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__pR(dp~, GI) + qR(dp~,, G2) = pR(GI) + qR(G2). TIIEOREM I. If
(5)
sup sup L(A, 0) = M < oo Ae~IOen
then every ¢~with property (A) is an asymptotically optimal empirical Bayes rule.
1965]
THE COMPOUND DECISION PROBLEM IN THE OPPONENT CASE
121
Proof. Let ~G denote the strategy by which the At are independent, identically distributed according to G. Then for every real t Err K.(t) = G(t), and it follows by Lemma 1 and by use of the usual approximation by step functions that
ErGR(Kn) <=R(G).
(6)
Now whatever be 0, we have that conditionally on A = 0 the Xt's are independent. Hence from property (A) it follows that conditionally on A = 0 lim sup [R(~., 0.) - R(G.)] < 0 n...-~ O0
and hence by the Fatou-Lebesgue theorem, since by (5) L and R are bounded functions, (7)
lira sup E~[R(~b,,A,) - R(K,)] _< 0. n-*O0
(6) and (7) imply
(8)
lim sup R(O,, G") < R(G), I1"~ O0
whereas from Theorem 1 of [13] we have infR(~,, G") = R(G). Hence the theorem follows. We remark that if ~bhas property (A) and the At's are distributed independently of the X's, and At has (marginal) distribution G, but the At's need not be mutually independent, then (8) is still correct, but one can easily obtain examples where strict inequality holds in (8). Theorem 1 states that it is at least as easy to find asymptotically optimal empirical Bayes rules as it is to find rules with property (A). In [6] p. 147 Robbins suggests that one should try to find rules with property (A) by acting as though one was treating the empirical Bayes problem. 3. Maximin strategies for the opponent. Suppose the component problem is such that there exists a G* for which supR(G) = R(G*). G
If the opponent has malicious intentions, in that he wants to select a strategy 7 which maximizes lira sup E~L((~,,A,) n-~O0
then if the statistician uses a (~ with property (A0) an optimal strategy for the
122
ESTER SAMUEL
[September
opponent is to select Ai independently, each with distribution G*. (This is simply a sequence of independent repetitions of the maximin strategy for the component problem.) Clearly by (A0) we have lim sup E~L(d~n,An) =
whereas from Theorem 1 it follows that with the above strategy lira E~L(~n,An) = R(G*). n--~ O0
Thus, when (Ao), holds there exists an optimal strategy for the opponent which does not take advantage of the fact that all the past observations are known to the opponent before he selects Ai. 4. Some theoretical results. In [12] Lemma l, a pointwise inequality, valid for all the sequences 0 is given. We restate it here for the random sequence A. LEMMA 2. With P~-probability o/1e,for every/1, 1 ~ R($~,,A,) < R(Kn). /1
i~1
Let ~'i-1 denote the a-field generated by Ai, Xi_ 1, Zi-1, and let R(c~i[5 i_ t) denote the conditional expected loss incurred on the i-th decision, given ~ - t . Corresponding to Lemma 3 of [12], we have LEMMA 3. If(5) holds [L(¢n, A n ) - / 1 - 1 ~ R(~il~,_l)] i=1
)-0 a.e. P~. n~OO
Proof. For/1 = 1, 2, ... let Yn = ~ [L(q~,,A,) - R(~b, I ~ , - ~)]" i=l
Then (Yn} is a martingale relative to {~-,}, and since Var(Yn - Yn-1) <--M2 it follows (see [12] Lemma 2, or [5] p. 387E) that Yn/n --*0 a.e. P~. From Lemmas 2 and 3 it follows that when (5) holds it suffices to show that (9)
[R(qbn I~-n- 1) - R(~PKn,An)]
>0 n'-* O0
a.e. P~, or in probability in order to prove that ~ has property (Bo)a.e. Pr, or in probability, respectively. 5. Testing a simple hypothesis against a simple alternative. This problem is treated for the fixed sequential compound case in [9] and [11], where rules t* and t are exhibited having uniform property (A) and property (B). We shall
1965l
THE COMPOUND DECISION PROBLEM IN THE OPPONENT CASE
123
here consider these rules in the opponent situation. One could shorten the discussion somewhat by making more use of the results obtained in [9] and [11], but we believe it is for the benefit of the reader to give a complete proof of our statement concerning the rule t*, and only state the results concerning i'. The interested reader should not find it too difficult to substitute the proofs. (See also [14].) In this problem there are only two possible distributions, to be denoted by P0, 0 = 0 , 1 , and two actions 9.I={0,1} where A = j means "say 0 = j " j = 0 , 1 . The loss function considered is L(A,O)= aO(1- A)+ b ( 1 - O)A, with a > 0, b > 0, and (5) holds with M = max(a, b). Without loss of generality we let f(x, 0) denote the density of Po, 0 =0,1, with respect to some dominating a-finite measure /t. An apriori distribution G over f~ = {0,1} is entirely specified by ~/= P(A = 1), and for any vector 0i of zeros and ones its empirical distribution Giis completely specified by the proportion of ones i -1 ]E~.= l 0j. Here any (randomized) decision function ~b(x) is completely specified by t(x) which denotes the probability by which the actions A = 1 is taken, when X = x is observed. We write t, instead of q~, and R(q) instead of R(G). It follows easily (see e.g. (3) of [9]) that 1
if qaf(x, 1 ) - ( 1 - q ) b f ( x , O )
> 0
0
if
"
< 0
L arbitraryin [0,1] if
"
= 0.
I (10)
t,(x) =
Let h(X) be an unbiassed estimator of 0 with finite variance, and let Pi = pi(x,) equal O, i -1 ~j~l h(xj) or 1, as i-1 ]~]=1 h(xj) is less than 0, between 0 and 1 or greater than 1, respectively. Let 2~ = i- 1 ]~j = xAi. By a martingale convergence theorem I i-x ]~J =~ h(Xj) - 2i ] converges to zero a.e. Pr, and since 0 < 2, < 1 and 12,_ - 2, I ___ i - a.e. Pr, it follows that also
[ P i - l ( X t - t ) - 2il-*O as i ~ ~ a.e. P~.
(11)
The rule t* is defined by
t*(xi) =
I
1 if Pi_l(Xi_l)af(x,,1)- (1 - Pi_l(Xt_O)bf(xi, O) > 0
L
0
otherwise.
A motivation for t* is given in [ 9 ] , and is also obtained by comparing t* with (I0), We shall prove THEOREM 2. If RO1) is differentiable for 0 < rI < 1 then t* has uniform property (Ao), and has property (Bo) a.e. Pr" Proof. Let e > 0 be given. If R(t/) is differentiable then by Lemma 1 of [9] there exists a 6 > 0 such that for any versions of (10) (12)
max 0=0,1
IR(q,o) -
R(t¢, 0) l < e/3 for
[ r / - r/* [ < 6.
124
ESTER SAMUEL
[September
Let co denote a generic point in our measure space and let
Let NI be a fixed integer greater than 2/6 + 1. For i > NI we have by Chebychev's inequality (13)
P , [ [ P , - x - 4,[ ->__6] Z P~[IP,-1 - '~,-11 >- 6 - i -1] =< V / [ ( i - 1),~z - 26]
whero V =. max{Eoh(X) 2, EI[h(X) - 1]z}. With our definitions we have (14)
R(t* ] ~-,_1) = R(t°,_t ,A,)
where t~o is defined through (10) with the arbitrary part taken to be 0. Thus for co ~ Sl- 1 we have from (12) and (14)
(15)
JR(t71
-
I < 5/3
and hence for i > NI, it follows from (13) and (15) that (16)
E,/.(t?, Ai) = E,R(t* I ~ ,_,) < E,R(t~,,A,) + , / 3 + M V/6[(i - 1)6 - 2].
Now for n ~_ 3 M N 1 / s n - 1 ~N~ 1 L(t*, Ai) < e/3 a.e. Pr and for n > N2 sufficiently large n - t ~E~ffiN.+ 1 [(i -- 1)6 -- 2]- 1 < e6 ]3MV. Thus for n > max(N2, 3 M N I / e ) we have from (16) g , L ( t * ~ , a . ) - e , n -1 ~ L ( t L A . ) < E ~ n - i ~. R(ta,,ai ) + ~, i=1
iffil
and uniform property (Ao) for t* follows from Lemma 2. We shall show that (9) holds a.e. Pr for t*. This follows immediately from (11), (12) and (14), since here R(~^,,A~) = R(ta,,Ai). Thus t* has property (Bo) a.e. Pr, and the proof is complete. We remark that uniform property (A) and property (B) for t* was proved under the same condition on R(~), as in Theorem 2 above. The rule "/is randomized. Let Pi = Pi(Xi) be defined as before, but with h(x) a bounded estimator. For i = 1,2,..- let Z+=(zil,z~2) be independent random variables, uniformly distributed on the unit square, Z~ independent of X+,AI. For i ffi 2, 3,... let mi-x =~ rn(Zt, xi_l) = Pi-x(x~-x) + i -1/+
Zt2
1 + i-1/'*(z+1 + z+2) and define mo arbitrarily. ! is defined by means of ?i(x+) = t,,,_, o (xi), i = 1,2,..., where the right hand side is defined through the nonrandomized version of (10) mentioned earlier. We state without proof the following
1965] THE COMPOUND DECISION PROBLEM IN THE OPPONENT CASE
125
THEOREM 3. For any two distributions Po,P1, t has uniform property (Ao) and has property (Bo) in probability. In connection with Theorem 2 and 3 we remark that, as has been pointed out to the author by Professor H. Robbins, the third paragraph on p. 1084 of [9] is incorrect, in that there the opponent case sould have been considered. 6. The estimation problem. Rather than discussing the opponent-case properties of all the compound estimators considered in [12] in full generality, we shall consider a particular example. Let the component problem be that of estimating the parameter 0 of a Poisson distribution, where 0eg2 = {0:0 < ~ < 0 < fl < oo} (where ~,fl are otherwise arbitrary). Let 91 = f2, L(A, 8)= ( A - 8) 2, so (5) holds with M = ( f l - ~t)2 . We shall consider nonrandomized estimators only, and let ~b(x) denote the value in 9I chosen with probability one. Let ]¢j(x) be 1 or 0 according as Xj = x or X 1 ve x, and for x = 0,1,... let |
(17)
q i ( x ) = i -1
t
Z Y~(x), q~:,(x)=i -~ Z P(Xj=xIA,).
j=l
j=l
The sequential compound rule ~ considered in [12] (see also [7]) is defined by (I8)
dpi(xi) = (xi + 1)qi-l(xi + 1)/qt-l(xl), ~, or fl
according as the first term on the right hand side of (18) is between 0t and fl, less than 0t, or greater than fl, respectively. We shall prove THEO~M 4. The rule ~ defined by (18) has property (Bo) a.e. P r Proof. We shall show that (9) holds a.e. P r It is easily checked that (cf. e.g. [7], (18) or [12], Section 5.) (19)
q~r,(xi) = (x, + 1)qr, (x i + 1)/qr,(x~)
and ct __
dPK,(x)- dpi(Xi_l,x)
>0 a.e. P r l~o0
(20) follows immediately from (18) and (19), since similarly to (11) one has by the definitions (17) and the martingale convergence theorem, that
[ qi-l (x) -- q~,(X) [ ~ 0 a.e.
P~,
and for fixed x qK~(x) is a.e. bounded away from zero. That (20) implies (9) is seen as follows: For every e > 0 there exists a finite N~ such that Pe(X <-_N,) > 1 - ~/2M, for all 0 e ft. Thus (20) holds uniformly in 0 =< x =< N,, i.e. for every 6 > 0 there exists 1(8,6) such that
126
ESTER SAMUEL
(21)
Pr(lqbK,(x) - q~(X,_ l, x) I < e / 6 f l f o r all i > I(e, a n d all 0 < x < N~) > 1 - 6.
But s t r a i g h t f o r w a r d a p p r o x i m a t i o n s show t h a t (21) implies P~(IR(q 5, I ~ ' , - i) - R((°r,,A,)I < ~ for all i > 1(5, 6)) > 1 - 6, which is j u s t an explicit f o r m o f (9).
REFERENCES 1. D. Blackwell, Controlled random walks, Prec. International Congress Math. 3, Amsterdam (1954), 336-338. 2. D. Blackwell, An analogue of the minimax theorem for vector payoffs, Pacific J. Math. 6, (1956), 1-8. 3. J. F. Hannan, The dynamic statistical decision problem when the component problem involves a finite number, m, of distributions. Abstract, Ann. Math. Statist, 27 (1956), 212. 4. M. V. Johns Jr., The parametric and nonparametric compound decision problem in the sequence case. Abstract, Ann. Math. Statist. 34 (1963), 1620-1621. 5. M. Lo~ve, Probability Theory. 2nd ed. Van Nostrand Comp., (1960). 6. H. Robbins, Asymptotic subminimax solutions of compound statistical decision problems, Proc. Second Berkeley Symp. Math. Statist and Prob. Univ. of California Press, (1951), 131-148. 7. H. Robbins, An empirical Bayes approach to statistics, Prec. Third Berkeley Syrup. Math. Statist. and Prob. 1, Univ. of California Press. (1955), 157-163. 8. H. Robbins, The empirical Bayes approach to statistical decision problems, Ann. Math. Statist. 35 (1964), 1-20. 9. E. Samuel, Asymptotic solutions of the sequential compound decision problem, Ann. Math. Statist. 34 (1963), 1079-1094. 10. E. Samuel, An empirical Bayes approach to the testing of certain parametric hypotheses, Ann. Math. Statist. 34 (1963), 1370-1385. 11. E. Samuel, Convergence of the losses of certain decision rule~for the sequential compound decision problem, Ann. Math. Statist. 35 (1964), 1606-1621. 12. E. Samuel, Sequential compound estimators, Ann. Math. Statist. 36 (1965), 879-889. 13. E. Samuel, On simple rules for the compound decision problem, J. Roy. Statist. Sec. (]3) 27 (1965), 238-244. 14. E. Samuel, On the sequential compound decision problems in the fixed case and the opponent case. Unpublished (1964). THE HEBREW UNIVERSITY OF JERUSALEM
THE KERNEL OF THE COMPOSITION OF CHARACTERISTIC FUNCTION GAMES(1) BY
BEZALEL PELEG ABSTRACT
The structure of the kernel of a composition of two games is investigated; a comparison with the results for Von-Neumann and Morgenstern solutions is included. 1. Introduction. In this paper we investigate the kernel of the composition, in the sense of yon N e u m a n n and Morgenstern, of two games. Sections 2, 3, and 4 contain the necessary definitions, preparatory lemmas and an example which shows that the phenomenon of transfer occurs in the kernel. In Section 5 we give a complete description of the kernel of a composition and obtain bounds on the transfer. The case of composition of two ordinary three-person games is studied in full detail in Section 6. In Section 7 we apply the results of Section 5 to constant-sum games; it turns out that the results obtained are similar to the results of yon N e u m a n n and Morgenstern for solutions. 2. Definitions. Let N = {1, 2, . . , n} be a set with n members. A characteristic function is a non-negative real function v defined on the subsets of N which satisfies
(2.1)
v(~b) = 0, and v({i}) = 0, for i = 1, ..., n.
The pair (N; v) is an n-person game. The members of N are called players. Subsets of N are called coalitions. The game (N; v) is an ordinary game if (2.2)
v(S) < v(N),
for all
S c N.
Let G = (N; v) be an n-person game. An individually rational payoff vector (i.r.p.v.) is an n-tuple x = (xl, "", x,) of real numbers which satisfies: Received June 21, 1965 (1) The research described in this paper was partially supported by the U.S. Office of Naval Research under Contract Number N62558-4355. 127
128 (2.3)
BEZALEL PELEG xi > O, i = 1, ..., n
[September
(individual rationality),
and (2.4) i=l
x i = v(N).
The set of all the i.r.p.v's is denoted by A(G). Let i a n d j be two different players. We denote by J i j the set of all the coalitions which contain player i but do not contain player j; i.e., (2.5)
J't~={S: ScN,
ieS
and j ¢ S } .
