric) between the curve and a circle enclosing the same area when the latter is free to move in the plane. Precisely, the problem we deal with in the present paper is the following: Among all the convex plane domains of fixed area enclosed by a curve whose isoperimetric deficit does not exceed a fixed constant, find that one at maximum distance from the disk of the same area. The distance we shall use here is either the translative Hausdorff distance or the translative version of the symmetric difference distance (i.e. the area of such a difference), both of interest for their immediate geometrical meaning. The case of the Hausdorff metric will be discussed in Section 1, where we shall show that the problem can be solved by

using classical results of J. Favard and T. Bonnesen. Section 2 is devoted to a discussion of the solution of the problem for the other distance, which does not appear to be reducible to such classical results.

*Work supported by Istituto di Analisi Globale ed Applicazioni of C.N.R. Firenze.

Geometriae Dedicata 43: 71-81, 1992. © 1992 Kluwer Academic Publishers. Printed in the Netherlands.

72 S T E F A N O C A M P I

It should be pointed out that for the case of the translative L2-metric a similar problem was solved by using Fourier analysis; for a complete description of the solution see [6]. Our approach here is somewhat different. Owing to the special meaning of the distances we are considering, the tools that we use will be mainly geometric.

For simplicity we shall assume throughout the paper that the fixed value for the area of the sets involved is normalized to one. With this normalization

let us introduce, for any a > 2V/~ -, the class F~ of all plane convex sets of unit area and perimeter I ~ a; in other words, F~ consists of all convex sets of unit area whose isoperimetric deficit does not exceed a 2 4n. In F~ two sets each obtained from the other by a rigid motion will be considered identical.

.

For any E e F~, let us define the outer translative Hausdorffdistance of E from the disk of unit area in the following way:

h*(E) = m i n [ m a x ( m i n l x - - y l ) l , teR 2 I xeE \ yeS ( t )

where S(t) denotes the disk of unit area centered at t. The number h*(E) (equivalent to the girth coefficient defined in [5]) measures how much E protrudes out of the disk of unit area when the latter is set in the minimizing position. It is immediate that such a minimizing position exists. To see that it is also unique, one can start from the disk in a minimizing position, blow up the disk keeping fixed the center and stop as soon as it contains E. The final disk is just the circumscribing disk for E; otherwise the starting position would be not minimizing. Thus, the uniqueness of the circumscribing circle implies the uniqueness of the minimizing position. Moreover, by the same argument we infer also that

(*) h*(e) = R--p ,

where R is the circumradius of E and p the radius of the disk of unit area, i.e.

p = 1/x//-~. By (*) the set in F~ maximizing h* is that one whose circumradius is as large as possible. Thus, we have reduced our problem to one solved by J. Favard, and we can state the result as follows:

T H E O R E M 1. There exists maxeero h*(E) and the unique maximizing set is the symmetrical lens (i.e. the convex set bounded by two circular arcs of the same radius) whose perimeter equals a.

Proof In [3], Favard has shown that the smallest disk which can contain

SETS OF M A X I M U M T R A N S L A T I V E D I S C R E P A N C Y 73

every plane convex set of fixed area and fixed perimeter is the disk circumscribed about the symmetrical lens of that area and that perimeter. The theorem immediately follows from this result, by taking into account that, for a fixed value of the area, the diameter of the lens increases as the perimeter increases.

It should be pointed out that Besicovitch [2] has given an alternative proof of Favard's result, by showing that the Favard problem is equivalent to finding, among the convex sets of given perimeter and given circumradius, the one of maximum area, and that the lens is just the unique solution of the latter problem.

Let us consider now the inner translative Hausdorffdistance of E E F~ from the disk of unit area:

h ( E )= min Fmax (min ,x-- y,) ] , t~R z Lx~S(t) \ y~E / A

which measures how much E intrudes into the disk when the latter is in a minimizing position (see the content coefficient of [5]). By an argument analogous to that used for the outer distance, one can see that

~(E) = p - r,

where r is the inradius of E. It has been shown by Bonnesen (see [3], [4]) and later by Besicovitch [1] that among the convex sets of given perimeter and given area that one having the minimum inradius is the biscuit of that perimeter and that area. Throughout this paper, the term 'biscuit' is used to refer to a figure formed by capping the opposite ends of a rectangle by two semicircles (the shape of such a set reminds one of a baby's teething biscuit). It is easy to check that, for a fixed value of the area, the inradius of the biscuit decreases as the perimeter increases. Therefore, the following result follows immediately from Bonnesen's theorem:

T H E O R E M 2. There exists maxEero h(E) and the unique maximizing set is the biscuit whose perimeter equals a.

