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Measuring flatness in linear spaces
OF FUNCTIONAL
27, 203-214
ANALYSIS
Measuring
Flatness
of Mathematics,
Department
in Linear
DAVID
A.
State
University
Communicated Received
(1978)
December
Spaces
SENECHALLE
College,
New
Paltz,
New
June
13, 1976
York
I2561
by the Editors
31, 1976;
revised
Let L be a finite-dimensional normed linear space and let M be a compact subset of I, lying on one side of a hyperplane through 0. A measure of flatness for M is the number D(M) = inf{sup f (x)/f( y): x, y EM), where the infimum is over all f in L* which are positive on M. Thus D(M) = 1 if M is flat, but otherwise D(M) > 1. On the other hand, let E(M) be a second measure on M defined as follows: If M is linearly independent, E(M) = 1. If M is linearly dependent, then (1) let Z be a minimal, linearly dependent subset of M; (2) partition 2 into mutually exclusive subsets U = (q ,..., u,} and I’ = {wi ,..., v,} such that there exist positive coefficients ai and bi for which XT=, aiuS = d,, biwi ; (3) let r = max{ET==, a&& b, , CT=, b,/&,, a,}; (4) let E(M) be the supremum of all ratios r which can be formed by steps (l), (2), and (3). The main result of this paper is that these two measures are the same: D(M) = E(M). This result is then used to obtain results concerning the Banach distancecoefficient between an arbitrary finite-dimensional normed linear space and Hilbert space.
1.
INTR~DUC-TION
AND
NOTATION
Until Section 3, we are concerned with a finite-dimensional linear space L with norm Ij * 11.L* is the space of all linear functionals on L. The theorems concern closed and bounded point sets lying on one side of a hyperplane through 0, so we let W denote the collection of all such sets. For each set M in W we make the following definitions. (i)
6)
f in L* such that f (x) For eachfin CM, 4f, Ml = su~~.~~~f(~)If(~). CM is the set of all
d(f, M).
(iii)
D(M)
(iv)
(relative to D) if D(M’) < D(M) whenever D(M) > 1 if M is minimal and nonsingleton. If 2 is any minimal linearly dependent subset of M, a number e(2) M
= inff,,--
> 0 for all x E M.
is called
minimal
M’ E W and M’ E M. Thus
(v) is defined as follows: Let U = {ur ,..., ZJ,} and V = {q ,..., v*} be mutually exclusive subsets of 2 with U u V = 2 and such that for some positive coefficients
203 0022-1236/78/0272-0203$02.00/0 All
Copyright Q 1978 rights ofreproductionin
byAcademicPress,Inc. any form reserved.
204 ai and max(r,
DAVID
A.
SENECHALLE
bi , CzP_r aiui = Es=, b,ui . Let r = xi”=, ai/~~=i bi and let It is easily seen that e(Z) is uniquely determined by Z.
e(Z) =
l/y).
(vi) If M is linearly dependent, define E(M) = sup e(Z) where the supremum is over all minimal linearly dependent subsets Z of M. We define E(M) = 1 in the trivial case where M is linearly independent. Thus for ME W, two measures are defined on M: D(M), which has a clear geometric meaning as a measure of flatness but does not lend itself to computation, and E(M), which is computationally nice but has no apparent geometric meaning. Our main result (Theorem 10) states that these two measures are, in fact, equal: D(M) = E(M). Theorems l-9 are merely steps useful in obtaining this result. In Section 3, this theorem is applied to obtain results on the Banach distancecoefficient. In Section 4, an earlier result of this author for 2-dimensional spaces is derived in a much easier way using the results developed herein.
2. THEOREM
1.
If ME
PROOF
OF
MAIN
W, then there is an
RESULT
f
in C, such that d(f, M)
= D(M).
