RICE UNIVERSITY
SOME RELATIONS AMONG ORLICZ SPACES
by
Max August Jodeit, Jr.
A THESIS SUBMITTED IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF
MASTER OF ARTS
Thesis Director's signature:
Houston, Texas
May, 1965 ABSTRACT
In this paper it is shown that for the study of Orlicz spaces the condition that a Young's function A be convex
can be replaced by the more general (and more convenient)
condition that A(x)/x be non-deereasing. Some properties of the lattice of Orlicz spaces ordered by inclusion are given. R. O'Neil has shown that the condition
is sufficient in order that
GO (f*g)(x) - f f(y)g(x-y)dy
belong to L^, whenever f € L^, g € Lg. The condition is
also sufficient when the convolution is formed over the in¬
tegers or (0,2TT]. It is proven here that the condition is also necessary; all triplets A, B, C of Y-functions for which the condition holds are determined. ACKNOWLEDGMENT
The author wishes to thank Professor R. O'Neil for un¬ derstanding, insight, great patience, and valuable suggestions while this thesis was being written. The research contained in this thesis was supported in part by the Air Force office of Scientific Research. Thanks are tendered to Mrs. Nancy Singleton and to Miss Janet Gordon for their skillful typing and helpful suggestions. And to my wife Jane go thanks and gratitude. INTRODUCTION
In this paper we study the partially ordered (by inclusion) set of Orlicz spaces over a measure space (X,S,n). Background results and definitions are given in Section 1. Young's functions (among which are N-functions) are equi¬ valent if they always determine the same Orlicz space. If A
is a Young's function the condition A(kx) ^ B(x) ^ A(Kx) is that of equivalence expressed in terms of functions. This re¬
lation holds for many functions B, not all of which are convex.
In Section 2 we introduce the Y-functions, which are among these, and show that they are, for our purposes, entirely equivalent
to Young's functions. Each Y-function A satisfies the condition that A(x)/x
be increasing. This is our substitute for convexity. In particular, Y-functions are convex at 0 (a function is convex at p if the chord from the point p to the graph lies on or above the graph). We find the dual notion of concavity at
0 to be useful, and discuss inverse Y-functions. Given a Y-function A we can find a convex function C equivalent to it. Indeed we may choose C so that C(x)^A(x)SC(2x).
For a measurable function f we define ||f||^ as in Section 1, and in a similar fashion the quantity |f|^, using A in place
of C. We then have ||f||c^|f |A^2||f|(c. A does not determine a metric, however, since the triangle inequality breaks down. How¬ ever since Y-functions are non-decreasing and 0 at 0 the 2
function A determines the metric
p(f,g) - inf {K>0: f A( | f-g |/K) dpi =5 K) X
(see [5;p.l09]) which turns out to be equivalent to the norm- metric given by C, so that A determines a uniform topology for !,£ equivalent to that given by C.
The use of Y-functions instead of Young's functions in many cases simplifies proofs. In some cases involving natural constructions convexity at 0 is preserved but convexity is not. An example of the first kind is the function (used in
[l;Th.9]) defined for xgl by
Q(x) » sup M(y)N(x/y) , lSy^x where M,N are convex. It is not easy to show that Q is convex (for xsl), but if 0sl
Q(0x)/0x = sup M(y)N(0x(y) 5 sup M(y)N(x(y) 5 Q(x)/x, l^y^0x l^yS0x so that Q is a Y-function. The pointwise minimum of two Y-functions is a Y-function, but that of two Young's functions is not necessarily convex. We derive results on the structure of the set of Y-func¬ tions and of equivalence classes of such functions, and interpret these in terms of Orlicz spaces. These form a distributive sublattice of the lattice of subspaces of measurable functions (modulo equality almost everywhere). This lattice is not CT-complete. A natural complementary ordering on classes of Y-functions induces an order relation among Orlicz spaces in 3 which pairs of spaces are comparable.
Section 3 is based on properties of the operation of forming the Young's complement of a Y-function : convexity at 0 is again a satisfactory substitute for convexity. Complementation solves the problem: given the Orlicz
space L^, to find B such that f £ L^, g 6 Lg => |J* fg d|a|<«>. X The mapping A -* A induces a dual automorphism on the
lattice of equivalence classes of Y-functions, and yields the result that the complementary ordering is isomorphic to the original one.
A second dual automorphism relates the behavior of a Y- function for large values and for small values of the argument. Further information on the structure of the lattice follows.