Finally let us remark that, if B~ denotes the biscuit in F , whose perimeter equals o-, it is easy to check that

h(B~) < h*(B~),

for every a. This fact implies that for the complete translative Hausdorff distance, i.e. the maximum between h*(E) and h(E), the solution of our problem is the lens whose perimeter is a.


.

The question we deal with in this section is the same as in the previous one, but here, instead of using Hausdorff distances, we choose to measure the discrepancy of E from the disk of unit area by the following distance:

q~(E) = min [area (EkS(t))]. t~R 2

This minimum exists because g( t ) - -a rea (EkS(t)) is a continuous function,

constant for it[ sufficiently large. We point out that, for every t, 29( 0 measures the distance between E and S(t) in the so-called symmetric difference metric; thus q~(E) can be considered as the translative symmetric difference distance from the disk of unit area.

By standard arguments (use, for instance, the Blaschke selection theorem and the fact that ~b is continuous in the Hausdorff metric) one can easily

deduce that, for any fixed a > 2w/~-, ~b has a maximum value #~ on F~. For which sets E , in F , is this maximum value attained; that is, which

E, e F , satisfy ~b(E,) = #,?

An ad hoc tool for answering this question seems to be the Bonnesen semicircular symmetrization, described in [3] (see also [4]; for the circular symmetrization see [8]). Given a set E (not necessarily convex), fix a straight line s and a point P on s. For any r > 0, let Cr be the circle centered at P with radius r and 2(r) the total length of Cr n E. On Cr take two opposite arcs, each of length ½2(r), so that s passes through the midpoints of both arcs. The set formed by the collection of all such arcs, for r assuming every positive value, is just the semicircular symmetrized set of E with respect to s and P. It is easy to verify that each symmetrized set has the same area as E. Although nothing can be said in general about the perimeter of the symmetrized set, further information is available in the case where E is convex. In such a case we can consider, among all the circular annuli containing the boundary of E, that one of minimal thickness; let O be the center of this annulus. Every circle

centered at O and contained in the annulus intersects E at least in two distinct arcs; hence, if we assume O as the center of the symmetrization, then the perimeter does not increase when we pass from E to the symmetrized set (the so-called annular symmetrized set). For further details see [3]. Henceforth zE will denote the annular symmetrized set of E (as axis s we can choose the horizontal line). Unfortunately it is not true in general that zE too is convex; however we are able to prove a weaker property, expressed by the following lemma, which will be sufficient for solving our problem.

L E M M A 1. I f we choose the line s as the x-axis and the perpendicular to s at 0


as the y-axis, then the boundary of ~E, for x >>. 0 and y >1 O, is the oraph of a non-increasing function y(x).

Proof Let us denote by Crl and Cr2 respectively the inner and the outer circle of the minimal annulus for E. For r 1 < r < r 2, let 71(r), 72(r) . . . . . 7,(r ) , . . . be the arcs of C~c~E and 01(r), 02(r ) . . . . . O,(r) . . . . the corresponding ampl i tudes (with respect to the center O). The quant i ty

(**) 0(r) = ¼[01(r) + 02(r) + --. + 0,(r) + .--3

decreases as r increases; therefore, to prove the lemma, it is sufficient to show that r sin O(r) is not increasing with respect to r. To do this, let us fix r = r o and consider the tangent lines to OE at the endpoints of each arc of C~o c~ E and all the lines joining two consecutive endpoints not belonging to the same arc. All such lines bound a convex polygon T, o containing the center O. Let pl(r),