Proof. Let (fi) be a sequence in C, such that d(fi , M) ---f D(M) as i+ co. Let L, be the subspace of L generated by M. For each i, let gi be the restriction of fi to L, and we may assume that each g, is on the unit sphere S,, of L,*. By the compactness of S, , a subsequence (g,u,) of (gJ converges to a member g of Lo*. Suppose g(x) = 0 for some x E M. Then if y EM, g&y)/g&x) < d(g,ci, , M) -+ D(M) as 1 + co. Therefore, gnci)(y) -+ 0 as i--f 00 and g(y) = 0. g is 0 on M, and M generates L, , so g is 0 on L, . However, g E S, , a contradiction. Therefore g is positive on M. g has an extension f in CM. d(f, M) = D(M). THEOREM
2.
If M E W, then there is a closed subset M, of M which is minimal.
Proof. Let G denote the set of all closed subsets Z of M such that D(Z) = D(M). Suppose G’ is a monotonic subset of G and let N = nsEG,g. Suppose D(N) < D(M). Let f be a member of C, such that d(f, N) = D(N), u = inf,,, f (x) and let b = SU~,,~~(X). Then D(N) = b/u. There exists an E such that 0 < E < a and such that (b + ~)/(a - l ) < D(M). Let C = {X EL : f (x) < a - E or f(x) > b + e}. Every member of G’ intersects C, so N intersects C, a contradiction. Therefore, D(N) = D(M) and NE G. By Zom’s Lemma, G has an element M, which is not a proper subset of any member of G. M, is minimal relative to D. 3. If M E W, ,f E C, , M is minimal, id,,, f (x), and b = sq,,Mf (x), th en f OY every point OY f (x) = b. THEOREM
d( f, M) = D(M), x of M ,either f(x)
a = = a
MEASURING
Proof. Suppose a < c <
FLATNESS
IN
LINEAR
SPACES
205
b and there is a point x of M such that f (x) = c.
Let U={xEM:(a+c)/2
L* which is >0 on 7J and <0 on
V.
(2) There exists a function fz in L* which is >O on U and
Proof. Let f be a member of C, such that d(f, M) = D(M) and f has a value a at every point of A and a value b > a at every point of B. Supposea member of L* separatesA from B. There exists a g EL* which is positive on A, negative on B, and such that 1g(x)] < 6 - a for all x in M. Then f + g E CM , and for all x and y in M, (f + g)(y)/(f + g)(x) < b/u = D(M). Hence d(f + g, M) < D(M), a contradiction. THEOREM
5.
No minimal
set in W has more than one pair of inner and outer
subsets. Proof. SupposeM is a minimal set in W and that (A, I?) and (A’, B’) are pairs of inner and outer subsetsof M. Let f, g, and r be such that f, g E CM,
206
DAVID
A. SENECHALLE
d(f, M) = d(g, W = W4), f has value 1 at every point of A and Y > 1 at every point of B, and g has value 1 on A’ and Y on B’. Suppose there is an element x of A not in A’. Then f(x) = 1 and g(x) = r. Let h == (f + g)/2. For all y in M, 1 f h(y) < I, so d(h, M) < D(M) which implies d(h, M) = D(M). By Theorem 3, either h(x) = 1 or h(x) m: Y, but we also have h(x) = (1 + r)/2, a contradiction. Therefore A CA’. A similar argument shows A’ C A. Therefore A = A’ and B = B’, completing the proof. THEOREM 6. Suppose M is a minimal element of W with inner and outer subsets A and B respectively. Suppose A’ C A, B’ C B, A’ and B’ are closed, but A’ u B’ # M. Then some f in L* separates A’ from B’.
Proof. Let M’ = A’ U B’. Let g be a member of C, such that d(g, M) = D(M) and g has values 1 and r > 1 on M. Let h be a member of CM, such that d(h, M’) = D(M’) and infz,,bl, h(x) = 1. Then d(h, M’) < D(M) = Y. If x and if x E B’, g(x) > h(x). Therefore, the functionf = g--h x E A’, g(x) < 4 >, separates A’ from B’. THEOREM
7.
Every minimal
element of W is$nite.