Let x be an i.r.p.v, and let S be an arbitrary coalition. The excess of S with respect to x is(z) ; (2.6)
e(S, x ) = v(S) - ~ xi
The maximum surplus of i over j with respect to x is (2.7)
si~(x) = max {e(R, x) : R e,Y'o}
i is said to outweigh j with respect to x if (2.8)
s~(x) > s~(x)
and
xj > 0.
x is balanced if there exists no pair of players h and k such that h outweighs k. The kernel(3), (4)of G,~Y'(G), is the set of all balanced i.r.p.v's. Let c be a real number which satisfies the inequality c < v(N). We define a new game G(c) = (N; vc), with the same set of players N as in G, whose characteristic function vc is defined by
(2.9)
Oc(S)
[ v(S),
s ~N
1 (v(N)-c,
S=N.
The core (s) of the game G is the set of all i.r.p.v.'s x which satisfy (2.10)
e(S, x) < O, for all S = N.
The strict core is the set of all i.r.p.v.'s which satisfy (2.11)
e(S, x) < 0, for S = N, S ¢ ~ and S ¢ N.
(2) e(~, x) is taken as equal to zero. (3) The reader is referred to 12] for a comprehensive introduction to the kernel theory, and to [4] for recent developments. (4) For the coalition structure (N} (see 12]), (5) See [7, page 11 for the origin of this concept.
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THE COMPOSITION OF FUNCTION GAMES
129
3. The kernel of the composition of two games. Let Gt-----(Nt; vt) and G2 = (N2;v2) be two games with disjoint sets of players, (i.e., N t N N2 =J25). The composition(6) of G t and G2 is the game G = (N; v) which satisfies (3.1)
N = Nt k) N2,
and (3.2)
v(S) = vt(S N Nt) + v2(S N N2), for all S c N .
EXAMPLE 3.1. Let G1 and G2 be two three-person constant-sum games in (0, 1) normalization, with disjoint sets of players. In G, the composition of Gt and G2, each pair of players of GI, or of G2, are symmetric. Hence, by [4] Theorem 9.3, the i.r.p.v's of the kernel of G are of the form (x,x,x,2/3 - x , 2 / 3 - x , 2 / 3 - x ) , where x is limited by the requirement m a x ( 1 - 2 x , 2 / 3 ) = max(2x-1/3,2/3), i.e., 1/6<x<1/2. We see that X'(G) contains i.r.p.v's outside the cartesian product .¢d (GI) x ,¢d (Gz)=(1/3, 1/3, 1/3, 1/3, 1/3, 1/3). Thus, as in the set of yon Neumann and Morgenstern solutions of a composition of games, the phenomenon of transfer occurs; it will be the subject of our following investigations. 4. Auxiliary lemmas. We shall prove in this section two simple results on balanced i.r.p.v's, which will be needed in the following section. Let G = (N; v) be an n-person game and c < v(N). For x ~ A(G(c)) (see (2.9)) the following functions are defined (4.1)
f(x) = max{e(S, x) : S c N, S # ~ and S # N},
(4.2)
gi(x) = max{max{e(S, x) : i ~ S, S # N}, c} and
(4.3)
hl(x ) = max{e(S, x) : i 6 S},
where i = 1, ..., n. Observe that hi(x) is non-negative since e(j~, x) = 0. The first lemma is concerned with unmodified ordinary games. LEMMA 4.1. I f G is an ordinary game (see (2.2)) then hi(x) >=gi(x) for all x~.Yd(G),for all i e N . Proof. Assume, per absurdum, that there exist x eX'(G) and i e N such that g~(x) > hi(x). It follows that there exists a coalition S which satisfies i e S and g~(x) = e(S, x) > hi(x). Let j ~ N - S. sij(x) > e(S, x) > hi(x) > sj~(x). Since x is balanced, x~ = 0. Hence
e(S,x)=v(S)-
~, X k = v ( S ) - k~S
Z Xk < v ( N ) k~N
~, Xk = O . k~N
Thus g~(x) < O. But we have assumed that g~(x) > h~(x) > O. Hence, our assumption cannot hold, and the proof is complete (6) See [5], Chapter IX, for the origin of this definition.
130
BEZALEL PELEG
[September
The Second lemma applies to modified games. LEMMA4.2. Let x ~o,~F(G(c)) and i ~ N. I f f (x) > 0 and x i > 0 then gi(x) ~ hi(x). Proof. f ( x ) < max(g~(x), hi(x)). Assume hi(x ) > gi(x); then f ( x ) ~_ h~(x). Our assumption f ( x ) >- 0 implies that f ( x ) > hi(x). Thus f ( x ) = h~(x). It follows that h~(x) = e(S,x) for a coalition S which satisfies S # ~ and i ¢ S . Let j e S'sji(x) > e(S, x) > gi(x) > sij(x). Thus j outweighs i, which is impossible. Hence our assumption hi(x)> gi(x) cannot hold. 5. The Structure of the kernel of a composition of two games. Let G 1 ----(N 1; vl) and G2 = (N2, v2) be two ordinary games with disjoint sets of players, and let G = (N; v) be the composition of G 1 and G2. An i.r.p.v, z ~ A(G) is a real function defined on N which satisfies (2.3) and (2.4). The restrictions of z to Nx and to N2 will be denoted by x and y respectively. The amount of transfer exhibited by z is given by
(5.1)
t = t(z) = vl(Nt) -
~, x~ i~Nj
The games Gl(t ) and G2(-t) (see (2.9)) are related in a natural way to z. Clearly x e A(GI(t)) and y ~ A(G2(- t)). In denoting function of x and y we adopt the following convention: functions depending only on x[y] will bear the superscript " 1 " [ " 2 " ] . Thus, e.g., if i , j ~ N l then s:j(x) is the maximum surplus of i over j with trespect to x (see (2.7)) in the game Gl(t). LEMMA 5.1. Let z ~ A ( G ) and i, j e N l [ i , j e N 2 ] , i outweighs j with respect to z (see (2.8)) in G, if and only ifi outweighsj with respect to x[y] in Gt(t ) [ G 2 ( - t)], where t is defined by (5.1). Proof. Clearly it is sufficient to consider only the case i,j ~ Nt.
su(z ) = max{e(S, z) : S ~ Y',j} = max{e(S, z) : S n NI e ~'ij} = max{e(S, z) + e(R, z) : S ~ NI, SEY-ij and R c N2} = max{el(S, x) : S c NI, S e ~iy} + max{e(R, z) :R c N2}
= s~(x) + max{e(R, z) : R c N2} Similarly sj,(z) = sJ,(x) + max{e(R, z) :R c N2}. Hence s~(x) > s~,(x) if and only if slj(z) > sji(z). Since Xy = z j, the proof follows. COROLLARY 5.2. I f Z E.)[" (G) then x ~ )F ( G l ( t) ) and y e :,'f" (G2(- t)).
LEMMA 5.3. I f z ~ A(G), i ~ N 1 and j ~ Nz then
sij(z ) = g~(x) + h~(y)
(Sy,(Z)= g~(y) + h](x)).
1965]
THE COMPOSITION OF FUNCTION GAMES
131
Proof. Again only the first half of the lemma will be proved.
sil(z) = max{e(S, z) : S ~ }
= max{e(S, z) + e (R, z) : i ~ S, S c N I ,
and R ~ N21j ~ R} = max{e(S, z) : i ~ S, S ~ N~} + max{e(R, z) :j ~ R, R c N2}
= g~(x) + h~(y), (see (4.2) and (4.3)). We remark that g~(x) [hff(y)] is computed here with respect to Gl(t) l-G2(-t)]. THEOREM 5.4. ~ ( G ) n rA(G1) x A(G2) ] = ~3~"(GI) x ~ ( G 2 ) . Proof. Corollary 5.2 implies that -,~(Gl) x X'(G2) ~ oE'(G) ~ rA(Gt) x A(G2)]. Let now x ~ J~(Gt), y E)E'(G2) and z = (x, y). If i and j both belong to Nt, or both belong to N2, then, by lemma 5.1 j does not outweigh i with respect to the game G. So let MeN 1 with xi > 0 and j ~ N 2. By L e m m a 5.3 so(z ) = g~(x) + hi(y) and sj,(z) = g](y) + h~t(x).We shall show that g~(x) > hi(x). To prove this we observe that if f l ( x ) < 0, (see (4.1)) then gtl(x ) = h~(x) = O, and i f f t(x) > 0, then by Lemma 4.2, g~(x) > h~(x). By I_emma 4.1 h2(y) > g~(y). Thus so(z ) > sji(z ) and j does not outweigh i. Clearly we can interchange N t and N2 in the above argument. Thus the proof is complete. The following example shows that our assumption that both G t and G2 are ordinary games is essential for the validity of Theorem 5.4. But since constantsum(7), superadditive(7), or even monotonic(7) games are ordinary games, (2.2) is not too restricting assumption. EXAMPLE 5.5. Let us consider the following two games: ({1,2};u1) , where ul({1, 2}) = 1, and ({a, b, c}; u2), where u2({a , b})= 1 + e , 0 < e <1, u2({a,c}) = O, u2({b, c}) = 0, and u2({a, b, c}) = 1. The second is not an ordinary game. The kernel of the first game is (½, ½), and the kernel of the second is (½, ½, 0). The kernel of the composition of these two games is
,1 4'2
4'2
.1
,)
+ 4'2
+ 4 '0
"
We proceed now to determine the structure of ~ ( G ) . LEMMA 5.6. Let z~o,~(G). I f t < 0 then f l ( x ) >=O. Proof. Suppose, per absurdum, that f l ( x ) < 0. Choose MeN 1 who satisfies x~ > 0. This is possible since t < 0. Our assumptions imply that g~(x)< O. (7) A game (M; u) is constant-sum if u(S) + u(M-S) = u(M), for all S c M. It is superadditive if u(S) + u(T) <=u(S k3 T), for all pairs of disjoint coalitions S and T. (M; u) is monotonic if u(S) <=u(T) whenever S c T.
132
BEZALEL PELEG
[September
Let S c N2 be a coalition with the greatest excess, i.e., e(S, z) >=e(R, z) for all R c N 2. t < 0 implies that S # JZL Let j ~ S.
sj,(z) = g](y) + hi(x) >=g](y) >=h~(y) > h](y) -I- g~(x) = so(z ) . Thus j outweighs i, contradicting our assumption z ¢,XC(G). Hence fl(x) ~_ 0 must hold. COROLLARY 5.7. l f z e X ( G ) and t > 0 then f2(y) > O. LEMMA 5.8. If z~3F(G) and t > 0 then hi(x) > g~(x) for all i ~ N t . Proof. Assume that there exists an i such that g~(x) > h~(x). Let j ~ N 2 be a player who gets a positive payment in z, i.e., zj > 0. Since t > 0 such a player exists. Since - t < 0, G 2 ( - t) is an ordinary game. By Corollary 5.2 y ¢ : ~ ( G 2 ( - t)). Hence I_emma 4.1 implies that hi(y) > g~(y). Combining the above inequalities we have
s,.i(z ) = g~(x) + h2(y) > hi(x) + g](y) = s.u(z) i.e., i outweighs j with respect to z, contradicting our assumption that z ~ :g'(G). Hence hi(x) >= g~(x) must hold for all i e Nt.
COROLLARY 5.9. If Z ~ X(G) and t < 0 then hi(y ) >=g~(y) for all i e N 2. COROLLARY 5.10. If z e J:(G) and t > 0It < 0] then if(x) ~_ t[f2(y) ~_ - t].
Proof. Supposeff(x) < t. If i e N1 then g~(x) = t. Hence
hi(x) ~_max(0,ft(x)) < g~(x), which contradicts Lemma 5.8. Thus fl(x) ~_ t must hold. Although we do not use this result, it is included since it provides an interesting bound on the transfer. LF.MMA 5.11. Let vI(N1)> t > 0, xeOF(Gl(t)), yeoF(G2(-t)) and z = (x,y).
If h~(x) >=g~(x) for all i ¢ Nt and f2(Y) >=O, then z e.~(G). Proof. If i and j both belong to NI, or both belong to N2, then by Lemma 5.1 j does not outweigh i. We have to distinguish two additional possibilities: (a) l e N t a n d j ~ N 2 . If xt = 0 then clearly j does not outweigh i. If x~ > 0 we claim that
d(x) To see this observe that h~(x) >=g~(x) _>-t > 0; hence fl(x) ~_ t > 0. Thus, by I.emma 4.2, g~(x) ~_ hi(x). In addition we have, since G 2 ( - t) is an ordinary game, hi(y ) ~_ g](y) (see Lemma 4.1). Combining the above inequalities
THE COMPOSITION OF FUNCTION GAMES
1965]
133
sij(z) = gii(x) + hff(y) >=hl(x ) + g2(y) = sit(z) and thus j cannot outweigh i. (b) i e N 2 a n d j E N 1. If Yi > 0 then, since f2(y) => 0, g2(y) => h2(y), (see Lemma 4.2). Adding our assumption that h~(x) >= g~(x)
sij(z) = g~(y) 4- hi(x ) >=h2(y) 4- g~(x) = sji(z ) . Hence j cannot outweigh i. COROLLARY 5.12. Let 0>t__>--v2(N2) , xe~r(Gl(t)), y E ~ g ' ( G 2 ( - - t ) ) and z = ( x , y ) . Ifh2(y) > g~(y)for all i~N2 andf~(x) >=O, then ze,,Y'(G). In order to be able to formulate and interpret our results in a satisfactory way we need the following: DEFINITION 5.13. Let H = (M; u) be an ordinary game. For c <=u(M) we define
[ {x : x ~ JT'(H(c)) and f(x) > 0},
c< 0
l
~*(H(c)) = ~ ~ ( H ) ,
c=0
I
L {x : x ~ o,Y'(H(c)) and hi(x) > gi(x) for all i ~ M), u(M) > c > 0 For c < 0Jd*(n(c)) consists of the part of JY'(H(c)) which is outside the strict core of H(c) (see (2.11)). When c = 0 Jl*(H(c)) coincides with.X~(H). When u(M) > c > 0 the interpretation is less intuitive. It seems that in this case :JY'*(H(c)) contains those i.r.p.v's of o,Y'(H(c)) which preserve certain properties which are characteristic of i.r.p.v's of kernels of ordinary games (see Lemma 4.1). We emphasize that when u(M) >=c> O, H(c) is not and ordinary game, and ~Y'*(H(c)) may be strictly contained in ~(H(c)), or even be empty. We now denote (5.2)
D(H) = {c : :¢f*(H(c)) ~ ~ } ,
(5.3)
D(H) = {c : 0 < c <- u(M) A (3x) (x ~ ~ ( n ( c ) ) A hi(x) > gl(x), i e M)}
(5.4)
D(H) = {c : c < 0 A (~x) (x e ,¢d'(H(c)) A f ( x ) > 0)}.
By Lemma 4.1 0 ~ / ) ( n ) ; hence o ( n ) = D(H) t3 D(H). LEMMA 5.14. /3(n) and D_(H) are compact. Proof. /)(H) is clearly bounded. To see that _D(H) is bounded observe that if c < - ( M 2 _ 1)u(M) then JT'(n(c)) is contained in the strict core of (8) H(c) (here M denotes the number of players in M). Since X'(H(c)) is uniformly (8) Let x c.~d'(H (c)) and let i be a player who gets a maximum payoff in x. x, > [ M[ u(M). Let j # i. If S ff 37j then e(S, x) < -- ( [M I -- 1) u (M). Since s,j(x) ~_ sj,(x), we must have xj > u(M). Thus it is clear that x must be in the strict core of H(c).
134
BEZALEL PELEG
[September
bounded when c is restricted to a bounded set, and is an upper semicontinuous function(9) of c, both/)(H) and D(H) are closed. Lemma 5.14 renders the following definitions natural (5.5)
~(H) -- max{c : c e/)(H)}
(5.6)
c(H) = min(c : c ~ D(H)}
LEMMA 5.15. /5(H) [D(H)], and hence ~(H)[c(H)], can be effectively determined. Proof. To determine/)(H) one has to eliminate x out of the formula (5.7)
( 3x)(x ~ .~F(H(c)) A hi(x) > gi(x), i ~ M)
The nature of the inequalities involved enables us to carry out such an elimination (See [31 for a detailed account of such a procedure). Thus (5.7) is equivalent to a set of linear inequalities (with integral coefficients) in c and the u(S), S c M. Adding the inequality 0 < c < u(M), we can represent/3(H) as a finite union of closed intervals(1°) with end points which are linear functions (with rational coefficients) of the values u(S), S c M. LEMMA 5.16. I f n is a two person game then D(H)= {0}. Proof. b ( n ) = {0} and D_(H) c {0} (see (5.3), (5.4)). We now return to the description of the kernel of G, the composition of the ordinary games G1 and G2. THEOREM 5.17. Let G be a composition of ordinary games GI and G2; then the kernel of G is given by (1~) (5.8)
y c ( c ) = u { ~ * ( c l ( t ) ) × y r * ( G ~ ( - t)) :t ~ o ( 6 1 ) c~ - o(G2))
(see Definition 5.13 and (5.2)). Proof. Corollary 5.2., Theorem 5.4., Lemma 5.6, Corollary 5.7, Lemma 5.8, Corollary 5.9, Lemma 5.11 and Corollary 5.12. COROLLARY 5.18. 7he maximum amount, according to ~r(G), which(is given by N1 to N2, is (5.9)
i = min(?(Gl), - c(G2)) (see (5.5), (5.6)),
and the maximum amount which is given by N2 to NI, is (5.10)
t = min(~(G2), - c(G1)).