~2(r) . . . . . ~n(r) . . . . be the arcs of C~c~T~o, i l ( r ) , i 2 ( r ) , . . . , i n ( r ) , . . , the ampl i tudes and i(r) the relevant function, defined as in (**). If we prove that

r sin i(r) is not increasing, this holds for r sin O(r) too; in fact O(ro)= i(ro), O(r) <<. if(r), for r > ro and r o is arbi trary. Now, for any fixed ~ > to, let us assume that the arcs of C~ c~ Tro are more than two (otherwise the following procedure clearly can be simplified) and choose three consecutive ones f rom

them, say Pi- 1(~) = Pi- 19--~i-- 1, ~i(~') = PiQi, ~i+ 1(p) =/5i+ 1(~i+ 1; moreove r let us suppose that Pi-l(r) = Pi 1Qi- 1, Pi(r) = PiQi, 7i+l(r) = Pi+lQi+l do not vanish for P ~< r < 7+6 , 6 > O, and work with r in such an interval. If P~+I

and Qi-1 are the closest points (from opposi te sides on C,) to Q~ and Pi, respectively, and if rl(r) denotes the ampl i tude of that arc P~+IQ~-I not

c o n t a i n i n g PiQi, then it is easy to see that

~(r) = ~ ( r ) - Y ij(r) j ¢ i

increases as r increases. Since

i(r) = ¼[g,(r) + ~(r)-- ~o(r)],

it will be sufficient to prove that r sin[(i~(r) + q(r))/4] is not increasing. To this end consider the region Z bounded by the lines Pi+lQi, PiQi_ 1 and containing the arc p~(~). We have that C s ~ Z is made by two arcs y*(r), 7*(r) whose ampli tudes O*(r), O*(r) satisfy

0.(~) + 0"(~) = 6(~) + ~(~),

O*(r) + O*(r) > Oi(r) + rl(r)

Moreove r

for r > ?.

O*(r)-I-O*(r)= 2 (a rcs in dl ~ ) - - + arcsin r


where dl and d 2 a r e the distances of O from the lines bounding Z; hence, if

- ~ 0 1 + 02) ,

r sin O*(r) = ~22 ( r 2 - ~ ~ + d~d2) '/z,

which is a non-increasing function. Therefore, owing to the fact that ~ is arbitrary, the lemma is proved.

The strategy we are going to use for solving our problem is the following: We find the maximizing sets for q~ in an auxiliary class of sets (containing zE for every E ~ F,) and we show that they are the solutions of the problem also in F, .

For fixed o- > 2 ~ - , let f~, be the class of the plane simply connected sets Z such that:

(i) area(Z) = 1; (ii) 10ZI ~< a, where 10ZI denotes the perimeter of Z;

(iii) Z has two orthogonal axes of symmetry; (iv) if O is the center of symmetry for Z, then every circle centered at O

intersecting 3Z has in common with Z only two (symmetric) arcs; (v) if the axes of symmetry for Z are chosen as the x-axis and y-axis, then

0Z, for x ~> 0, y >~ 0, is the graph of a non-increasing function y(x).

As we have seen, if E ~ F~ then zE E f~,. Moreover, one can verify that f~ is closed with respect to the Hausdorff metric.

If we give in f ~ the same definition of the distance ~b as in F , , it is easy to verify, for Z ~ f~,, that

~b(Z) = area(Z\S),

where S = S(O) is the disk of unit area centered at the center of symmetry. Using standard arguments one can show that q5 has a maximum in f~; thus, let us set

#'~ = max qS(Z) ZE~

and find the structure of the sets Z . ~ ~ such that

¢(Z~) = K .

LEMMA 2. I f Z~ e ~ and ¢(Z~) = #'~ then the perimeter of Z~ is maximum, i.e. IOZ~l-- a.

Proof Let us suppose that IOZ~l < a. By slight symmetrical modifications of OZ~ outside every possible S(t) for which area(Z~\S(t)) is minimum, one

S E T S O F M A X I M U M T R A N S L A T I V E D I S C R E P A N C Y 77

can obtain a new set Z* keeping all the properties (ii)-(v) of Z , , while

area(Z*) > 1 (for instance, one can add two symmetrical sharp ends to Z~ in

the parts most distant from O). If the variations are sufficiently small, then the set Z~, of unit area and homothetic to Z*, belongs to fl~ and satisfies area(Z~\S) > area(Z~\S), which is a contradiction since Z~ gives the max-

imum of ~b.

The above lemma enables us to find the shape of the maximizing sets in f~. For simplicity we shall term oval any set Z ~ ~ such that 8Z is composed of two pairs of equal opposite circular arcs, with the endpoints on 8S and with common tangent line at each endpoint. The following theorem holds:

T H E O R E M 3. Let Z~ be a set maximizing c~ in f~ , i.e.

and let

¢ ( z o ) = ~'., z o ~ ~

2(% 2 + 8)

ao -- x/re(re2 + 16)

(i) I f 2v /~ < a < a o, then Z~ is an oval; (ii) if a >i go, then Z~ is a biscuit.