Proof. Suppose that M is a minimal set in W which is infinite. Let L’ be the subspace of L generated by M. Let x be a limit point of M and let (RJ be a monotonic decreasing sequence of open balls with center x whose radii converge to 0 and such that R, does not contain all of M. For each i, let M(i) = M\Ri . is monotonic and increasing and each set Then the sequence of sets (M(i)) M(i) is a closed proper subset of M. For each i, let fi denote a member of C,M(i) such that d(fi , M(i)) = D(M(i)) and such that 11fi llL, = jl fi /I = 1. (ilfi [jL, denotes sup{f(y)l y EL’, 11y/j =-= l}. Th e f unction fi can be obtained by (1) letting fi,l be a member of Cnrfi) such that d(fi,, , M(i)) = D(M(i)), (2) letting fi,z = fi,J] fiSl jjL, , (3) letting fi,3 be the restriction of fi,z to L’, and (4) letting fi be the function obtaining by extending fi,3 to L without increasing its norm, by the Hal-n-Banach theorem.) By compactness of the unit sphere of L*, the sequence (fi) has a subsequence which converges to a function f in L*, and for convenience, we will assume that the sequence (f<) itself converges to f. Then
llfll = 1. Suppose y E M\x and f(y) = 0. Let z also be a point of M\x. There is a positive integer n such that for i > n, y, z E M(i). For i > n, fi(z)/fi(y) < D(MJ < D(M), and since the sequence (fi(y)) converges to 0, so also must the sequence (fi(z)). Therefore f (z) = 0. f is 0 at every point of M\x. Since f is continuous and x is a limit point of M, f(x) = 0, also. f is 0 on all of M and is then 0 on ,!,’ because M generates L’. However, IifliL, = 1, a contradiction. Therefore, f is positive on M\x. If p, q E M\x, then f(p)/f(q) = limi.+, f4pYf4q) and for each i, h(p)lf(s) f
MEASURING
FLATNESS
IN
LINEAR
207
SPACES
m >~Jw)) < D(W,sof(P)/f(!?) Gww. since
x is a limit point of Mix, the last inequality holds for all p, q E M. Therefore, f E C,\r and d(f, M) = D(M). For each i, let Ni be a minimal subset of Mi and let Ai and Bi be the inner and outer subsets, respectively, of iVi . Let A and B be the inner and outer subsets, respectively, of 111. If Ai C A and Bi C B, then, by Theorem 6, some member of L* separates Ai from B, . However, since Ai and Bi are the inner and outer subsets of the minimal set Ni , Theorem 4 asserts that no member of L* can separate them. Therefore, for each i, either Ai intersects B or Bi intersects A. Since f is positive on M, there is an Y > 0 such that r
< f4q)f(p) f,(P) f (9) <--
This
is a contradiction,
r T--E
_
r-kc Y completing
f(p)
fi(P)
fi(4) f(q)
r+c = -E
(
f(p)
f(P)
- E
f(q) + E f(q)
< D(Mi).
the proof.
THEOREM 8. Suppose that M is a minimal element of W with inner and outer subsets A and B, respectively. (1) If A = A, v A, , A, n A, = ia, A, # o and A, # 5, then a member ofL* separates A, from A, u B. (2) If B = B, u B, , B,nB,-a,B,#~,andB~f~,thenamemberofL*separatesAuB, from B, . Proof. (1) Suppose A, = {x} and A, = A\A, , where {x> ,C A. Let f be a member of C, such that d(f, M) = D(M) an d such that the values f assumes on M are 1 and D(M). Let M’ = M\x. By Theorem 7, M is finite. Thus M’ E Wand D(M’) < D(M). Let g b e a member of CM, such that d(g, M’) = D(M’) and such that infDEM*g(p) = 1 and supVEM, g(p) = D(M’). Either g(x) < 1 or g(x) > D(M). Suppose g(x) > D(M). Since f(x) = 1 and g(x) > D(M), there is an s such that 0 < s < 1 and h(x) = D(M’), where h = (1 -s)f+sg. Th en f or all points p of M, 1 < h(p) < D(M), so d(h, M) < D(M), a contradiction. Therefore g(x) < 1. Thus f(x) > g(x) and if p E B, then f(q)
208
DAVID
A.