(9) See [1] Theorem 3.1. (10) We conjecture that/)(H) ID(H)], and hence D(H), is an interval. (11) -D(G2)= { c : - c~D(G2)}. See also the footnote above, from which it would follow that the range of t is an interval.
1965]
ON THE COMPOSITION OF FUNCTION GAMES
135
COROLLARY 5.19. I f G1 (or G2) is a two person game then (5.11)
~f~(G) = ~ ( G a ) x ~¢F(G2)
Proof. Theorem 5.17 and Lemma 5.16. We remark that the results for non-ordinary games may be drastically different, and many degeneracies may occur. 6. Composition of three-person games, The first non-trivial case of a kernel of a composition of games is the case of two three-person games (see Example 3.1 and Corollary 5.19), which will be studied in this section. Let 13 = (N; v) be an ordinary 3-person game. To compute c(G) we recall that the kernel of a three-person game(12) consists of a unique point(J2], Section 4) which is contained in the strict core, if the strict core is not empty ([4, Theorem 5.4]). The condition that the strict core of the game G(c), c <__O, is empty, is (6.1)
v(N) - c < v({1, 2}) + v({l, 3}) + 0({2, 3}) =
2
Hence, if D(G)lis not empty, i.e.,(6.1) is satisfied for a non-positive value of c, then (6.2)
c__(G)= v(N) _v({1, 2}) + v({1, 3}) + v({2, 3}) 2
To compute e(G) observe that if c > 0, c~D(G), and x z~(G(c)), then gl(x) = max(e({id}, x), e({i, k}, x), c) > 0, where N = {i,j, k} ; hence hi(x) = e({j, k}, x). Since h (x) >=gi(x), i e N, we have (6.3)
e({i,j}, x) = e({i, k}, x) = e({j, k}, x) >=c
A straightforward computation shows that there exists an i.r.p.v, x ~ A(G(c)) that satisfies (6.3) if and only if (6.4)
c = v({1,2}) + v({1, 3}) + v({2, 3}) - 2v(N).
Hence (6.5)
e(G) = max(0, v({1,2}) + v({1,3}) + v({2, 3}) - 2v(N))
Recalling that c(G) < 0 is a necessary and sufficient condition for G to have an empty core, we have THEOREM 6.1. Let G be an ordinary three-person game. I f the core of G is empty then D(G) consists of the interval
if the core of G is not empty then D(G) = {0}. (12) Not necessarily ordinary.
136
BEZALEL PELEG
[September
6.2. Let G t be an ordinary three-person game with a non-empty core. I f G 2 is an ordinary game, then the kernel of the composition of Gt and G2 is equal to oV'(Gt) x Jd(G2). COROLLARY
Proof. Theorems 5.17 and 6.1. We remark that Corollary 6.2 cannot be generalized to games with an arbitrary number of players. The counter examples that we have at present require long computations, and therefore will not be given here 7. Comparison with the results for solution theory. In this section we shall compare the results we have for the kernel of a composition with the results for the yon Neumann Morgenstern solutions ([5], Chapter IX). In order to carry out our program we need the following. LEMMA 7.1. Let H = (M;u) be a superadditive(la) constant-sum game, and let 0 < c <- u(M). I f x ~ ~r(H(c)) then hi(x) > gi(x) for all i ~ M. Proof. Suppose there exists an i ~ M such that gi(x)>h~(x). Since u ( M - { j } ) = u(M) for all j e M , there is a coalition S ~ M such that gt(x) = e(S,x) (see (4.2)). Furthermore, f ( x ) = gi(x) (see also (4.1), (4.3)). Let j e M - S. sij(x) ~_ e(S, x) > hi(x) > sji(x). Hence xj = 0. Thus g,(x) = e(S, x) = u(S) - ~, Xk = u(S) k~S
,
~, Xk < u(M) - (u(M) - c) = c. k6M
Let now h ~ S with(14) xn > 0 and R = M - {h}. e(R, x) = u (R) -
~. x~ = u ( M ) keR
Z
xk > c.
keR
Since f ( x ) >--e(R, x), we have a contradiction and the proof is complete. COROLLARY 7.2. Let H = (M;u) be a superadditive constant-sum game. For 0 ~_ c ~_ u(M)JY'(n(c)) = ~ * ( H ( c ) ) (see Definition 5.13); furthermore , D(H) is equal to the interval [0, u(M)] (see (5.3)). We now recall a definition of yon Neumann and Morgenstern ([5], p. 369). Let H = (M;u) be a superadditive constant-sum game. An i.r.p.v, in A(H(c)) is (fully) detached if it belongs to the (strict) core of H(c). We remark that (fully) detached i.r.p.v's exist only if c < 0 [c < 0]. Let now GI = (Nt;Vl) and G 2 ---"( N 2 ; v 2 ) be two superadditive constant-sum games with disjoint sets of players, and let G be the composition of GI and Gv Combining Theorem 5.17 and Corollary 7.2 we get (10 (13) This assumption is not used in the proof. (t4) If no such player exists then c = u(M) and the proof is immediate. (Is) See Section 5 for the notation.
1965]
THE COMPOSITION OF FUNCTION GAMES
t37
COROLLARY7.3. An ix.p.o, z e X'(G) if and only if the following three,conditions are statisfied : (7.1)
x e X'(Gi(t)) and y e J~'(G2(- t))
(7.2)
x and y are not f u l l y detached
(7.3)
- vz(N2) < t < vt(N1)
Corollary 7.3 is to be compared with the results of von Neumann and Morgenstern for solutions ([5], p. 401). In our opinion the similarity deserves notice, since it is seen that the i.r.p.v's of the kernel and the von Neumann Morgenstern solutions obey the same composition rules. The comparison seems to us legitimate since both the balanced payoffs and the solutions specify the payments to the players(16). We remark that the situation for non constant-sum games may be different. To take an extreme case let us consider the following example. EXAMPLE 7.4. Let Gl = ({1,2,3};vl) be a constant-sum three-person game in (0,1) normalization and let G2 = ({a, b, c} ; v2) be given by v2({a, b}) = v2({a, c}) = v2({a,b,c})= 1, and vz({b,c}) = 0. If G is the composition of G1 andG2 then by Corollary 6.2, o,~d(G)= X'(G1) x o,~f(G2), while the solutions of G are obtained by forming all possible cartesian products of solutions of Gt(t) and G z ( - t ) for - ½ < t < 0. A discussion of the solutions of a composition of two general sum games( t 7) can be found in [6]. The main result is: THEORE~ 7.5. Let G = (N;v) be a composition of two (general sum) games G1 = ( N t ; v t ) and G2 = (N2;v2); V is a solution of G i f and only if it has the following f o r m : V = V 1 x 1"2, where Vi and V2 satisfy: (7.4)
V1 is a solution of Gl(t) and V2 is a solution of G2(-t).
(7.5) Gl(t)
and G 2 ( - t ) have no f u l l y detached i.r.p.v's;
and
(7.6)
- v z ( N 2 ) < t < vt(Nt).
Example 7.4 shows that the range of the transfer determined by the kernel may be strictly contained in the range determined by solution theory. We remark that the opposite case does also occur. So, no general conclusions can be made and a detailed investigation is called for. (1~) The determination of the payments by a solution is, of course, generally not unique. (17) Not necessarily ordinary.
138
BEZALEL PELEG REFERENCES
1. R. J. Aumann, B. Peleg and P. Rabinowitz, A method for computing the kernel of n-person games, Research Program in Game Theory and Mathematical Economics, The Hebrew University of Jerusalem, R-MS, 1964. 2. M. Davis and M. Maschler, The kernel of a cooperative game, Econometric Research Program, Princeton University R-M 58. 1963. 3. L. Hodes, Dense linear order and linear inequalities, Research Paper, IBM Corporation, Thomas J. Watson Research Center, Yorktown Heights, N. Y., 1963. 4. M. Maschler, and B. Peleg, A characterization, existence proof and dimension bounds for the kernel of a game, to appear in Pacific Journal of Mathematics. 5. J. yon Neumann, and O. Morgenstern, Theory of games and economic behavior, 2nd edition, Princeton University Press, 1947. 6. B. Peleg, Composition of general sumgames, Econometric Research Program, t'rinceton University, R-M 74, 1965. 7. L. S. Shapley, and M. Shubik, The core of an economy with non-convex preferences, The Rand Corporation, Santa Monica, California, R-M = 3518-PR, 1963. THE HEBREW UNIVERSITY, JERUSALEM, ISRAEL
U N I F O R M L Y NON-1 t" ORLICZ SPACES* BY
KONDAGUNTA SUNDARESAN ABSTRACT
A characterisation of uniformly non-l(nI )Orlicz space is obtained intrinsically in terms of the Young function determining the Orlicz space. It is shown that a uniformally non-10 )Orlicz space is reflexive. In a recent paper James [1] conjectured that a uniformly non-l~(t)Banach space is reflexive. Here it is proposed to establish the conjecture when the Banach space is an Orlicz space. We also obtain an intrinsic characterisation of uniformly non-l(,1) Orlicz spaces. We start with the basic terminology and definitions required in what follows. Let (X, S, it) be a non-atomic measure space and • be a non-zero Young function. We adopt the convention O(u) = 0(I u I ) if u is a real number. The Orlicz set L~ is the set of all real valued #-measurable functions f such that M(f) = SxO(f)d# < oo. It is known, Weiss [5], that L® is linear if and only if • satisfies the growth condition (I)(2u) < Kt~(u) for large values of u((1)(2u) < kO(u) for all u > 0) if #(X) is positive finite (if/~(X) is infinite). Further the linear space L® can be equipped with a norm It II by setting Iif II = inf {1 ~ i~>0
and M(¢f) < = 1} .
The normed linear space (L¢, II ll) is indeed a Banach space and we denote this Banach space by L~. For a detailed account of this class of Banach spaces we refer to Luxemburg [2], Weiss[5], and Zaanen [6]. REMARK 1. We assume throughout thae paper that • satisfies one or the other of the growth conditions ensuring that L® is linear. Thus, in particular (1)(u) is finite for all real u. Since (I) is convex continuous function it follows that if u > v > 0 and O(u) ~ 0 then O(u) > ¢~(v). REMARK 2. With regard to the functions M ( f ) and f we note that M(f) < 1 if and only if [[fl[ 1 and M(f) = 1 if and only if f = 1. Remark 2 is an easy consequence of the definitions. DEFINITION 1. (JAMES [1]). A normed linear space B is uniformly non-l(~~) (n > 2) if there exists a positive number 6 such that for any n elements xl, "",xn in B with [[ x, I1 < 1 it is true that Received September 6, 1965 • This work was supported in part by Air Force Grant G2A-152015. 139
140
KONDAGUNTA SUNDARESAN
[September
- ( x l _+ x2_+'"_+x.)[ _-<1 - / i n
for some choice of signs. REMARK 3. If a positive number ~ exists satisfying the above definition then it is clear that 1 > 1 - t5 >_ 1 In. In the case when the normed linear space B is the space L* the definition 1 may be reformulated as follows. DEFIr~mo~ 2. L* is uniformly non-l~t j) l<~
M(~fl+-f2+'"t-f"
if for some positive number ~,
) <
for some choice of signs where f l , "",f, are n functions in L~ such that M(fi) < 1. The equivalence of definitions 1 and 2 in the case of Banach spaces L~ follows from Remark 2. With regard to expressions of the form u 1 +__u2 +__"" + u, where {ui}~'=l are n real numbers we adopt the following notation. For a given n-set of reals {u~}in__1 once for all we enumerate the 2"-1 possible expressions of the above form and designate them as Ea,E2,'..,E2.-,, and for K, l < K < 2 "-~ we denote by Ux the n-vector (ux,__+u 2 , ' " , + u,) where the signs in front of the u~ are the same as those occurring in front of u~ in E x. We define the permutations P , 1 < t < n, of any n-vector Y=(y~,...,y,) by setting PtY=(p,yl,ptyz,...,pty,) where PtYj = Yi+t-l(mod,) for 1 < i < n. Further TtUk = signrtPtU K where sign rt = 1 or - 1 according as the sign in front of ut in E K is + or - . We define the functions S and S~ on the n-dimensional space R "by setting S(U)= ~ = ~O(u~) and S~(U;~)= ~'t~ ( ~ u' +- ''' q-u" where ~ is a real number, U = (ul, "", u,) and the summation in the definition of $1 is over the 2 " - t possible choices of signs. We denote by CIU the ith coordinate of the vector U. We proceed now to obtain characterisations of uniformly non-l~ 1) L* spaces. We present our characterisations separately in the cases (i)#(X) is infinite and (ii)#(X) is positive finite. THEOREM 1. If (X,S,p) is an infinite non-atomic measure space then L*~ is uniformly non-l~l)if and only if there exists a real number ~, ~ > 1, such that 2 n- 1
S(U) if ui>=O.
S~(U;~)__< n
1965]
UNIFORMLY NON-1 gl) ORLICZ SPACES
141
Proof. Suppose the function ~ satisfies the inequality in the theorem for some > 1. Let {fi}i"=~ be n functions in L * such that M(fi) < 1. The inequality above clearly implies
(
2 M ~ f~ +-f2 +"" +.fn
~
n
M(f~) <=2~-~ n
i=1
where the summation of the left side of the inequality is over the 2 " - l possible choices of signs. Thus there is a choice of signs for which
M(,
fl+J2+'"+f")
= 1.
Hence by definition 2, L~ is uniformly non-l(.1) . Conversely let L* be uniformly non-l(. 1). Let ¢ be a real number assured by definition 2 for such a space. If possible let there be a n-vector U = (u~,-..,u,) with coordinates nonnegative reals such that 2 n- 1
sl(u;~) > n
s(u).
Since • is convex function satisfying the growth condition ¢ ( 2 u ) < K ¢ ( u ) for u > 0 there exists a positive number 2 such that ~(~ u) < 2¢(u) for u > 0. Hence the inequality above implies S(U)> 0. Since (X, S,p) is an infinite non-atomic measure space there exist 2"-1 pairwise disjoint measurable sets {Ai}2=~ 1 such that
p(Ai)- 2 . _ 1nS ( U ) • Let tSAm1 i j .n, = l , for 1 - < i < 2 " - 1 , be a measurable partition of A~ such that 1
p(AT) - 2._ 1 S(U) Now we define n functions {f~}i%1 in L* by setting 2n-1
fi = Z K=I
~ CIT~UK~At K. t=l
It is verified that
M@ f' + f2"4-'" +--f" ) n
nS,(U;~) = 2,_aS(U)
> 1
for any combination of signs while M(f,) = 1 for 1 _< i < n. Thus a contradiction on the choice of ~ is obtained and the p r o o f of the Theorem is complete. We proceed to the case when 0 < p(X) < oo. We state first a definition and establish some auxiliary lemmas.
142
KONDAGUNTA
SUNDARESAN
[September
Following the terminology in Nakano 13] the Banach space L* is said to be uniformly finite if sup
M ( K f ) < oo
M(f)< 1
for any positive real number K. LEMMA 1. If L* is uniformly finite then if (f,},_>_i is a sequence of functions intheunit ballofL* such that 1 a s n ~ oo then M ( f , ) ~ 1 as n ~ oo. The proof of the lemma is an immediate consequence of Th. 4 on p. 224 in Nakano [-3].