Remark. The theorem implies that, for fixed a, the set Z~ is uniquely determined, by the conditions area = 1, perimeter = a. At the transition value

ao between ovals and biscuits we have q~(Z,o ) = 0.267. Proof For fixed a, let P1P2, Q1Q2 be the arcs of Z~nSS and PIQ1 one of

the two pieces of t?Z~ inside S. If P1Q1 is not a circular arc or a segment we can replace it by a circular one with the same endpoints such that the area enclosed by the new arc and the segment PIQa is the same as before. Let us repeat it symmetrically for PzQ2. The new set Z* so obtained belongs to f ~ and ~b(Z*)= q~(Z~); thus Z* too is a maximizing set in f~ . But this is impossible, by Lemma 2, since the perimeter of Z* is less than a. Let us consider now the pieces of 8Z~ outside S; let PIP2 be one of them. Denote by

the area that it encloses together with the segment P1P2. Among all the arcs with endpoints at P1 and P2 contained in the strip between the lines P1QI and P2Q2, let us consider the arc of minimal length enclosing, with the segment PxP2, an area e. It is well known that such an arc is circular or is

composed by two parallel segments connected by a half-circle, according to the value of e. Now, if the arc of 8Z~ is not of such a form, let us replace it by the arc of minimal length and do the same symmetrically on the opposite side: We obtain another maximizing set Z * e f ~ , that is impossible, since

I,~Z*l < a.

78 STEFANO CAMPI

Now, to complete the proof, it is sufficient to show that at each point of 8Z~ n 8S the tangent line to 8Z~ is unique. To do this, let us fix one of the points, say P, of 8Z~ n 8S and suppose that two tangent half-lines q and s issue from P such that the angle ~g (the one containing Z~) is different from re; here q will be the half-line out of S. Let us distinguish two cases: (i) ~'g < re; (ii) ~ > re.

In case (i) we use a standard argument (see, for example, [-2]) to obtain a contradiction. Namely, consider a half-line p from P, exterior to S, such that q'g < p"g < re; for fixed e > 0, let AB be the chord of OZ~ (closest to P) parallel to p whose length is e and let P ' be the point on p equidistant from A and B. Finally, replace the piece AB of 8Z~ by the segments AP' and P'B and repeat symmetrically this replacement for the other three points of 8Z,, n 8S. If al(e ) denotes the area of the part of Z~ cut off by the segment AB and a2(8 ) the area of the triangle AP'B, then it is easy to see that

0 < al(e)-a2(~) = 0(~3).

On the other hand, if ll(e) is the length of the piece of 8Z~ between A and B and 12(e) the length of the segments AP' and P'B then

0 < /1(~)--12(e) = O(e).

Setting Al = ll(e)-12(e), we can attach to the circular parts of Z , outside S two symmetrical regions (for instance, of triangular shape) in such a way that the variation of length does not exceed 4A/and the variation of area is of the same order as Al, that is 0(8). By a suitable choice of e and by the help of a similitude restoring the unit area, the previous construction leads to a set Z* e f ~ for which q~(Z*)> ~b(Z,) (in fact some area has been suppressed inside S), which is a contradiction.

(ii) If ~g > 7r, then 8Z,, c~ S must be formed by two circular arcs. For fixed small e > 0, let A and B be two points close to P on 8Z,,, with A outside S and B inside S, such that the segment AB has length e; let e be the area enclosed by the segment AB and 8Z,, Now let us replace the piece AB of 8Z,, by the segment AB and repeat symmetrically such a replacement for the other three points of 8Z, n OS; moreover, by two lines parallel to the axis of symmetry, cut off from Z , n S two symmetrical regions, each of area 2e. If e is sufficiently small, we obtain in such a way a set Z* e f ~ for which ~b(Z*)> q~(Z,), which is impossible, since Z , must give the maximum of ~b. Thus the proof is complete.

The above theorem shows that, for any ~r, the unique maximizing set for 4) in f ~ is a convex set of F , ; therefore


In the final step we prove that actually

#" = / ~ ,

by showing that the maximizing set in f ~ is also the unique maximizing set

in F. .