SENECHALLE
part (1) also holds if A, has m elements, and, by induction, this will complete the proof to part (1). Suppose A, has nz elements x1 ,..., N,, . For each xi E A, there is a linear functional which is negative on A, u {xi} and positive on (A,\xJ u B. Suppose no member of L* separates A, from A, u B. For distinct elements xi and xi of A, , there exist linear functionals j’ and g with values as follows: Positive
Negative
f
A, u {xi>
(As\4
u B
g
A, u {xj>
(Az\xd
u B
There is a linear functional h = (1 - t)f + tg, 0 < t < 1, such that h(xj) = 0. Then h(q) > 0, f or otherwise h would separate A, from A, u B which we have assumed does not happen. Therefore, for any two members xi and xj of A, there is a linear functional which is positive on A, U {xi}, 0 at xi , and negative on (A,\bi , $3 u B. Now suppose xi , xj , and xg are three distinct points in A, . There exist linear functionals f and g (different from the f and g above) with values as follows: Positive
f g
0
Negative
A, u {xi>
Xl,
(A,\{xi t ~1) u B
4, u {Xj}
Xk
(A,\h
9 ~1) u B
There is a linear functional h = (1 - t)f + tg, 0 < t < 1, such that h(+) = 0. Then h(q) > 0 because otherwise h would separate A, from A, u B. Therefore, h has values as follows: Positive
0
A, u bil
xj > x7c
Negative (A,\{xi
, xi , 4)
u B
Continuing in this manner, for each point xi of A, , there exists a linear functional h, which is positive on A u {xi}, 0 on A,\xi , and negative on B. Then, h, + ... + h, separates A from B, contradicting Theorem 4. This proves (1). Part (2) can be proved by similar arguments. THEOREM 9. Suppose M E W and D(M) > 1. Then the following stafements are equivalent: (1) M is minimal relative to D; and (2) M is minimal relative to the property of being a linearly dependent subset of L.
Proof. If (2) is true, then every proper subset M’ of M is linearly independent, so D(M’) = 1. Therefore (1) is true. The rest of the proof shows that (1) implies (2).
MEASURING
FLATNESS
IN
LINEAR
SPACES
209
Suppose (1) is true. Let A and B be the inner and outer subsets of M, respectively, and let y E B. Let L, be the subspace of L generated by M, and let L, be the subspace ofL generated by M\y. SupposeL, &L, . D(M\y) < D(M), so there is a linear functional g,, defined on L, such that infz6:EM,Vg,,(x) = 1 and su~zE,\f\g g,,(x) = WWY). S ince y $ L, , there exists an extension g of g, to all of L such that g(y) = 1. Then g E C,,I and d(g, M) =-= D(M\\y) < D(M), a contradiction, Therefore M\y spans L, . For every pointy’ of L, , let B(y’) = (B\y) u {y’) and let M(JJ’) = A U B(y’). Suppose there exists a sequence (p,) in L, converging to y such that for each i, either M(pi) is not minimal or it is minimal but A and B(pi) are not its inner and outer subsets, respectively. For each i, let Mi be a minimal subset of M&Q), let Ai and Bi be the inner and outer subsets, respectively, of Mi , and let fi be a member of L,* such that 1 .
210
DAVID
A.