IIf, I1-
LEMMA 2. I f L* is uniformly finite and if there exist real numbers t and tl 0 < t, tl < 1, such that if (f~},"=1 are any n functions in the unit ball of L* with M(fi) >__1 - t for 1 < i < n then for some choice of signs M ( fx + ' " + - f "
) < 1 -
L* is uniformly non-l~ 1) . Proof. If L* is not uniformly non-1,(t), then for any sequence of reals ~i such that ~j > 1 and ~j ~ 1 there exist n sequences {fiJ}j____1, 1 < i < n in the unit ball of L* with the property 1 --
f / +- "" +-f~
<
Cj
-
< 1 =
n
for all choices of signs, Thus lim j~oo
f~ +- "'" +~ ~
~
1
n
for all choices of signs. Since IIf II =< ~ each of the sequences {f~J }j---l admits a subsequence {giJ}j~l 1 < i < n, such that lim 11g,J"II-' 1 for i, a ~ i _ n, and j~oo
lim ]lgl ±"" --+g2 II- 1 j--, oo
n
for all choices of signs. Since L* is uniformly finite, by Lemma 1 M(g•) ~ 1 as j ~ oo for 1 < i < n, and lim M ( g / - I - . . . - I - g ~ ) = 1 j--, oo
n
for all choices of signs, a contradiction on the inequality in the hypothesis.
1965]
UNIFORMLY
NON-ltn 1) O R L I C Z S P A C E S
143
I f (X,S,I~) is a positive finite measure space and if there are two positive numbers 2, ~ and v with ~ > 1 such that ~(~u) < ~,¢~(u)for u > v > 0 then L* is uniformly finite. LEMMA 3.
Proof. Let K be a positive number. I f f is in the unit ball of L* and t is a positive integer such that e l > K and if e
(xlx x, If(x)l_-
=
M(KT) < M(~tf)
= f o(¢7)d. +
fx.y )d.
< O(~tv)p(E) + 2tM(f)
=< ~(~'v) ~(x) + ).'. Thus sup
M ( K f ) < ~ . Hence L* is uniformly finite.
M(/') ~ 1
Tnv.OR£M 2. I f (X,S,#) is a positive finite non-atomic measure space then L* is uniformly non-l~,1) /f and only if (i) There exist positive numbers C and ~, 1 < 4, such that 2"- 1
S~(U;~)< (ii)
~-
n ¢
S(U) if S ( U ) > C > O a n d q~(vo)if ~(Vo) -
C ~ U > 0 for l < i < n .
#(X) "
ProoL Let us assume that L* is uniformly non-l~(1). Then definition 2 guarantees the existence of real number Go > 1 such that if {f~}~ 1 are any n functions in the unit ball of L* then for some choice of signs M
o
n
We shall prove that ~ fulfills the condition (i) with the choice of ~ = Go. Assuming n = 1 such that the contrary there exist non-negative numbers {U~}~ n
2"- 1
S(U) > /z(X---~ and SI(U;~) > ~ n Since
S(U).
n
~(x) >= s(u) there exist 2"-1 pairwise disjoint measurable sets tYAij~ = 1l such that ~2,n
/~(A~) = 2,_1S(U) for 1 < i < 2"-1
144
KONDAGUNTA SUNDARESAN
[September
Now we can complete the proof by constructing the functions {fi}7= i as in theorem 1 contradicting the choice of ~o. Next we shall show that
di)( v° n ¥ ) # !,(Vo). If possible let Vo) +
n
= 1¢(%).
Since*(v o) =
n
n
and X is a nonatomic measure space there exist n pairwise disjoint measurable sets {~}i=lA ~ such that da(Vo)l~(Ai) = 1. Let f i = vozA, for 1 _< i _< n. Then clearly M(f~) = I[fi I[ = 1 for 1 -< i < n and for all choices of signs
since
/')0 by Remark 1. Hence a contradiction arises on the choices of Go. Next we shall prove that if • satisfies the inequalities (i) and (ii) then L* is uniformly non-l~ 1). If • satisfies inequality (i) and ~(v~) = C then by choosing the vector U such that CiU = u > Vl it is verified that ~(¢u) < 2 " - l ~ ( u ) for all u > Vl. Thus L* is uniformly finite by Lemma 3. Since Vo)
* n
#n
1 ¢(0o)
and • is continuous it follows by Remark 1 that there exists a real number vl such that 0 < vl < Vo, ~(v~) > 0, and # nl ~(01) # ~1 ~(Oo). Let~(vl) = 0~(Vo) = 0--/,(X) where 0 < 0 < 1. In the condition (i) we can assume that C > n [#(X). Let K = U I U e R" and/~(.~) < S ( U ) < C . We note that S I ( U ; 1)<(2 n-1/n) S ( U ) for U ~ K as a consequence of Remark I and of the inequality ~(v 1/n) < (1/n) ~(ol). Since • is continuous and K is a compact subset of R n there exists a real number ~1 > 1 such that
UNIFORMLY NON-1(1) ORLICZ SPACES
19651
2"-- 1
S~(U;~I) < ~
n
S(U)
145
U~K.
forall
Let Go = min(~l,¢). Then for all U such that
S(U) >_
0/7 ,
n-1
SI(U;¢o)< 2 n S(U).
Let t be a real number such that 1 - t > O. We shall prove that there exists an q, 1 > r/> 0 such that if {f~ }~'=t are n functions in the unit ball of L~ with M(f3>l-t
for l < i < n
then for some choice of signs
M ( fj+f2+'''+-f~n Let E = {x I I2~'=1 ~(f,(x)) > ~M
( fl+f2+'"n
)<
1-
q.
Oq/p. Then we obtain the following inequality. ---f~ ) = ~
fr ~( fl + f2 + ''' +
) d#
Lo( :,+-..+-:.. )..
+I2
<= Go t_2"-' n ~"fr ~(f~)du + 2"-' fx n
< 2" - 1 -
2"-1 1 - 1 ]~o]
(1
- t-
~E
tb (fi)d#
0)-..(I)
since L
0~/ • (fi)d# > n(1 - t) - -~-X~-#(X ,,~ E) n(1
-
t)-
nO.
Setting 1 - q = ( 1 - t - 0 ) ( 1 - 1 / 4 0 ) we obtain from inequality (1)that for some choice of signs
M(fl+-f2+-'"+-f"n
)<
1-
q.
Hence L* is uniformly finite and the associated function M(f) satisfies the inequality in Lemma 2. Thus L* is uniformly non-l,(, 1) and the proof of Theorem 2 is complete. James [1] has shown that a uniformly non-1 °) Banach space is reflexive if it has an unconditional basis. Since it is not assumed that ~uis separable, the Banach space L* is not necessarily separable, Luxemburg [2]. However, it will be shown that
146
KONDAGUNTA SUNDARESAN
every uniformly non-l~ 1) Orlicz space L* is reflexive. We observe first a consequence of the inequalities in the statements of Theorems 1 and 2. If for some ~>1, 2 n- 1 S I ( U ; ~) ~_
n
S(U)
for every n-vector U with non-negative coordinates then choosing U such that C~U = u > 0 and CiU = 0 for 2 < i < n it is verified that (A) for u > 0,
o ( u ~) ~=l @(u). Similarly if n-1
SI(U ; ~) <=2
S(U)
n
for U such that S(U)__>C > 0 then there exists a v > 0 such that (B) for all
u>=v>O,
() If • is the Young's complement of O, and if • satisfies (.4) or (B) then there exists a constant K > 0 such that ~(2u) < K~(u) for all u > 0 or ~(2u) < K~(u) for large values of u according as (A) or (B) is true. T~OREM 3. IlL* is uniformly non-!~1) then it is reflexive. Proof. Since • and • satisfy the growth conditions ensuring that/.,0 and L , are linear it follows from theorems 4 and 5, Rao [4] that L~ is reflexive.
REFERENCES 1. James, Robert C., Uniformly Non-square Bannch Spaces, Ann. Math. 3, Vol. 80 0964), 542-550. 2. Luxemburg, W., Banach Function Spaces, Thesis., Technische Hoge School te Delft, 1955. 3. Nakano, H., Topology and Linear Topological Spaces, Maruzin and Co., Tokyo, 1951. 4. Rao, M. M., Linear Functionals on Orlicz Spaces, Nieuw Archief Voor Wiskunde 3, XI (1964), 77-98. 5. Weiss, G., A Note on Orlicz Spaces, Portugaliae Math. 15 (1956) 35-47. 6. Zaanen, A. C., Linear Analysis, North Holland Publishing Co., Amsterdam, 1956. CARNEGIE INSTITUTEOF TECHNOLOGY PITTSBURGH, PA, U.S.A.
ON FINITE DIMENSIONAL SUBSPACES OF BANACH SPACES BY
A. LAZAR A N D M. ZIPPIN* ABSTRACT
The common Banach spaces are investigated with respect to some properties of their finite dimensional subspaces. 1. It is well-known (by the Hahn-Banach theorem) that for each element x in a Banach space X one can find a functional feX* such that Ilfll--1 and x ll. The following natural question arises: Given a finite dimensional subspace E c X, is it possible to find a finite dimensional subspace F c X* such
f(x)--II
that for each x e E IIx II -- sup,, s~ I f ( x ) I ? In this paper we show that the answer is negative, and investigate some similar properties concerning finite dimensional subspaces. We prove that the spaces co(S) and real Lp, where p is an even integer, satisfy the above-mentioned condition, while C and Lp for all other p's do not satisfy it. The terminology and notations are generally the same as in [1]. Sx denotes the closed unit ball of the Banach space X. If F is a subspace of a conjugate space X* then F.L = {x:xeX,f(x) = 0 for all f~F}. If E c X then E ~-= {f:f~X*, f(x) = 0 for all x e E}. 2. Let us discuss the following conditions on an infinite dimensional normed space X, concerning its finite dimensional subspaces: (1) For every finite dimensional subspace E c X there exists a subspace F = X* such that for each x e E 1[x II -- supy~ sF If(x) l and F_L is infinite dimensional. (2) For every finite dimensional subspace E c X it is possible to find a finite dimensional F ~ X* such that for each xEE llxll= supsos lf(x)l. (3) For every finite dimensional subspace E ,-- X there exist a finite dimensional subspace G = X such that E c G and a projection PG of X onto G with II II = 1 Denote by ~¢~ the class of all normed spaces satisfying condition (i) i = 1,2,3. It is obvious that if we require in (1) that F is w*-closed ~¢1 will not change. Of course, d l -~ ~ 2 ~ ~¢3. In §4, 7 we shall show that ~¢1 ~ ~¢2 ~ ~¢a. The following two lemmas give equivalent conditions to (1) and (2) respectively. Received July 15, 1965. * The contribution to the paper of the second author is part of his Ph.D. thesis prepared at the Hebrew University of Jerusalem under the supervision of Professor A. Dvoretzky. The authors wish to thank Professor A. Dvoretzky for the interest he showed in the paper and for his helpful advice. 147
148
A. LAZAR AND M. ZIPPIN
[September
LEMMA 1. Let X be a normed space; then X e ~ 1 if and only if for every finite dimensional subspace E ~ X there exists an infinite dimensional closed subspace G c X satisfying the following conditions: (a) G n E = {0}. (b) If H is the subspace spanned by the elements of E and G, then there exists
a projection Plz of H onto E along G.
(c) IIe~ll : J. Proof. Necessity: Take for G the subspace F L. If x e G O E then Ilxll=supsos~lf(x)l=0, hence x = 0 ; if x = e + y , eeE, y ~ G then I l e + y i l > s u p s ~ s ~ l f ( e + y ) l = s u p f ~ s ~ l f ( e ) l = l l e l I . Hence, the transformation P~ defined by Pr.(e + y) = e is a projection of H onto E along G and since IIe II --< II e + Y II, II P~ II -- 1 Sufficiency: For every f e E* define f(e + y ) = f(e) for each e + y e H. Now, f e H*, and f l y ) = 0 for every y e G. By (b) and (c) II/ll~ = suPlle+,ll=l f(e + y) = SUPllell__
flY
I[x[l=sup~s~lf(x)[
LEMMA 2. Let X be a normed space. Then X e ~ 2 if and only if for every finite dimensional subspace E there exists a closed subspace G ~ X satisfying the conditions (a) (b) and (c) of Lemma 1 and condition, (d) The co-dimension of G in X is finite. Proof. The necessity is clear by the proof of Lemma 1. Sufficiency: By [1] p. 25, Lemma 1, if T is the natural mapping of X onto X/G, defined by T x = x + G, then T* is a linear isometry of (X/G)* onto G 1 n X* = F. Since, by (d), X/G is finite dimensional, so is F. By (b) and (c), if x e E I x + G = inf,, a x + y > [ x [. On the other hand I1x ÷ c z x, hence x I = x + G . For every x ~ E there exists a functional ge(X/G)* such that I[gl[ = 1 and g(x + G) = [1x + G IIBut T * g e F , and
T* is isometric
so
II Z*g I[ -- IIg 11 - 1 ~t follows that for each
x ~
E
Ilxll = suplf(x)l. f e SF
REMARt~ 1. The class ~¢a is not empty; every Hilbert space space belongs to ~¢a. REMARk: 2. According to I3] if X e ~¢2, and if for every n dimensional subspace E c X the dimension of F is also n then X is a Hilbert space.
1965]
ON FINITE DIMENSIONAL SUBSPACES OF BANACH SPACES
149
REMARK 3. If X e "~2 then every subspace Y ~ X also belongs to .~2. But as shown in §4 there exists a Banach space X and a closed subspace Y c X such that X e d I while Y~ ~ 1 , 3. We have mentioned that 12 ~ ~ . Let us discuss now other examples of members of d 3 . A Banach space X is called polyhedral (see [4]) if every finite dimensional subspace of X has a polyhedron as its unit ball. J. Lindenstrauss proved in [6, p. 100, Corollary 2] that every polyhedral Banach space X for which X** is a PI space belongs to ~t 3 . In fact he proved that every finite dimensional subspa~ of X is contained in a finite dimensional subspace which is a PI space, that is, isometric to l~ for a suitable n. The following converse implication is also a consequence of the theory of J. Lindenstrauss [6]: LEMMA 3. I f X ~ a~ta and X** is a Pa-space then X is polyhedral. Proof. Let E c X be of finite dimension. There exists a projection P of X onto a finite dimensional subspace G which contains E, such that IIP -- 1 From [6] p. 16 Corrollary 3 it follows that G is a PI space, hence, it is isometric to a space lo~, so the unit ball of E is a polyhedron. In connection with polyhedral spaces, let us prove
II
LEMMA 4. I f a Banach space X is polyhedral than X ~,~t 2. Proof. Let E c X be a finite dimensional subspace. Denote by f l , f 2 , "",fk the extreme points of Sr,, and by f j any Hahn-Banach extension of f j to the whole space X; 1 __<j < k. Obviously, any element of E attains its norm on the unit ball of the subspace spanned by fl", f2 ,"" ,fk. In [4] V. Klee proves that Co is polyhedral. Since m is a Pl space, it follows from the preceding remarks that Co e ~ a . We shall give here a direct proof of a slightly stronger result for Co. Let {ek}k°°=1 be the unit vectors basis in Co. Let us denote by Et the closed subspace spanned by the first k unit vectors: Ek = [el, e2,'", ek]. E k will denote the closed subspace [ek+l, ek+2, ""]. Denote by Pk the projection of c o onto Ek, defined by Pk( Y?~=x~iei) = zk=l ~,e,. It is known that II Pkll = I1' - ek 11= x for all k. TrmOREM 1. Let e < 1/2 be a positive number, and E a finite dimensional subspace of c o. Then there exists a finite dimensional closed subspace G of co with the following properties: (a) E = G = co . (b) I f dim(G) = n then there exists a linear isometry T orE n onto G. (c) The transformation Pc = TP~ is a projection of c o onto G and ]JPc ]l = 1.
(d) Ill- Poll = 1 + (e) ( I - Pc)(Co) = E".
150
A. LAZAR AND M. ZIPPIN
[September
Proof. If E is k-dimensional, then by [7, Theorem 2] we can find a basis Xl,X2,'",Xk in E and functionals f~,f2,'",f~ in X* such that f/(x~)= tSi~ and i1~,II = II~,11-- 1 for 1 _<e_< k If x j = ~,~to~{ei is the representation of xj with respect to the usual basis {ei}i~t then an integer n can be found, such that the following conditions are satisfied :
I. Io~/[ < e . k - ~ f o r all i > n and l < j < k. II. The elements x] = P,x~ = Y~'~=~ ct/e~ 1 < j < k are linearly independent. Let us denote by E' the subspace spanned by x;, x~,.-., x~. Let U be the transformation of E to E' defined by U(Xikl~iXi) = ~EiklfliXi '. U is a linear one-to-one transformation from E onto E'. Let us show that U is isometric.