T H E O R E M 4. For every f i xed a > 2x /~ , the maximizing set for 4) in F . is

unique and it is precisely the same as in f~ ; that is, the oval o f perimeter a if

2 x / ~ < a < ao, the biscuit o f perimeter a if a >. a o. Proof The proof consists in showing that if E . is in F . and satisfies

q~(E.) = #. , then E . must be in f~.. Let S be the disk of unit area with the same center as the minimal annulus

for E.; then

(***) ~b(E¢) = area(E.\S).

In fact

and

~b(zE¢) = area(zE¢\S) = area(E~\S)

~b(G) = m / > ~ ; > / ~ ( ~ G ) .

Equality (***) implies that the above inequalities are equalities indeed, that is

m = ~ ; , O ( G ) = O(~G) .

In other words, zE~ is the maximizing set in f ~ and it is maximizing also in F~. It remains to show that E~ = zE~. To this end let us remark that E~ and

rE~ have the same perimeter; in fact

but I~zE,~[ = a, since zE~ is the maximizing set in fl~. Now, in order that a convex set E and its annular symmetrized set zE have the same perimeter, it is necessary and sufficient that both the following conditions hold:

(i) if Cr denotes the circle of radius r centered at the center of the symmetrization, then E c~ Cr (when it is neither empty nor the whole C,)

is formed by two arcs; (ii) if, in polar coordinates r and (p, q~l(r)~ (p ~ ~p2(r) and (p3(r)

(p ~ q)4(r) are the two arcs of E n C r , then for every r

~ i ( r ) = - ~ ( r ) = ~ ; ( r ) = - ~ ( r ) .

For further details on these conditions see [3], [8]. Because of (i), E¢ ¢~ 08 is

80 STEFANO CAMPI

formed by two arcs; we shall prove below that they are equal and symmetrical

with respect to the center of S. Therefore, by (ii), bo th the arcs of E , c~ Cr are

equal and symmetrical for every r, that means E , = zE, .

In order to see that E , c~ 8S is formed by two equal symmetrical arcs, let us consider, for t E N2, the function

u(t) = a rea [E , n S(t)].

We have already seen that u(t) attains its max imum value when t coincides with the center of S. One can verify that the gradient of u at that point equals

to

~1~,~2 n(s) ds,

where 71 and 7z are the arcs of E . ~ 8S and n is the exterior normal. Since the gradient must vanish, we have that

f , n(s)ds = - f,2 n(s) ds,

which implies that ~1 and 72 must be equal and symmetrical with respect to

the center of S. Thus, the p roof is complete.

The author wishes to thank Prof. C. Pucci for suggesting the problem and for

many helpful discussions.

REFERENCES

1. Besicovitch, A. S., 'A variant of a classical isoperimetric problem', Quart. J. Math. Oxford 20 (1949), 84-94.

2. Besicovitch, A. S., 'Variants of a classical isoperimetric problem', Quart. J. Math. Oxford Ser 2, 3 (1952), 42-49.

3. Bonnesen, T., Les problkmes des isopkrim~tres et des is~piphanes, Gautier-Villars, Paris, 1929. 4. Bonnesen, T. and Fenchel, W., Theorie der konvexen K6rper, Springer-Verlag, Berlin, 1934. 5. Burago, Yu, D. and Zalgaller, V. A., Geometric Inequalities, Springer-Verlag, Berlin,

Heidelberg, 1988. 6. Groemer, H., 'Stability properties of geometric inequalities', Amer. Math. Monthly 97 (1990),

382-394. 7. Osserman, R., 'Bonnesen-style isoperimetric inequalities', Amer. Math. Monthly 86 (1979), 1-

29. 8. Polya, G., 'Snr la sym&risation circulaire', C.R.Acad. Sci. Paris 230 (1950), 25-27.

SETS OF MAXIMUM TRANSLATIVE DISCREPANCY 81

Author's address:

Stefano Campi , D ipa r t imen to di Ma tema t i ca 'U. Dini ' , Universitfi di Firenze, Viale Morgagn i 67A, 1-50134 Firenze, Italy.

(Received, July 8, 1991; revised version, October 14, 1991)

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Isoperimetric deficit and convex plane sets of maximum translative discrepancy