SENECHALLE
\Ve now prove A linearly independent. Let A = {x1 ,..., x,j. There exists fe CM such that d(f, AZ) = D(M) and f(x) = 1 for x E A and f(.v) =~ D(M) for x E B. Suppose A is linearly dependent. There exist cr ,..., c,, not all 0 such that ~~=r cjx$ == 0. Therefore, there is at least one ci < 0 and at least one cj > 0. Let C = {xi E A : ci < 0), and let I’ = {xi E A : ci > O}. No linear functional separates U from I, contradicting Theorem 8. This proves =1 linearly independent. Now suppose B is a singleton set, B = {y]. M is linearly dependent because D(M) > 1, so y = ~~=, aixi . Since f(y) = D(M) and f(zi) = 1, i :- 1 ,..., m, we have ~~~, ai = D(M), implying that at least one of the ai :, 0. If one of the ai < 0 then the set (xi : ai > O> cannot be separated from the set B u {xi : ai < 0}, contrary to Theorem 8. Therefore ai >, 0 for i =-- I,..., n. This means that M is a minimal linearly dependent set. Suppose B is nonsingleton. Let yr E B. Suppose yr = ~~~, avvl . Since f(y) = D(M) andf(q) = 1, i == I,..., n, at least one coefficient a, > 0. If they are all >0 then, contrary to Theorem 8, {yr] cannot be separated from A u (B\y,}. Therefore, there is a coefficient aj < 0. But then {s, : a, G, O$u B cannot be separated from {xi : ai > 01. Therefore, A u [?,rj is linearly independent. Let B = {yl ,..., y,> and let p denote the least positive integer i such that {Xl I.‘., xvi > yr ,..., pi} spans the same subspace of L that M spans. Then (x1 ,..., x,, , yr ,..., y,} is linearly independent. Suppose q >,p + 2. Then ykll = = xr:, Cpi + Cfzl diyi . Let a = zyzl ai, Z:Zl uixi + CL bi~i and yk+z b = g=, bi ) c = g, ci ) and d = X:=1 di By the first section of this proof, ye+r can be moved slightly in any direction in L, without altering the inner and outer subsets of M. Or, to state it precisely, there is an open subset U of L, . . contammgy,,, such that if u is any point of U, then the set M(u) =- (-II’ ~9~:~~~)U (u} is minimal relative to D and has inner and outer subsets -4 and B(u) =z v 1 < R, (B\~k+d ” {uf, respectively. There exists R ;b 0 such that for ykJ.r + rxr E U. For I Y ! < R, let s(r) = D(M(y,,+* + rxr)). Then for Y < R, there is a linear functional g,, having value 1 on A and having value S(Y) on B(y,+, + YXJ. Evaluating g,. at y,+r + YX, vve get S(Y) == a -- h(r) -q- r. It follows that b -;i 1 and S(Y) = (a + Y)/( 1 - b). Evaluating g,. at yJc. c yields S(Y) = c + ds(r). Just as it has been shown that h # I, parallel arguments beginning with an open set u’ containing yA,~a will show that d # 1. Then S(Y) = c/(1 - d). Th ere f ore, if j Y I < R, (u + Y)/( 1 - b) = c/( 1 - d). The left side varies with Y and the right side is constant, a contradiction. Therefore B = h >..., ~n+d. Because M is minimal relative to D, it follows that every point of 31 is a linear combination of the other points in M. Suppose there is a point x of M such that M\x is linearly dependent. Let y be a point of M\x which is a linear combination of the other points of M\x. Then x and y are linear combinations of the points of M\{x, y]. However, M has m + p + 1 points and 112+ p of
MEASURING FLATNESS IN LINEAR SPACES these are linearly linearly dependent THEOREM
10.
(See Section
independent, set.
a contradiction.
Therefore
211
M is a minimal
If M is a set in W, then D(M) = E(M).
1 for the definition
of E(M).)
Proof. First suppose M is minimal, let A and B be the inner and outer subsets of M, respectively, and let A = {x1 ,..., .x~,) and B = (yl ,..., yn}. Because M is linearly dependent, there exist coefficients a, ,..., am, b, ,..., b, not all 0 such that a, > 0 and ~~~, aixi = xr=, biyi . None of these coefficients = 0 for then M would not be a minimal linearly dependent set. Thus a, > 0. Suppose that 2
0 and large enough that gr(xp) > 0. Then g, separates A from B, a contradiction. Therefore, ai > 0 for i = l,..., m. A similar argument shows all bi have the same sign. Let f be the linear functional in C,,, such that d(f, M) = D(M) and f (x) = 1 for all x E A and f (y) = D(M) for ally E B. Evaluating f at both sides of the equation x2, a$,~~ = Cr=, biyi , we get CL, ai = D(M) Ci”=, bi . Th ere f ore, we get that all the coefficients bj are positive and
Finally,
if M is not minimal,
E(M) = sup{e(M’) = sup{D(M’)
: M’is
then a minimal,
linearly
dependent
: M’ is a subset of M minimal
subset of M]
relative to II}
= D(M).