IfII
,P,x,[I = 1 then
'
(
0t~e, i=1
=
)t
:
j=l
~, fljo~{
sup l
)ira
~ /~j~{ e, i
= sup l~_i
j=l
•
flj~[
.
j=l
The last equality follows from the following considerations: Since ItfJ II-- 1 and II xL,~,x, II = 1 it follows < 1. Hence, by I, for i>nt~p=z/~ia/l<<supl~j___~l/~ i .(]Ey=,lc~ / ) < k . e-k-3= e'k-2< 1/2. k j But supl__.,<~{I X;=l/~jcq l} = 1, therefore the sup is attained from 1 < i < n. We have just proved that
thatll&l=lfj(Xf=,~,x,))
1 = sup
~ Pjo~i
l~_t<=n
=
j=l
1 i=1
e,
"=
k
It follows that U is a linear isometry from E onto E'. E' is a k-dimensional subspace of E,, so there exists a projection Q of E, onto E' with 11o 11---
x'
n
fie
n
k
+ a,/) -- sup,_~,<~ (I x;=,~j~/+ ~,11, where for i < n
t 0 for i > n
j
1965]
151
ON F I N I T E D I M E N S I O N A L SUBSPACES OF B A N A C H SPACES
(The last equality follows from II). But sup 1 ~ i < oot,
X fljct{+S,I j=l
=
X fljxj+~,
3,e,
=
T
i=1
~, 7iei
,
i=1
so we have proved tllat if II X,~=l~,eill = 1 then II T( •;=,7,e,)ll = 1. A similar method yields the converse implication. An immediate conclusion is the following: G = T(E.) is n-dimensional and T is a linear isometry from E11 onto G. The definition of T ensures that E = T ( E ' ) c G. n e n ~o If, again, ~.i=17i i = ~,.,j=tej~ x ' ~ + ~i=ltSiei then P11TP11(~,t=l~te~) = 11 I1 (~ k t tl P.T( Zi: 1 riei) = P.( E~: 1 pjxj + ~,i= 1 ie~) = ~'~:lfljXj ÷ ~ l = l tSie~ = ~,'~=1)'iei = P11( ~,i~ 1 Yiei), hence P~TP. = P., and so T P ~ T P , = T P . . It follows that P c = T P , is a projection of Co onto G = T(E.), with IIeo/l-(for 1 < 11 <= 11zll" I1P1111: 1). Let us remark that the definition of T and the same methods yield the following inequality: For each ~,~%:y~e~eE11 11 e ~-T(~7=~Y~e,) < e ~'=~y,e, . It follows that I1~,=1T' 1-TP11 = I-Pn +P11-TP~ < 1 - P ~ + P11 " 1 1 o - T
zeoll
COROLLARY 1. Theorem 1,with slight modifications in its formulation, remains true if we replace e o by co(S), where S is an infinite set. In fact, if E c co(S ) is of finite dimension, then there exists a sequence S o c S such that x(s) = 0 for every x e E and s E S - So. Denote by the subspace {x : x ~ co(S), x(s) = 0 for all s ~ S - So}. The natural projection P of co(S) onto H is of norm 1, and so is I - P. H is isometric to co , so by Theorem 1 there exists a projection PG of H onto a finite dimensional closed subspace G which contains E, with IIPG II = 1. PoP is a projection of norm 1 of co(S) onto G. By the same Theorem 1, G is isometric to a space l~. It is easy to see that since 111 - Po II 1 ÷ (Io is the identity on H) tl I - P~PII --11¢Io - Po)P + 1 PII-- max{H(Io - Pc)PI], ] 1 - P I I } < m a x { 1 + e,1} = 1 + e. REMARK 4. The properties (a) to (e) do not characterize c o. It is easy to prove Theorem 1 for the space (R1 @ R2 I~)R 3 "")co = X instead of Co. (Rk denotes the k-dimensional euclidean space.) J. Lindenstrauss proved in [5] that X ts not isomorphic to Co. 4. Now we shall show that the space c has not the property (1). From this will follow that no infinite dimensional space C(S) with S compact Hausdorff has the property (2). This proves also that, as was expected, none of the properties discussed here is invariant under isomorphisms.
152
A. LAZAR AND M. ZIPPIN
[September
Let a ={at}t°21 be an element of c such that at ~ aj if i ~ j and 0 < at < 1, ao = lim~_.~ at ~ aj for j = 1,2,3,..-. We shall consider the plane determined by the null vector, the given a and the vector {(1 - -~ _2xl/2~ ) st= I. The family of vectors [{ana t + ( 1 - a ~ ) I / 2 ( 1 - a,2)t/2}L~] ,, = 0 , 1 , 2 , - - - i s contained in the plane and the only point of the unit ball of 11 at which the nth vector o f the family attains its norm is the n + lth unit vector of 11 . This follows from the fact that the nth coordinate (the limit)of{anai + (1 - a~)x/2(1 - a2Xl/2"~°°t J st = l is 1 where n >= 1 (n = 0) and the other are positive but less than 1. Hence, the smallest closed subspace of l~ on the unit ball of which every vector from our plane attains its norm is It itself. The same plane taken this time in the space m shows that this space also has not the property (1). (We use the fact that 1 is w*-dense in m*). Considering in C[0,1] the subspace spanned by f l ( x ) = x, f2(x) = 1 - x 2 it can be proved that C[0,1] does not satisfy condition (1). Now we are able to display a space which belongs to M1 but fails to belong to M2. If X is a space which does not belong to ~¢2, the space Y= (/2 ~ X ) t~ is the desired one. Indeed, the conjugate of Yis (/2 • X*)m and if E is a finite dimensional subspace of Y any element of E attains its norm on the unit ball of (P(E) ~ X*)m c (12 ~ X*)m, where P denotes the natural projection of Y onto 12. The annihilator of P ( E ) ~ ) X * is the orthogonal complement of P(E) in 12 which is obviously infinite dimensional. But Y fails to have the property (2) since its subspace X has not this property. (See Remark 3). If we suppose that X does not belong even to . d t, like the space c for instance, we see from the above example that property (1) of a space is not inherited by its subspaces. 5. We shall prove now that the property (1) is not valid in 11. Thus no infinite dimensional L-space has property (2). Let a = {at} ~ 1, b = {bt}~-°=1 be two elements of l~ such that
at a~+~ b--~> bt+l
,
bi > 0
for every i and consider the subspace generated by them. For a given natural number n let ~n be a scalar such that (a~/bn) > ctn > (an + 1)/(bn + 1). The vector a - ~nb attains its norm on the unit ball of m at the point {sign(at-ctnbi)}i=l and we have: sign(at - ~bt) = 1 sign(at - ~bi) = - 1
l
Choosing a scalar ~o such that ~o > (al/bl) the vector { a t - ~obt}t~l will attain its norm only at the point ( - 1, - 1, - 1, ...) of Sin. The smallest w*-closed sub-
1965]
ON FINITE DIMENSIONAL SUBSPACES OF BANACH SPACES
153
space of m which contains all the points of m found above is m itself, and this proves our statement. The space L l [ 0 , 1 ] fails also to have property (1). To see this it is enough to take the subspace generated by f l ( x ) = 1 and f2(x) = x and to use the fact that the characteristic functions of intervals form a total family over LI[0,1]. One may ask if the completion of a normed space satisfying one of the conditions (1), (2), (3) must satisfy this condition. The answer is negative since the linear subspace of 11 generated by the unit vectors is a member of d 3 while its completion fails to belong even to ~¢tTHEOREM 2. The spaces lp 1 < p < oo, where p is not an even integer, do not belong to the class ,~2. THEOREM 3. / f (S,E,p) is a measure space, the real space Lp(S,Y~,#)= Lp with p an even integer belongs to d 2 . In proving these theorems we shall constantly make use of the known facts that the function f ( t ) of Lp(p > 1) atains its norm on the unit ball of Lq(1/p+ 1/q = 1) at the point I f -P " f ( t ) p-lsign f(t) and only at this point. We shall refer everywhere to f ( t ) P - l s i g n f ( t ) since this function belongs to the same subspaees of Lq as f l 1-p. f ( t ) P-lsignf(t) does. Proof of Theorem 2. Let n be any integer greater than p. (If p is not an integer n may be any integer greater than 1). We shall consider the two dimensional subspace E n of l~, spanned by an = (1, 1,... 1) and bn = (1,2,3, ...,n) and we shall show that there is no proper subspace of lq which contains all the points of the unit ball having a support hyperplane generated by an element of En. Assume that n is an odd integer. Denote P,(2) = (I + ).i) p-I = I 1 + 2liP-1
1 < i < n.
Since a, + 2bn e E, for any real 2, in order to prove the assertion it will be enough to display n independent vectors of the the form {Pl(2) sign (1 + 2), P2(2) sign (1 + 22),...,P,(2)sign(1 + n2)). We shall give to 2 n different values submitted to the conditions:
2i>0
l
1 .<2p+~< n -j
1 n-j+l
l <j
154
A. LAZAR AND M. ZIPPIN
[September
The corresponding vectors are the rows of the matrix: P1(2,),
P2(21),""
, P.(21),
P1(~2),
P2(22), "'"
, P.(22),
Pl(2p),
P2(2p),""
, P~(2p), ,P.- i(2p+i)),-
P.(2p+I),
P1(2p+2), :
P2().p+2),"',P.-2()~p+2),-P.-1(2p+2), -
P.(2p+2) ,
PI(2.),
P2(2.),'",Pn(2.),- Pp+1(2.) . . . . .
P1(21,+ 1), P2(~p+ I),"" :
P.(2.),
If its rank is less than n, there exist n scalars Pk 1 --- k -< n such that
[ II
~ [.lkPk(~Li)= 0 k=l n-j ~, ~lkPk(2p+j) -k=l
] <-- i <- p
~.~ l=n--j
Since the degree of Pk(2) is p III
-
-
~ l P l ( 2 p + j) = 0
1 < j < n -- p
1, we deduce from I that
~, /~gPk(2) -~ 0 k=l
for any 2. Taking in II j = l, in III 2 = 2p+ 1 and subtracting one equality from the other we get #p÷l =0. Continuing in this way we get pp+i-:O for l < j < = n - p . Hence, III may be written as P
IV
~, #kPk(2 ) = O. k=l
But it is easy to see that these polynomials are linearly independent, and this proves our assertion about IpII where p is an odd integer. Now suppose that p is not an integer. The proof of our assertion in this case will be achieved if we show that n positive numbers 2 1 , 2 2 , " - 2 , can be found such that D,(21,22,...,2,) = det(1 + i2j) p-1 ~ 0. For n = 2 it is easy to check that if 21 ~ 22 then D2(21,22)~ 0. Suppose that n - 1 positive numbers 21,2. 2,-..,2._ 1 were found such that
V
D,_1(21,22,,..,2,_1) # 0
but VI
Dn(~.1,22, "",~.,_ 1,2) # 0
for any 2 > 0. Differentiating VI n - 1 times we get
1965]
ON FINITE DIMENSIONAL SUBSPACES OF BANACH SPACES (1 + ).1)p-I, (1 + 221) p-l, ...
(1 + n21) p-1
(1 + 2~_~)P-~,(1 + 22,_~)P-1, .-.
(1 + n2,_ ~)P- '
(1 + 2) p - l - J , 2J(1 + 22) P - l - { ...
n J(1 + n2)p - I - j
155
=0
j = 1,2,...,n - 1. These equalities together with VI form a linear system of equations the coefficients being the elements of the last rows of the determinants. From V we deduce that this system has non-trivial solutions. Therefore (1 + 2)P-1, (1 + 2)~)p- 1,...
,(1 + hA)p-I
(I + 2)P-2,2(1 + 22) p-2, ...
,n(1 + n2) p-2
(1 + 2)P-', 2"-1(1 + 2) p-n, ...
n "- 1(1 + n2) p-~
=0
for any 2 > 0. This equality is obviously false and, consequently VI cannot be true for any 2 > 0. Now we shall construct a two-dimensional subspace of Ip with the property that its elements attain their norms on the unit ball of l~ at points which span an infinite dimensional subspace of lq. This plane is spanned by the vectors a and b constructed as follows: The nth block of n coordinates in a is n -2
• ( 1 -k
2 p + 3 p + ... + nP)-l/p- (1,1, 1, ..., 1),
and in b is n-2.(I+2P+3P+...nP)-I/P.(1,2,3,...,n), for n = 1 , 2 , 3 , . . . . From the above discussion we know that this plane has the required property. P r o o f of Theorem 3. We shall prove that for any n-dimensional subspace F of Lp there exists a subspace F of L~ of dimension Cn+p-2 n-, such that Ilxll =supf~sFlf(x)l for any xeE. Let xl(t), x2(t),...,xn(t ) be a basis of E. We have to detemine the dimension of the space spanned by the functions ( ~El~12~xi(t)) p- 1 = [ ]~,= 1~.,x,(t) Ip- 1. sign ( ~'=~ 2~x,(t)). These functions are linear combinations of the functions x k~' Xzk=""Xk" for all possible non-negative integers kl, k2,..., kan which satisfy ~ =n 1 k ~= P - 1. Any function of this type belongs to L~ (this can be seen if we use the fact that Lq is a lattice) and there are n--I Cp+n-2 such functions. Consequently all the functions (]~'=1 ix~(t))p-1 lie in a n-- 1 subspace of L~ which has the dimension Cp+n-2. An immediate consequence of Theorem 2 is the fact that no space Lp (#) with p ~ 2k, k = 1,2, 3,.-. satisfies condition (2).
156
A. LAZAR AND M. ZIPPIN
7. Let us discuss now reflexive Banach spaces X for which Sx, is strictly convex. By ['8] and ['2], the last property is equivalent to the following one: Every bounded linear functional f defined on a subspace E of X has one and only one HahnBanach extension f. LEMMA 5. I f a reflexive space X belongs to ~ a , and Sx. is strictly convex then X * • ~ a .
Proof. Again by [7], given a finite dimensional subspace F c X*, one can find a basis f l , f z , ' " , f k for F and functionals X l , X 2 , . . . , x k in X such that 1 and fi(xj)=tS,j,1 <_ i<_ k 1 < j < k. Denote by E the subspace of X spanned by x l , x 2 , . . . , XR. Since X • ~¢a, there exists a subspace G c X satisfying the following conditions: (a) G is of finite dimension, (b) E c G, (c) There exists a projection Pc of X onto G.
llx, ll--IIJ, II--
(d) IIP II = 1. Denote (I - P~) (X) = H. Each x • X has a unique representation x = y + z where y • G and z • H. For every element ~b• G* define:f~,(y + z) = q~(y). f~ is an extension of 6, it is linear and by (d) = lY,(y + z)l
llf ll
= suPll,ll----11 l = [I I[. Denote by M the subspace {fa: ~b• G*} of X*. It is easy to see that M is isometric to G* and so it is of finite dimension. Moreover, there exists a projection Q of X* onto M, with [[ Q 1[ = 1. For 1 _< i _< k f~ • M, because otherwise the restriction ~bi of f~ to E would admit two Hahn-Banach extensions - f~ and f~,. But this is impossible by the hypothesis of the lemma. It follows that F c M. COROLLARY. For p = 2n (n an integer >=2) L p • ~¢2 but L p q ~ d a . Proof. If L p • ~ a then L q • ~ a where 1 / p + l / q = 1 . But by Theorem 2 Lq ~ ~¢a, since q = pip - 1 = 2n/(2n - 1) is not an integer. REFERENCES M. M. Day, Normed linear spaces, Springer Verlag, 1958. 2. S. R. Foguel, On a theorem ofA. E. Taylor, Proc. Amer. Math. Soc. 9 (1958), 325. 3. S. Kakutani, Some characterizations of euclidean space, Jap. J. Math. 16 (1939), 93. 4. V. Klee, Polyhedral sections of convex bodies, Acta Math. 103 (1960), 243. 5. J. Lindenstrauss, On some subspaces of co andll, Bull. Res. Council of Israel, 10F (1961), 74. 6, - ., Extension of compact operators, Memoirs Amer. Math. Soc. 48 (1964). 7. A. E. Taylor, A geometric theorem audits application to biorthogonal systems, Bull. Amer. Math. So•. 53 (1947), 614. , The extension of linear functionals, Duke Math. J. 5 (1939), 538. 1.