3. RESULTS CONCERNING BANACH DISTANCE-COEFFICIENT We call a norm F on a finite-dimensional linear space L euclidean if L is an inner product (pre-Hilbert) space relative to this norm. There are many different properties of euclidean norms which characterize them among the set of all norms. A summary of these results may be found in [2]. In case F is noneuclidean it is interesting to somehow measure the “euclideanness” of F and to relate this measure to other properties of the norm F. Our means of measuring “euclideanness” is the following definition: K(F) is the least number r > 0 such that for some euclidean norm G on L dominated by F, the norm rG dominates F. Then K(F) = 1 in case F is euclidean and otherwise K(F) > 1.
212
DAVID
A.
SENECHALLE
Our measure K(F) ammounts to a special case of the Banach distancecoefficient. If Xand Y are finite-dimensional Banach spaces of the same dimenson their Banach distance-coefficient is d(X, Y) = inf I/ T I! /I T-l I/ where the infimum is taken over all linear operators T from X onto Y. In case Y is an inner product space and F is the norm on X, d(X, Y) = K(F). Theorems relating K(F) to specific properties of the norm F are contained in the author’s papers [3, 41. Dvoretzky [l] p roved the remarkable theorem that given any E > 0 and positive integer n, there is a positive integer m > n such that every m-dimensional normed linear space contains an n-dimensional subspace with norm F, such that K(F,) < 1 + E. Our results from the previous section have application to this subject as follows: Let F be a norm on Rn and consider the points of Rn as column vectors. If X E Rn, XXT is a symmetric n x n matrix and all matrices so formed lie in a subspace of M,,, , the linear space of all real n i< n matrices. Let S = {X E R* : F(X) = I>, the unit sphere for F. Let 112 = {XXT : X E S}. Then K(F) = (E(M))l12. To prove the above theorem we prove that the following expressions are equivalent: (1) K(F) = inf A, where A = {r > 0: for some F < rG}. (This is the definition of K(F).) (2) K(F) = inf B, where B = {sup(G(X)/G(Y): euclidean).
euclidean X,
G,
Y E S}:
G < G is
(3) K(F) = inf C, where C = {sup{ T(X, X)/( T( Y. I’))‘/” X, Y E S}: T is an inner product on R” x R”}. (4) K(F) = inf D where D = {sup{(f(x)if(y))“‘: x, y E M}:~E M&, and f is positive on M}. (This says K(F) = (D(M)li2.) Proof. Let Y E A and let G be a euclidean norm such that G < F < rG. Then for X, YE S, G(X) < 1 < rG(Y), so G(X)/G(Y) < r and r> a member of B. Therefore, inf B < inf A. On the other hand, suppose r E B. Then r = sup(G(X)/G(Y): X, Y E S) = sup(G(X): x E S)/inf{G(Y): Y E S} for some euclidean norm G. Then the norm H = G/sup{G(X): X E S} is dominated by F and rH dominates F, so r E A. Thus B CA so inf A < inf B. Therefore inf A = inf B and (1) e(2). That (2) o (3) follows from the fact that every Euclidean norm derives from an inner product. The equivalence between (3) and (4) is easily seen once it is realized that the transformation (X, Y) - XYT has the universal factorization property; i.e., B is bilinear functional on Rn x R’b if and only if there exists a linear functional f on Mnxn such that B(X, Y) = f(XY’) for all X, Y in Ii”. Since we proved D(M) = E(M) in the previous section, we now have K(F) = (E(M))l12.
MEASURING
FLATNESS
4. THE
IN
&DIMENSIONAL
LINEAR
213
SPACES
CASE
We now assumen = 2. Reviewing the notation of previous sections,F is a norm on R2, S is the unit sphere for F, and M = {XX’ : XE S}. Thus M is a set of 2 x 2 matrices lying in a 3-dimensional subspaceof Mzx2 . From Section 3 we have that K2(F) = E(M). Th is author [2] proved by geometric methodsthat K2(E)
= sup
cdF2(ax + by) + abF2(cx - dy) cd(u2 + 62) + ab(c2 + d2)
’
(*I
the supremum over all X, YE S and all real numbers a, b, c, d > 0. We now prove this result using the theorems proved in previous sectionsof this paper. We first note that if in the above fraction a, b, c, d are replaced wherever they occur by r,u, rib, r2c, r,d, respectively (rr , r2 # 0), the value of the fraction is the sameas before. Thus, an equation equivalent to (*) is
K2(F)
= sup
cd + ub cd(a2 + b2) + ub(c2 + d2)
taken over all a, b, c, d > 0 such that for some X, YE S, UX + bY and cX - dY E S. Let W, be the set of numbers over which the supremum is taken in (**). Let YE WI and let a, b, c, d, X, Y satisfy the above conditions with
’ =
cd f ub cd(u2 + b2) + ab(c” + d2)
*
Let U = UX + bY, V = cX - dY and let a,, 6, , ci , dl be the solutions (which are positive) to the equations Y = u,U + b,V, X = clU - d,V. Let rl = (c,d, + u,b,)/(c,dl(u12 + b12) + u,b,(c12 + d12)). By a simple computation y1 = l/r. rr is also in WI . Th iss howsthatiftEW,,thenl/tEW,. On the other hand, let W, = {e(M’): M’ is a minimal, linearly dependent subset of M}. The result (*) will be establishedif it can be shown that sup w, = sup w, . Let YE W, . There exist X, YE S and a, b, c, d > 0 such that UX + bY E S, CX - dY E S and Y = (cd + ub)/(cd(u2 + b2) + ab(c2 + d2)). Let M’ = {X;yT, YYr, (ax + bY)(uX + bY)T, (cX - dY)(cX - dY)T). Then it is easily checked that if r > 1, then M is a minimal, linearly dependent set and e(M’) = r. Thus sup W, < sup W, . To get that sup W, > sup W, , let e(M’) E W, . Since M liesin a 3-dimensional subspaceof Mzx2, M’ hasexactly two, three, or four elements.It turns out not to be possiblefor M’ to have only two or three elements.If M’ has only two elements XXr and YYT, then for some a > 0 and fl, XXT = uYYr and X = d2Y or -u~/~Y, soF(X) = a1j2 # 1. 5S+7/2-5
214
DAVID
A. SENECHALLE
Now consider the 3-element case. Let the three elements of M’ be XXr, YYT, and UUT chosen so that U = aX + bY, a, b > 0. By a linear transformation we may assumeX = (1, 0) and Y = (0, 1). Thus there exist c, , ca# 0 such that XXT + c,YYT + c,(aX + bY)(aX + bY)T = 0. In matrix form:
which is impossibleunder the assumptionsa, b, c, + 0. Let the elementsof M’ be XXT, YYT, UUT, I’VT chosenso that for a, b, c, d > 0, U = aX + bY, V = cX - dY. Again, by a linear transformation, we may assumeX = (1, 0), Y = (0, 1). There are coefficients cr , ca, ca # 0 such that XXT, c,YYT + C,UUT + C,VVT = 0. In matrix form
Solving, we get
Cl =
bd(ad + bc) ac(ad + bc) ’
c2 =
-cd ac(ad + bc) ’
” =
-ab ac(dd + bc) ’
Thus
e(“‘)
=
cd + ab ac(ad + bc) + bd(ad + bc) -
cd + ab cd(a2 + b2) + ab(c2 + d2)
or e(M’) is the reciprocal of this fraction, whichever is greater. In either case e(W) E WI, so sup WI 3 sup W, . This completesthe proof. REFERENCES 1. A. DVORETZKY, Some results on convex bodies in Banach spaces, in “Proceedings of the International Symposium on Linear Spaces, Jerusalem, 1960,” pp. 123-160. 2. J. A. OMAN, “Characterizations of Inner Product Spaces,” Ph.D. Thesis, Michigan State University, East Lansing, Michigan, 1969. 3. D. A. SENECHALLE, Euclidean and non-Euclidean norms in a plane, Illinois J. Math. 15 (1971), 281-289. 4. D. A. SENECHALLE, Equations which characterize inner product spaces, Proc. Amer. M&z. Sot. 40 (1973), 13061312.