A NOTE ON ERLANG'S FORMULAS BY D.
MEJZLER
ABSTRACT
A service system consisting of n = oo lines, is considered. Formulas (2) giving the probabilitiesof various states of the systemwere obtained by the assumption that the incoming stream is stationary, orderly, and without after-effects.It is proved in this note that these formulas hold as well when the incoming stream is not orderly. 1. Introduction. We confine ourselves to the terminology use in Khintchine's monograph l1 ]. Let us consider a service system consisting of n lines into which enters a random stream of calls. The durations of the call service are assumed to be identically distributed random variables which are independent both of each other and of the current of the incoming stream. Denote by Itk (t) the probability that exactly k lines are occupied by the service at the moment t. If the incoming stream is simple (a Poisson stream) then the limits 7~k = lim ~k(t), (k = 0, 1, .-., n) t'-'~ O0
exist and are independent of the initial probabilities ~k(0) (0-< k < n). These limits are given by the formulas (1)
7Zk =
(Us)" / ~ (u~) k--T- li i!
(k = o,1, "'" ,0
=0
where # is the intensity of the stream and s is the mean duration of the service of one call. Formulas (1) were first obtained by Erlang [2] under the assumption that the duration of the service of a call is exponentially distributed; however, it was shown after wards ['3]- [5] that they remain true for an arbitrary distribution of the service duration. Khintchine has also studied Erlang's problem for n = ~ . He proved that in this case the following meaning must be attributed to Erlang's formulas (1): (2)
~Zk= e-"~ (m)k k!
(k = 0,1,...) "
The aim of our note is to show that the last formulas are correct also for the case when the incoming stream is an arbitrary stationary stream without aftereffects (i.e. not necessarily orderly) with a finite intensity /~ and the distribution H(x) of the duration ~ of service of one call (3)
H(x) = P(~ > x),
H(O) = 1
Received October 23, 1965 157
158
D. MEJZLER
[September
satisfy the only natural restriction (4)
0< s= -
f3
dH(x) =
H(x)dx < oo.
2. An auxiliary proposition. We will need the following elementary proposition from the theory of limits: LEMMA. Let the function f(x) be differentiable in the closed interval [a, 1] and infinitely differentiable in the half closed interval [a, 1), and let in addition
f(k)(x) => 0
(5)
(k => 2; a =< x < 1).
Let dp (t) (t > O) be a function such that lim ¢(t) = s > 0.
(6)
t "* O0
Then as t--+ oo (7)
t{f(1) - f [ 1
¢(t) ]} ~
and for every positive integer n
f~"+l)[1-~blt)] /t"~O.
(8)
Proof. From (6) follows that for large enough t a <
1 -
¢(t___)) <
t
1.
Using the Mean Value Theorem we obtain
where 0 < 0 < 1. The expression (7) follows now, since by assumption (5) the derivative f'(x) is continuous from the left at x = 1. Let us prove (8) by induction on n. f"(x) is not decreasing by (5), hence f'(1)-f'[1-¢(t)]>t
=¢(t)f"[1-¢(t)]t
/ t >0;=
these inequalities prove (8) for n = 1. Again, from (6) follows that for large enough t and therefore
¢(20 < 2¢(t)
g i f t ) < 1 - ¢(2t) t 2t On the other hand, f~k+ 1)(X) is non decreasing by (5). Thus another application of the Mean Value Theorem will yield 1
ftk+x)[1- ~'~] 2t
t
=
2
"
t
> O, r
..
~
,
1965]
A NOTE ON ERLANG'S FORMULAS
159
hence afortiori
f(k) [1
t~) ]
f(k+~)[l_flp(t)] >
2~b(t)- ~b(2t) 2k
(2t)k- 1
t
> O.
tk
Thus we proved that if (8) holds for n = k - 1 then it holds also for n = k. 3. Proof of formulas (2). We prove the validity of the formulas (2) under assumption that the stream of calls began at the moment to = 0, i.e., no(0) = 1 (The proof would be only a little more difficult for arbitrary preliminary data). Let v(t) denote the probability that the service of a call that occurs in the time interval [0, t) will be determined in the same interval. It will be shown that
v(t) = 1 - -i-
(9) and therefore
H(x)dx
v(t) is independent of the number of calls which occur in the interval
[0, t). For the proof, let us assume that in the interval [0, t) exactly m calls occured and let r/be the moment that one of the calls, taken at random, appeared. Clearly
v(t) = PQI + ~ < t),
(10)
where ~ is the duration of service of a call. In [6] it was shown that if the stream is stationary and without after-effects then, whatever is m, the random variable r/is uniformly distributed in the interval [0, t). Since by our assumptions the variables t/and ~ are independent, equation (9) is easily deduced from (3) and (10). Denote by P,,(t) the probability that exactly m calls will occur in the interval [0, t). Then, by the above remarks, oo
n~(t) = ~,
Pm+k(t)("'~k)[1-v(t)]kv'(t)
m=O
or nk(t) _ [ 1 - v(t)] k £ vm(t). k[ m=O m!(m+k)!Pm+k(t)" Consider the generating function of the stream oo
F(t, x) = ~, Pm(t)x ra,
( I x ] ~ 1).
m=O
Since (m + k) lPm+k(t) =
8m+kF(t,O) 0xm+~
it is easy to see that the probabilities 7~k(t)
~m ( ~kF(t,x) ) -
~x m .
can also
dx k
[~=o,
be expressed by
160
D. MEJZLER
n (t)
[September
= [ 1 - v(t)] k t~kF[t,v(t)]
k!
Oxk
Finally, if we put (11)
q~(t) = f / H(x)dx,
we conclude from (9) that
qbk(t) 10kF[t,l--qS(t)/t] (12)
nk(t) =
k!
t~
~x k
It is known, [1], that the generating function of a stationary stream without after-effects has the form
r(t,x) = exp{2t[f(x) - 1]},
(13)
where 2 > 0 and p~ > 0, ( ~i~ z P~ = 1) are constants and oo
f(x) = ~, pix',
(14)
(Ix[ < 1).
i=l
Moreover, the intensity of the stream is given by GO
/t = ~ ipi = 2f'(1).
(15)
i=1
Assuming that the intensity is finite we get (16)
1 < f ' ( 1 ) < co.
It is easy to see that the partial derivatives of the generating function (13) are given by (17)
OkF(t'x) = F(t,x) {[2tf'(x)] k + k~- I (2t)m[ ~A(iD'",ir,) ~xk
m=l
~I s=l
fti')(x)]},
where the inner summation is over various systems of positive integers (il, "', i,) that satisfy (18)
il + ... +im = k,
and A(il, ..., ira) are constants that depend only on the indices (iz, "', ira). (Note that it ts possible to give the more detailed formula
k (2t)r~ ~ fti,)(X ) dkF(t.__._k_L,X)ox= k ! F(t, x),~=1 ~ . {Z ~=~1= is .' "' where the summation is over all the ordered systems of positive integers satisfying (18); for our purpose formula (17) is sutficient). Taking into account that in view of (18)
19651
A NOTE ON ERLANG'S FORMULAS
fl
16i
ti'- 1 = i k-ra,
s=l
we conclude by (12) and (17) that (19)
kt nk(t)=F(t,x) ~{[2f'(x)] k +'-'~___l~.rtt[~A(il,."" im)fi: : , f(~s'(X)II t'.-' JJ'
where we put for brevity x =
1
~(t) t
Our functions f ( x ) and ~b(t) satisfy all the conditions of the Lemma: f ( x ) by (14) and (16), and ~b(t) by (4) and (11). Since f(1) = 1, from (6), (7) and (13) follows that
limF[t,l--~-]
=e -~''°)
--- e - ' .
On the other hand, if 1 _< m _< k - 1, then every system of positive integers (il, "", ira) that satisfies equation (18) contains at least one integer i > 2. Therefore, because of (8), &
lira | 1
t-*oo s=l
tis-1
= 0
for all the systems (il, "--, ira) that paticipate in (19). Thus the second term in the braces of the expression (19), being a finite combination of terms that tend to zero, also tends to zero. This proves formulas (2), since lim ckk(t) = sk t--~ O0
and
REMARK. Formulas (2) can be applied to a slightly more general situation. Let a finite or countable number of stationary streams without after-effects xi(t) with intensity /~i (i = 1,2, ...) enter a service system consisting of n = ov lines. Assume that the distribution of the duration of the call service is the same for all calls in the same stream but this distribution depends on the index i and it may vary from one stream to another. Denote by st the mean duration of the service of a call which belongs to the stream x~(0, and assume that st < oo for every i, and ¢o
i=l
oo
i=I
I62
D. MEJZLER
Since the total stream x(t)= ~,~1 x~(t) is also finite, stationary and without after-effects, then formulas (2)hold also for it. It is clear that #and sare determined by co
1
oo
REFERENCES 1. A. Ya. Khintehine, Mathematical methods in the theory ofqueueing (translated from the Russian), Charles Griffin, London, 1960. 2. A. K. Erlang, Solution of some problems in the theory of probabilities of significance in automatic telephone exchanges, Post Office Electrical Engineer's Journal 10 (1917-18), 189-197. 3. A. Ya. Klaimchin¢, Erlang's formulas in the theory of mass service, Theory of Probability and its Applications (an English translation of the Soviet Journal) "/(1963), No. 3, 320-325. 4. B. A. Sevast'yanov, An ergodic theorem for Markov processes and its applications to telephone systems with refusals, Theory of Probability and its Applications (an English translation of the Soviet Journal) 2 (1957), No. 1,104--112. 5. R. Fortet, Calcul des probabititds, Centre National de la Recherche Scientifique, Paris, 1950. 6. D. Mejzler, On a characteristic property of stationary streams without after-effect J. Appl. Prob. 2, (1965), 2 455--461. THE HEBREWUNIVERSITYOF JERUSALEM
SETS OF CONSTANT WIDTH IN FINITE DIMENSIONAL BANACH SPACES BY
H. G. EGGLESTON ABSTRACT
In Euclidean space a set of constant width has the property that it is not a proper subset of any set of the same diameter. The converse implication is also true. Here we show that if Euclidean is replaced by n-dimensional Banach space the direct statement is true, but the converse statement is false. Attention is drawn to the problem of characterising those Banach spaces of finite dimension for which the converse is true. Introduction. In Euclidean n-dimensional real space, a set X is said to be of constant width, if and only if, it is convex, compact, and the distance apart of any two parallel support hyperplanes is constant (i.e. the same whichever pair of parallel support hyperplanes we consider). A number of equivalent properties are known of which the most important (in Euclidean space) is that a set X is of constant width if and only if it is complete, that is to say, if Y is a set of which X is a proper subset then the diameter of Yexceeds that of X. Instead of"complete", which is used in many other ways, we shall use the phrase "diametrically maximal".
[4] The object of this note is to develop analogous properties in n-dimensional Banach spaces [see 2.3]. In particular we show, in contradiction to accepted belief [see 1], that the concepts "constant width" and "diametrical maximality" are not equivalent in every Banach space. It is always the case that constant width implies diametrical maximality. The reverse implication is not necessarily true even when the unit sphere of the space is both smooth and rotund. It is not easy to see what conditions on the space will ensure equivalence of these two concepts but a simple example can be given of a non Euclidean space in which this equivalence holds. Many of these equivalent conditions have been known previously (See [2]). Notation. Denote an n-dimensional Banach space by B" and its unit ball by S. For two parallel hyperplanes nl, n2 in B n the width between nx and n2 is twice the largest number 2 such that a set obtained from 2S by translation is contained in the strip bounded by nl and nz-A convex compact subset X of B n is of constant width if and only if the width between each pair of parallel support hyperplanes of X is constant. We shall use the phrase " o f constant width" to imply that X Received October 12, 1965. 163
164
H.G. EGGLESTON
(September
is convex and compact. The vector domain of a set X, denoted by Xv is the set of all vectors of the form xl - x2 where Xx and x2 vary indopendently in X. We use + between sets to denote vector sum. Thus Xv = X + ( - 1 ) X . Also the width of A + B between two parallel support hyperplanes n~, It2 is the sum of the widths of A and of B between corresponding pairs of hyperplanes all parallel to ltl and it2. Finally (A + B)v = Av + By. Except where the contrary is expressly stated every set is assumed to be convex and compact and to contain interior points. The origin of Bn, the centre of S, is denoted by 0. We use the same symbol for points and for vectors. The symbol xy will be used either for the line xy or the segment xy or the length of the segment xy. We shall use B(S) to indicate the Banach space with the central convex set S as its unit sphere.
1. Some properties equivalent to "constant width". (A) X (assumed to be convex and compact) is of constant width if and only if for some 2 > 0 X v = 2S. If X is of constant width then so are ( - 1 ) X and Xv = X + ( - 1 ) X . In any case X v is central thus it is sufficient to prove that a central set of constant width is a sphere. These two properties of Xv mean that for some fixed 2 > 0, 2S is supported by each pair of parallel support hyperplanes of Xv. Since a convex compact set is uniquely defined by its support hyperplanes, X v is 2S. On the other hand if X v is 2S then X v is of constant width. But the width of X between any pair of parallel support hyperplanes 7q, n2 is equal to the width of ( - 1 ) X between support hyperplanes parallel to 7tI and 7~2, and thus is half the width of Xv between two support hyperplanes parallel to ~ and n2. Hence X is of constant width. (B) A subset X of B" has the support intersection property if, and only if, given any pair of parallel support hyperplanes n~ and n2 of X, and one of the support hyperplanes of S, say ~r, parallel to nl and It2, then to any point p ~ n n S we can find x ~ n l r ~ X and x2en 2 n X s u c h that line x~x2 or x2x~ is parallel to the line op. X is of constant width if and only if it has the support intersection property. If X is of constant width then for some 2 > OX v = AS. Given nl,n2 and n all parallel hyperplanes of which the first two support X and the last S, let p ~ n n S. Then 2p e Xv and thus 2p = x~ - x2 where xx ~ X, x2 e X. Also x~ and x2 belong one each to n~ and ~2 (for 2p is a frontier point of Xv and there is a hyperplane of support to Xv at 2p parallel to n). Thus X has the support intersection property. Support next that X (assumed to be compact and convex) has the support intersection property. Let p be a point of the frontier of S, and let the half ray terminating at 0 and containing p meet the frontier of X v in q. If the hyperplane
1965]
CONSTANT WIDTH IN FINITE DIMENSIONAL BANACH SPACES
165
supports S at p then 3 xt, x2 on the frontier of X such that line xlx2 is parallel to op and there exist hyperplanes supporting X at xl, x2 and parallel to n. By definition X v contains the point xl - x2 and it is a frontier point of X r (hence it is q or - q ) through which passes a hyperplane parallel to n supporting X v. Thus S and X v are two central convex sets such that any half ray meets their frontiers at p and q respectively and if p lies on support hyperplane n of S then q lies on a parallel support hyperplane, of X v. We show that this implies that for some 2 > 0 Xv = 2S. It is convenient to consider B" as represented in Euclidean space E" with an appropriate metric. It is sufficient to show that if 0 Plq t and 0 P2q2 are two half rays terminating at 0 with PD P2 on the frontier of S and qtq2 on the frontier of Xv then 0 Pl [0 ql = 0 P 2 / O q2. The plane , = plane of o PlP2, o qlq2 meets S in a set say S' and X v in a set X~. The sets S', X v as two dimensional convex sets have the same property as do S and Xv. For if p is any point on the frontier of S' in z and h is a support line to S' at p in z, then there exists a support hyperplane nx to S at p meeting z in h. If o p meets frontier of Xv in q then there exists a support hyperplane of Xv at q parallel to nl and thus meeting z in a line parallel to h. This line supports X~, at q. In what follows for the remainder of this paragraph we consider only subsets of z. Select a fixed half line ox through o and measure angles in a fixed sense from ox. Let h(O) be the half line through o making an angle 0 with ox. 0 is the angle in E" which with the appropriate metric represents B ". Let h(O) meet the frontier of S' in p(O) and that of X~, in q(O). Denote the length (in E") op(O) by f(O) and oq(O) by g(O). Then if 1 0 1 - 02[ < n and K = sup f(0), ¢
(1)
If(00 - f(02)I < p(OOp(02) <
f(0a)sin 101 - 02 I = Isin/0 p(01) p(02) l
K]Ot - 0 2 [ sin/op(Oa)p(02)l"
The angles [op(O1)p(02) are such that there exists 6 > 0 with the property that [ 0 1 - 0 2 [ < 6 implies [sin/op(OOp(02)l > 6 . It follows from (1) that f(O)is absolutely continuous. Similarly so is g(O) and finally f(O)/g(O) is absolutely continuous, f'(O) and g'(O) both exist for almost all 0. The condition of parallelism of the support lines at p(O) and q(O) implies that for almost all 0
f'(O)
g'(O)
f(o)
g(O)
166
H.G. EGGLESTON
[September
i.e. f(O)/g(O) has a derivative equal to zero almost everywhere. Thus f(O)/g(O) is a constant. This implies that Xv = 2S i.e. X is of constant width. (C) The coincidence normal property We say that a line h is normal to a hyperplane ~ if there is a sphere whose center lies on h, whose frontier passes through p, the unique point of intersection of h with n, and which is supported by rc at p.(It is always assumed that h does not lie in 7r). If p is a frontier point of the compact convex set X then h is a normal to X at p if and only if h is normal to at least one of the support hyperplanes of X at p.
X is of constant width if every two parallel normals of X coincide. The coincidence of every two parallel normals of X implies that X has the support intersection property and therefore is of constant width. The converse of this result is false. Indeed, if S is a non-rotund convex set then there exist non-coincident parallel normals to S in B(S) and S is certainly of constant width in B(S). On the other hand if S is rotund and smooth the reverse implication is true. I f S is rotund and smooth and X is of constant width then every two parallel normals of X coincide. Since S is rotund and 2S = X v it follows that X is rotund. Let NI, N2 be two parallel normals of X. Select p on the frontier of S such that op is parallel to NI and N2. Let n be the support hyperplane of S at p (n is uniquely defined because S is smooth). Since X has the support intersection property we can find two points of X, x~, x2 one each on the two support hyperplanes of X parallel to re, such that the line xlx2 is parallel to op and thus to N~ and N2. If the lines N~, N2 do not coincide at least one of them does not coincide with xl x2. Suppose that N1 is distinct from xlx 2. By the definition of normality and the smoothness of S, N~ is normal to a hyperplane of support of X parallel to zq and z~2, say rq. But then rq contains two distinct points of X and X is not rotund. A contradiction which establishes the required result. (D) The spherical intersection property Let S(p, r) be the sphere of B n obtained by applying the translation 0 to p to the sphere rS. For any set X (compact and convex) let the diameter of X be D(X) and define the set X s by X s = f~ S(p, O(X)). pcX
In any case X s ~ X. We shall say that X has the spherical intersection property if and only if X s = X.
I f X is of constant width then X has the spherical intersection property
1965]
CONSTANTWIDTH IN FINITE DIMENSIONAL BANACH SPACES
167
If p ~ X there exists a hyperplane n separating p from X. On the two parallel support hyperplanes of X, 7zl, Ir2 there is a point q of X on that one of these hyperplanes most distant from X. The strip bounded by nln2 contains a sphere of radius ½D(X). Thus S(q, D(X)) is separated from p by re. Hence p ~ X s and we have Xs c X . This establishes the required result. In Euclidean space the converse implication is known to be true see [4], but this is no longer the case in Banach spaces generally. For example let U be a regular tetrahedron in 3 dimensional space. Uv is a polyhedron with 8 triangular and 6 parallelogram faces. Now let K be a central polyhedron with centre 0, containing Uv but such that the 8 triangular faces of Uv lie in the frontier of K, and such that K is not Uv. In the Banach space B(K), U is not a set of constant width since Uv ~ 2K for any 2 > 0. On the other hand if the vertices of U are a, b, c, d then the set obtained from Uv by the translation 0 to a, contains U and lies so that the face bcd forms the translate of one of the triangular faces of Uv. It follows that v =
fl
Vv(p, 1)
p=a,b,c,tl
and the same is true with Uv replaced by K. Thus U ~ Ux, U has the spherical intersection property in B(K) but is not of constant width in B(K). A particular case is that in which K is a regular octohedron. It is natural to suppose that this situation arises because K is either not smooth or not rotund or both. A slightly more difficult example shows that this is not the case. Before explaining this example we introduce the last property. (E) The diametrically maximal property A set X is diametrically maximal if and only if p ~ X implies that D(p WX) > D(X).
A set X is diametrically maximal if and only if it has the spherical intersection property. If Xs = X and p ~ X then p (EXs and thus p ~ S(q, D(X)) for some q e X. Hence pq > D(X) and D(p L) X) > D(X). If X is diametrically maximal and p e X s then D(p u X ) = D(X) .'. p e X . Thus Xs c X and X has the spherical intersection property. I f X is of constant width then X is diametrically maximal. The converse implication is not universally true. This follows from (D) and the immediately preceding result. A set X is diametrically maximal if and only if every frontier point of X is distant D(X) from at least one other point of X. If X s = X then a frontier point x of X is a frontier point of S(p, D(X)) for some p e X for otherwise by compactness x would be an interior point of Xs. If Xs ~ X then 3p e Xs, p ~ X. Let q e interior of X. Then pq meets frontier
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H.G. EGGLESTON
[September
of X in a point x that necessarily belongs to the interior of X and hence of Xs. Thus x is distant < D(X) from every point of X. 2. An example of a space B(S) where S is smooth and rotund but there exist diametrically maximal sets that are not of constant width. We give next an example of B(S) where S is smooth and rotund and yet B(S) contains a subset X which is diametrically maximal without being of constant width. In three dimensional Euclidean space let a, b, c, d be four points all at the same (Euclidean) distance R from one another. Let X be the intersection of the four balls whose centres are a, b, c, d and whose radii are R. The set X is bounded by four portions of the surfaces of these balls and these four portions meet, in pairs, in six arcs each containing two of the four points a, b, c ,d and each of radius R,~/3/2. Of these arcs let that which joins a to b be v(a, b) and that which joins c to d be ),(c, d). Let p be the mid point of ~(a, b) and let q be the mid point of y(c, d). The points p, q, c, d lie in the plane which bisects perpendicularly the segment ab. Let line pq be perpendicular to the line cd. Thus the plane ~rt perpendicular to the line pq and passing through q meets plane pqcd in a line tangent to y(cd) at q. It follows because of the symmetry of X about the plane pdqc and the convexity of X that E~ supports X at q. Similarly a parallel plane ~2 supports X at p. Thus p and q are diametrically opposite points of X. Since the length of segment pq is R(~/3 - 1/a/2), and that of pd is R, the fact t h a t / d p q = lr/6 implies t h a t / p d q < 7r/2. Thus the plane through d perpendicular to pd does not separate p from q and the parallel plane which supports X is at a positive distance from q. Since a similar argument applies with d replaced by c we conclude that there are positive numbers p, 6 such that if the frontier point x of X is within a distance p of p then x is diametrically opposite to some point whose distance from q is greater than 6. Given x E FrX let P(x) be the set of points diametrically opposite to x and f(x) be the largest value of the acute angle made by pq with xy, y ~ P(x). (such largest value existing because P(x) is compact). Write 2( = i n f f ( x ) x efrX. Our aim is to show that Z > 0. Suppose on the contrary that X = 0 and that {x,} {y,} are sequences of points such that y.e P(x.) and the angle between pq and x,y, ~ 0 as n ~ oo. We can by considering appropriate subsequences assume that x, ~ x and y. ~ y as n ~ oo. Then xy is parallel to pq. There are parallel support planes to X through x and y. This is possible only if x, y lie on the support planes to X at p and q which in turn implies that x is p and y is q or vice versa. But then if the distance of x. from p is less than p that of y, from q is greater than 6: this contradicts the fact that x. ~ p and y, ~ q. Thus our assumption is false and in fact X > 0. Next consider Xv. Xv is rotund since X is rotund. Moreover Xv fails to be smooth if and only if there are two diametrically opposite points s~, s2 of X and
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CONSTANT WIDTH IN FINITE DIMENSIONAL BANACH SPACES
169
two pairs of parallel support planes of X lh, n2 and z~, z 2 such that sl lies on both n 1 and z~ whilst s2 lies on both 7~2 and %. Two such points would necessarily lie either at the points a, b, c, d or on the arcs such as ~(a, b) which connect these points. If s~ were a point of arc 7(a, b) not a or b, then planes nlzt would intersect plane abq in a line 21 tangent to ~(a, b) at sl. Now plane abq meets the plane containing ?(cd) in line pq. If s2 lies in this last plane then since n 2, z 2 intersect in a line which both lies in this plane and is parallel to 2t it follows that 21 is parallel to pq. This is impossible because the centre of ~(ab) lies on segment pq and its arc length is less than n. Similarly if s2 lies on the plane containing arc ?(bc), then 21 and 22 are parallel to the line joining b to the median point of triangle acd. This is not possible. Similarly s2 cannot lie on arc 7(ad). It is obvious that s2 cannot lie on arc 7(ab). Thus this case cannot arise and we are left with the case when each of sl, s2 is one of a, b, c, d. Suppose for example that sl is a and s2 is b. The planes nt and n 2 through a and b respectively and both perpendicular to ab are support planes of X. Consider parallel lines 21, 22 through a and b respectively lying in n~ and in n 2. On at least one site of the plane spanned by lines 21 and 22 there lies an arc ~(ax) terminating at a and lying both on the frontier of X and on the frontier of the ball centre b and radius R. The reflection of this arc in the plane that perpendicularly bisects ab also lies in X and terminates at b. But of two parallel planes z~, z 2 one through each of 21 and 22 one at least must cut one of the arcs ~(ax) or its reflection. Thus of these two planes one at least does not support X. Hence a, b do not have the properties required of st and s2. Thus finally Xv is s m o o t h . Next construct for every ~/> 0 a compact, smooth, rotund, convex, central set K , such that K, --~ Xv and the frontier of K~ contains the whole of the frontier of X V except possibly some of the points whose distance from p - q or q - p is less than ~/. Also suppose that K , is distinct from X. This is possible because of the smoothness and rotundity of X v. Now if ~/is sufficiently small every frontier point of X is distant D(X) from at least one other point of X where distance is measured in B(K~). (Because for some Z > 0 every frontier point x of X is diametrically opposite a point y of X such that the acute angle made by xy with greater than X). But then X is diametrically maximal in B(K~) without being of constant width. 3. Miscellaneous remarks. We say that a Banach finite dimensional space B(S) has (i) property (A) if and only if every set in B(S) that is diametrically maximal is necessarily of constant width, (ii) property (B) if and only if every set in B(S) that is diametrically maximal is necessarily a sphere, (iii) property (C) if and only if every set in B(S) that is of constant width is necessarily a sphere. It is not easy to see what simple property distinguishes the spaces B(S) that
170
H.G. EGGLESTON
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have property (,4) from those which do not have property (A), and in this paragraph a number of results are given which partially elucidate this problem. Let B(SI) and B(S2) be n-dimensional and m-dimensional Banach spaces respectively and let B(S) be the m + n dimensional Banach space whose unit sphere B(S) is the cartesian product of $I and of S 2. We prove the following lemma. LEMMA. I f X is a diametrically maximal subset of B(S) then X is the cartesian product of two sets X 1 , X 2 such that X1 is a diametrically maximal subset of B(S1) and X2 is a diametrically maximal subset of B(S2). Suppose ( as we may without loss of generality) that D(X) = 1. We have by the spherical intersection property
X = Xs =
~ S ( x , 1) x~x SI(XI, 1) x
=
~ XIXx2mxEX
=
(~')SI(Xl, 1))X(~S2(X2,
=
S2(x2,
1)
1)), X 1 X X 2 = x • X ,
fl Sl(xl, 1) x fl $2(X2,1) xt ~XI
x2eX2
where XI is the projection of X on B(S1) and X2 is the projection of X on B(S2). Now for any x e X , S(x, 1 ) ~ X and thus for x t e X l Sl(xt, 1)D X I. Thus
f~ Sl(xl, 1) ~ X i and similarly xl e x t
fl
$ 2 ( X 2 , 1 ) ) ~ X 2.
x2 e X2
To complete the proof of the lemma we have only to show that X c Xt x X 2. Bur this is trivial since X 1 and X 2 are two projections of X. The lemma is proved. COROLLARY 1. I f both of B(S1) and B(S2) have any one of the properties (A) (B) or (C) then B(S) has the same property. This is because the cartesian product of two sets X 1 in B(S~) and X 2 in (B(S2) which are both of constant width (or both spheres) is a set X in B(S) that is of constant width (or a sphere). COROLLARY 2. If S is a parallelopiped then B(S) has property (B). This follows from Corollary 1 because a linear Banach space has property (B). If the subset X is of constant width in B(S) and if P~ denotes projection orthogonally (in the Euclidean sense) onto a two dimensional space zr then P~(X) is of constant width in B(P~(S)). The converse of this is true if P~(X) is of constant width in B(P~(S)) for every plane lr then X is of constant width in B(S). However a set X can be diametrically maximal in B(S) without it being true that P~(X) is diametrically maximal in B(P~(S)) for every plane zc. For, as we shall see, in any plane property (A) holds and thus if the above statement were
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CONSTANT WIDTH IN FINITE DIMENSIONAL BANACH SPACES
171
false, property (A) would hold for any finite dimensional Banach space. Thus a space B(S) fails to have property (A) if and only if there exists a subset X of B(S) such that (i) for every x e X the frontier of S(x, D(X)) meets X and (ii) for some x ~ X and some plane rc P~(S(x, D ( X ) ) n X) is contained in the relative interior of P,(X), whilst x belongs to the frontier of P~(X).
Any two dimensional Banach space has property (A) We show first that if X is a diametrically maximal subset of the two dimensional Banach space B(S) then the set of points of X that are diametrically opposite to x (denoted by P(x)) either consist of a single point or form an arc of the frontier of X. This is because (i) for each x ~ FrX, P(x) ~ ~ , and (ii) if y ~ P(x) then there exist parallel support lines of X, one through each of x and y (for otherwise X would contain a segment parallel to and longer than segment xy, which is impossible as the length of segment xy is D(X)). Suppose then that y, z ~ P(x) and w lies on the arc yz of F r X that does not contain x. Since x lies on lines of support of X parallel to lines of support through y and z it follows that, every support line, parallel to a support line of X through w, must pass through x. Thus P(w) c x and by (i) P(w) = x. Thus w ~ P(x). Since P(x) is closed the stated result is proved. To complete the proof of the lemma let 21, 22 be any pair of parallel support lines of X. Take x~21 n X. If P(x) did not meet 22 n X we could find a point w of the frontier of X lying in that arc between 22 n X and P(x) which does not contain x. Then by an argument similar to that above w ~ P(x). It follows that P(x) does in fact meet 22 n X. Thus on 21, 22 there are points of X whose distance apart is D(X). This implies that the support lines of Xv coincide with those of D(X).S. Thus finally Xv is D(X).S.i.e. X is of constant width. The lemma is proved. In the first example of a space that was a finite dimensional Banach space and which did not have property (A), we considered B(K) where K was a central convex set obtained by modifying the vector domain Tv of a regular tetrahedron T. In fact the space B(Tv) itself does not have the property (A) because it has at least one vertex lying on four distinct edges of Tv. This follows from the lemma below. LEMMA. The space B(S) does not have property (A) when S is a 3 dimensional polyhedron of whose vertices at least one lies on at least 4 edges of S. Let a vertex as described be x. Then there are two 2 dimensional faces of S say rq and r~2 both containing x and meeting in a line 2 which meets S in the point x only. There is a support plane n o f S such that ~t ~ 2 and ~t n S is the single point x. A plane n(~) parallel to n, distant r/from n and lying on the same side
172
H . G . EGGLESTON
of n as S meets S in a polygon of which two sides are parallel to 2 (provided ~/ is sufficiently small). Select points x~ and x2 on the relative interiors o f these two sides. Let S~ be the set obtained from S by the translation x~ - x2. The set S~ n S has g(r/) as a support plane and in this plane a segment s parallel to 2. The segment s together with the points 0, x~ - x2 form a set of diameter 1 and therefore a subset of a diametrically maximal set Y. Now Y contains 0 and xl - x2. Thus Y is contained in S and in S~. Thus ~(~) is a support plane of Y. Hence Yv meets the support plane parallel to n in a set of dimension at least 1. Thus Yv ~ 2S and B(S) does not have property (A). REFERENCES
1. Kelly, P. J., On Minkowski bodies of constant width, Bull. Amer. Math. Sot:. 55 (1949), 1147-1150. 2. Petty, C. M., On the geometries of the Minkowski plane, Riv. de Mat. della U. di Parma 6 (1955), 269-292. 3. Hammer, P. C., Convex curves of constant Mir,kowski breadth, Convexity. Proc. Symposia in Pure Math. 7 (1961), 291-304. 4. Jessen, B., f)ber konvexe Punktmengen konstanter Breite, Math. Z. 29 (1928), 378-380. BilDFORD COLLEGE UNIVERSITY OF LONDON
GENERALIZED ABSOLUTELY MONOTONE FUNCTIONS BY S A M U E L K A R L I N AND ZVI Z I E G L E R ABSTRACT
The concept of al~solutely monotone functions is generalized by replacing the conditions (ptk)(t) ~__ 0, k =0, 1. . . . by an infinite sequence of differential inequalities ~b(t) _> 0, L~(9(t) => 0, k = 1, 2 . . . . . where the Lk are differential operators of a special type. It is shown that these functions have a valid series expansion in terms of basic functions associated with the operators Lk.
A function ~b(t) defined on (a, b) which satisfies q~(k)(t) > 0 for t e (a, b) and all k = 0,1,2, ... is called absolutely monotone. For a detailed discussion of the history and applications of the notion of absolute monotonicity with reference to areas of classical mathematics see [1, Chapter 4]. A multivariate generalization of the concept of absolute monotonicity is discussed in Bochner [5, Chapter 4]. It is a familiar fact that an absolutely monotone function can be expanded in a power series oo
(1)
~ ( t ) = ]~ ~b(k)(o+) ( t - - a ) k k=O k!
convergent for I t - a I < b - a. The purpose of this note is to generalize th e concept of absolute monotonicity and to establish the analogue of (1). Let {wi(t)}~= o be an infinite sequence of positive functions, each of class C ~l-a, b]. With these functions we associate the sequence of first order differential operators (2)
d 1 Dff(t) = at wi,t----- f ( t ) ,
i = 0,1,2,...
DEFINITION 1. A function if(t) defined on (a, b) is called a "generalized absolutely oo provided ~b(t) is of monotone" (abbreviated G.A.M.)with respect to {Wi}i=o class C ~ on the open interval (a, b) and satisfies the inequalities
(3)
~b(t) > 0 and (DkD,_ 1"'" Do)q~(t) > 0
for all t E (a, b), k = 0,1,.... The special choice wi(t) = 1, i = 0,1,... corresponds to the standard notion of Received, August 18, 1965 173
174
SAMUEL KARLIN AND ZVI ZIEGLER
[September
absolute monotonicity. A few remarks on other generalizations are in order. A homogeneous version of (3) was partly investigated by Hirschman and Widder [-2]. Their case corresponds to the circumstance where the differential operator Lk = DkDk-i ""Do reduces to a linear k + 1 st order differential operator with constant coefficients. The operator Lk for general {w~(t)) induces a linear k + 15t order differential operator with variable coefficients admitting a factorization into linear terms, i.e., Lk = (D + 21(x))(D + 22(x))... (D + 2k+l(X)) (see Karlin and Studden [3]). Differential operators of this type were first singled out by P61ya !4] in the course of developing certain generalizations of the mean value theorem. Turning to the task at hand we will first describe a geometric characterization of the class of G.A.M. functions involving certain convex cones. To this end we introduce the special functions (4)
Wo(Ofo'wx(¢)ff
It is straightforward to check that order differential equation (5)
~k-
w2(~2) "'"
1
fa
Wk(~k) d~k "'" d~. l, k = O, 1 , . . . , t ~ [ a , b ] .
Uk(t) is the unique solution of the k + 1st
LkU = ( DkDl~- 1"'" Do)u = 0
subject to the initial conditions u (o (a) = 6u i = 0, ..., k -I and U(k)(a):I"[ik:O w~(a). Notice that in the special case wi(t ) = 1, we have Uk(t) = tk/k! The functions Uo, u~,...,u~ constitute an extended Tchebycheff system on (a, b), i.e., any non-trivial linear combination ~-,i~oC~U~(t) with real coefficients can vanish at most n times counting multiplicities. We refer the reader to the book by Karlin and Studden [3] which contains an elaborate study of the theory of Tchebycheff system with emphasis on a geometric point of view. With respect to (Uo,-'-, un} we generate a convex cone ~(Uo,-", un) of functions in the following manner. DEFINITION 2. A function ~b(x) belongs to C~(Uo,..., un) (and is called "convex with respect to ( U o , ' " , u ~ )") if for every set of points Cx t ,~i :~+ ~ 2 satisfying a <x 1 <x2<
... < X n + 2 <
b
the determinant inequality Uo(XO, Ul(X
),
"'" Uo(Xn+ "'" ul(xn+2)
>0
(6) Un(XO,
"'" "'"
prevails.
Un(Xn+ 2)
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GENERALIZED ABSOLUTELY MONOTONE FUNCTIONS
175
For functions 4J(x) which are n + 1 times continuously differentiable, it is proved in [3] that ¢ belongs to W(uo, "-, Un) if
(D,D,_ i"'" Do)~(x) > 0
(7)
x ~ (a, b).
The converse is also valid provided the differential operator in (7) is suitably interpreted (see Karlin and Studden (3) chap. 11 and Ziegler [6] for further details). Consider now the intersection cone f f a = 5 + n [(~,=o~(Uo, U~,...,u,)]
(8)
where cg+ denotes the cone of continuous non-negative functions defined on (a, b). It is proved in [3] that ~ belongs to ~a if and only if q~ is infinitely continuously differentiable, ~(x) > 0 on (a, b) and (7) holds for n = 0,1,.... Thus, the convex cone ¢ga coincides with the class of G.A.M. functions. We quote the following Taylor-type formula needed later. Let f(x) be any n + 1 times continuously differentiable function defined on (a, b) such that limx.~+ [d"f (x)] /dx "exists for each n(*). Then (9)
f(x) =
~b,(x; t)L,f(t) dt + ~
pk(a+)uk(x)
k=0
where
po(a +) - f(a +) Dk- ,... Dof(a +) Wo(a) pk(a +) = wk(a) , k = 1,2,-.and
[0 (10)
~Pn(X;t) :
J I
I
a<x
wo(/)j,
x
nh(x; y) and Mi(x; y) be defined by (11)
O<mi(x;y)=
min x~_t<-y
wi(t)<
max
wi(t)=Mi(x;y), i=O, 1,...
x
* Observe that (7) together with q~(x) >= 0 implies that limx-.o+ [d" $ (x)l/dx" exists and is finite for all n = 0, 1, ... Therefore the formula (9) is applicable wheneverf(x) is G.A.M.
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SAMUEL KARLIN AND ZVI ZIEGLER
[September
I f for every c e [a, b] there exists a d, c < d < b and an ~ > 0 such that (12)
lim n-~oo
mi(c; d)
~n = 0
\ i=0
then each ¢ e W+ ~ []")k~ 0~(Uo, "'', Uk)] (i.e., ¢ is a G.A.M. function) possesses a representation oo
(13)
q~(t)= E pk(a+)uk(t),
te(a,b)
k=O
where dp(t) po(t) = Wo(t)
DR- "'" Do ¢(t) 1
pk(t )=pk(t;¢ ) =
'
w~(t)
'
k = 1,2,...
REMARK 1. The requirement (12) is manifestly fulfilled if wi(t ) are uniformly bounded from above and below. REMARK2. The convergence in (13) is uniform on every compact subinterval of(a,b). This follows immediately from Dini's theorem concerning monotone convergence of functions owing to the fact that all the terms including the limit function are non-negative and continuous. Furthermore, if ¢(t) is continuous at the end point b then (13) is valid also at b. To prove this statement we observe, since ¢(t)/Wo(t ) is continuous at b and ~b(t)/Wo(t) is non-decreasing, that for prescribed e > 0 there exists r/(e)> 0 and so small that _ < ¢(b) wo(b) =
_
_ ¢(b-t/)
wo(b_rl )
-4- / ~ .
Now the convergence of (13) at b-~/implies that for n large enough ¢(b-r/) wo(b-rl)
<
~, pk(a+) Uk(b--rl) k=O wo(b-rl) +
<
+
uk(b)
pk(a ) ~
+
k=O
<
~o , +, Uk(b) E pk~.a )W-o-(-~ + ~" k=O
the second inequality resulting since Uk(t) / Wo(t) is non-decreasing. Letting e ~, 0, we see that oo
.
+.uk(b)
¢(b) < ~, Pk~.a ) W - - ~ " wo(b) = k=O On the other hand, since ¢(t)/Wo(t) is increasing, we obtain ¢(b) > ¢ ( b - ~ ) => ~ pk(a+)Uk(b--e) wo(b) = w o ( b - e ) k=O wo(b-8) "
19651
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177
It follows by letting e~,0, that
¢(b) > ~, Pk(a+) us(b ) wo(b) = k=o Wo(b) and this inequality holds for all n. Therefore oO
]E Pk(a+)uk(b).
¢(b) =
k=O
For t = a the situation is much simpler since us(a ) = 0, k = 1,2,... Thus i f ¢ is continuous at t = a and t = b then the convergence in (13) is uniform over [a, b,]. Proof of Theorem 1. Let ~b belong to cd+n[r)~=oC~(uo,...,un),]. Then the functions pn(t), n = 0 , 1 , . . . , are non-negative, continuous and non-decreasing on (a,b). Thus, pn(a+), n = 0, 1,-.. exist, are non-negative and the generalized Taylor formula (9) applies. Since
s~(t) = ~
p~(a+)uk(t)
k=O
is a non-decreasing sequence bounded above by ~(t) we may infer that s,(O converges to s(t) < oo. We define
g(t) = ¢(t) - s(t) = lira
(14)
n ' - ~ O0
fb
Cn(t;x)dp~(x)
a
and it is required to prove that g ( t ) - 0 for t e [a, b). For this purpose the following lemma is useful LEMMA. Suppose for c e [a, b), the relation
each
(15)
lira
Cecda=c~+n[O,LoC~(Uo,...,un)]
f
and each
b
dpn(t;x)dpn(x;qb ) = 0
n-~oo i / c
is fulfilled for t in some non-degenerate interval [c,c + e) for e > 0 which may depend on dp. Then g(t) - 0 for t e [a, b). Proof.
By a result proved in [6'], rn
¢ ' ( t ; x ) e c~+ r3[k~--=o~(U0' "",Us)'],
for
n > m
and therefore
g(t) = lim
fb
n "-* oO ,d a
~n(t; X) dpn(x) e cg + n [
5 W(Uo,'", Uk)].
k = O
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SAMUEL KARLIN AND ZVI ZIEGLER
[September
This holds independently of m, and therefore g(t) ~ ~a. As pointed out previously every member of ~a is automatically of class Coo(a, b) and moreover p,(a + ;g) exists for all n. Now suppose to the contrary that g(t) ~ 0 on [a, b); then there exists a maximal interval connected to a on which g(t) = 0. We denote this interval by [a, c*], c* < b and c* exceeds a by virtue of the hypothesis of the lemma. Since g(t) = 0 for t e [a, c*] and g(t) ~ C°°(a, b), the representation formula (9) applied to g(t) (with respect to the interval (c*, b)) reduces to
b
g(t) =
fc
~b,(t;x) dp,(x; g).
Invoking the assertion of the lemma for g(t), we infer that
(16)
g(t) = lim n-+oo
~cbq~.(t;x) dp.(x; g) = 0
t~[c*,c* +6)
*
for some 6 > 0. This conclusion is in contradiction to the definition of c* and the proof of the lemma is complete. We now prove that the hypothesis of the lemma is satisfied. Consulting (9), we see that f o r a _ < x _ < t < d = < b
c)(d) :>
=
fx d(~.(d;Odp.(~;~b) fx ddp.(d;Ow.+x(Op.+x(Od~
> p.+l(x)
fx dc~.(d;Ow.+l(Od~
>-_ p,,+l(x) ft d(a.(d;Ow.+x(Od~. Moreover, from the definition of qb.(d; 0
(17)
f/
(Pn(d;Own+x(Od~ >-
(
I-I mi(t;d)
i=o
"
(n+l)!
where
mi(t;d)=
min t<-_z~_a
[wi(z),]
i = 0 , 1 , . . - , n + 1.
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GENERALIZED ABSOLUTELY MONOTONE FUNCTIONS
179
Hence, if a < c < t < d, then (18)
ff 4~.(t;x)dp.(x) = f)
(a.( t; x) w. +x(x) p. +x(x)dx
< q~(d). =
(n+l) l
f ~ .(t;x) w.+l(x)dx
n+l
( d - t ) "+1 1-] mi(t;d) i=0
< q~(d)
~
,=o m,(t;d)
where Mi(c;t) =
max wi(z) ,
i = 0,1,...,n+l
.
c~z~t
Using condition (12), it follows that 1 n
(a.(t;x)dp.(x) = 0 OOll C
for t in some non-degenerate interval [c, c + 6).
Q.E.D.
COROLLARY 1. I f wi(t), i = 0, 1,... are uniformly bounded from above and below, then the expansion (13) holds. COROLLARY 2. If wi(t), i = O, 1, ... are non-decreasing functions oft, then (13) holds. Indeed, the estimate in (18) reduces to (19)
~b,,(t;x)dp.(x) < (o(b)
since in this case mi(t; d) = Mi(c; t). THEOREM 2. I f there exist two sequences {e.}~°, {d.}~° and an integer N O such that a) The functions $n(t;x) for n >=No satisfy (20)
c. (t - x)"
n!
< ~.(t;x) <=d.
(t - x)"
b) There exists an e > 0 such that (21)
d lim ---~-"e" ~ 0 ; n.-+ oo C n
then the expansion (13) is valid.
n!
'
x < t.
180
SAMUEL KARLIN AND ZVI ZIEGLER
Proof. We observe first the formulae (see [7])
f:
q~.(fl; ¢) w,+ 1(4) d~ = qg.+ 1 (fl; t)
and
fc
'~b.(t; x) w.+ l(x) dx = ~.+ l(t;c).
Proceeding as in the proof of Theorem 1, and replacing the estimates in (17) and (18) by the estimate in (20), we obtain
dPn(t;x)dp.(x) < ~b(fl) ~b"+l(t;c) =
~.+~(/~;t) ----¢(/0
t- c Y-T
+1 d.+t c.+~
The validity of the theorem now follows by using (21). Q.E.D. With the aid of the above theorem, we can now characterize the dual cone to ~ea = ~e+ n [n.~= 0,(u0, ...,u.)]. DEFINmON 3. A signed measure # of bounded variation on (a, b)is said to belong to the dual cone cg* provided for every ~bE ffa that at least one of the integrals .f.~4 , d ~ or S~ 4,dv~ is finite (where t~=~l + , z represents the Jordan decomposition of V) and f b ~ d # > 0. THEOREM 3. Let {W,}o satisfy the requirements of Theorem 1. A signed
measure d# belongs to the dual of ~A if and only if (22)
u~d# > 0
i = 0,1, ...
Proof. We know that u~ e ~°a, i = 0,1, ... so that (22) is certainly necessary. The validity of (13) easily implies that the inequality (22) is also sufficient. REFERENCES 1. D. V. Widder, TheLaplace transform, Princeton Mathematical Series 6, Princeton University Press, Princeton, New Jersey, 1941. 2. I. I. Hirschman, Jr. and D. V. Widder, The convolution transform, Princeton University Press, Princeton, New Jersey, 1954. 3. S, Karlin and W. Studden, Tchebycheff Systems with Applications in analysis and Statistics, Interscience, New-York, to appear in 1966. 4. G. P61ya, On the mean value theorem corresponding to a given linear homogeneous differential equation, Tram. Amer. Math. Soc. 24 1922, 312-324. 5. S. Bochner, Harmonic analysis and the theory of probability, University of California Press, Berkeley and Los Angeles, 1955. 6. Z. Ziegler, Generalized convexity cones, to appear in Pac. J. Math. 7. S. Karlin and Z. Ziegler, Tchebycheffian spline functions, submitted to SIAM Journal. STANFORD UNIVERSITY, STANFORD, CALIFORNIA TECHNION--IsRAELINSTITUTEOF TECHNOLOGY, HAIFA~ ISRAEL