N-functions may be naturally introduced at this point. What might be called the "interior" of the lattice consists
of N-functions. In Section 4 we consider the problem: to express, in terms of Y-functions, the relation among Orlicz spaces L^, Lg,!^ over a topological group given by
f € L^, g € => f * g € LQ.
We assume that the group is locally compact and unimodular. The basic result in the solution is given by R. O'Neil [6;Th.2.5] the relation
_1 :L 1 A (X)B’ (X) ^ xC“ (x) is sufficient. It turns out to be necessary as well; we treat 4
the problem using Y-functions, and the associated problem of finding a third function, given two of A,B,C, for which the relation holds. In each case, if there is such a function, a best possible one can be found. We also consider the cases in which only large values, or only small values, are relevant. 5
(1.1) Definltion: The function A:[0,°°) -♦ [0,°°] is a Young's function if A(0) a 0, and A is left-continuous, convex, and non-decreasing. The two trivial Young's func¬ tions are those which on (0,°°) are respectively identically
0, identically +®. (1.2) Definition; The Orlicz space L^(X,S,|j) determined by A over (X,S,n) consists of those (equivalence classes of) measurable functions for which the quantity
||f||^ = inf {K > 0: J* A( I f (x) | /K)d|_i <; 1} (inf 0 = +») (1.3) X is finite. || • ||^, where finite, is called the L^-norm. The
functions x^, lsp<» are the Young's functions which determine the spaces Lp(X,S,n). L°°(X,S,|_i) is determined by the func¬ tion ( 0 0<;x<;l >^(x) = J (1.4) I CD 1 We use the following results, which may be found in [4] . 1° Two Orlicz spaces L^, Lg are comparable under inclusion if there is a positive constant k such that A(x) ^ B(kx), x s 0. In this case 2 Lg. We write A £ B for this relation. In case the underlying measure space has a a-finite, atom-free subset of infinite measure the condition A s B is necessary in order that 2 Lg . [4; Thm.5 p 52] 6 2° The Young's complement, £, of a Young's function A is defined by £(x) = sup(xy - A(y)) (1.5) y A,A satisfy Young's inequality xy <; A(x) + X(y), x,y > 0 , (1.6) from which is derived a pairing of L^, by means of the bilinear functional (defined on L^xL^) (f,g) -» 5 fg dpi . X Holder's inequality in the form ;x Ifgldn * 2||f||Allg|lK is immediate. (see [4;p 47] and [3;§14]) The map A -♦ reverses the order in 1° and "K = A. If A satisfies the Ap-condition A(2x) s KA(x), x ^ 0, K > 0 a constant, (1.7) then is the continuous dual of L^. In certain cases the condition is necessary. [4;pp 58-60] 3° If the measure space (X,S,n) is totally finite the values of Y-functions for small values of their arguments play no role; if (X,S,|_i) is purely atomic, and its atoms all have measure ^ 6 > 0 for some 6, the values of the Y-functions for large values of their arguments play no role. To see this, consider the following: Let Eu = {x € X:|f(x)| > u}, 7 u > 0. Then J A(|f(x) |/K)d(i(x) = J* A ( | f (x) | /K)d(j (x) + J* A(|f (x) |/K)dM(x). X E, X~E u u The second integral, in the presence of total finiteness, is dominated by |j(X)A(u/k). If all the atoms of a purely atomic (X,S,n) have measure ^ 6 > 0 the first integral, when represented as a sum, can be made to vanish by taking K sufficiently large. 8 (2.1) Definition: A function A:[0,®) -* [0,®] is a Y-function if A is left-continuous, convex at 0, and A(0) =0. A Y-function is non-trivial if it is not identi¬ cally 0 or identically ® on (0,®). For reference we have (2.2) Lemma: The function A defined on [0,®) is a Y-function if and only if 1° A(x) I 0, all x i 0; 2° A(x)/x is non-decreasing; 3° A(0) = 0; 4° A is left-continuous. Proof: We are concerned with 2°. If A is a Y-function then A(x) ^ A(Xq) • (x/xQ) for 0 ^ x ^ xQ, by convexity at 0, and the fact that A(0) = 0; conversely, if 1° - 4° hold A is convex at 0. In particular it follows from 2° that Y-functions are non-deereasing. If 0 < A(Xq) < ® then for all x > xQ, A(x) > A(Xq). Every Young's function is a Y-function. The 2 function A(x) = min(x,x ) is a Y-function but not a Young's function. (2.3) Definition: Let A, B be Y-functions. A is stronger than B, in symbols, A 5 B, if there is a positive constant k for which B(x) g A(kx) , x 5 0. (2.4) A is equivalent to B, in symbols, A ~ B if A § B and 9 B 1 A. The relation § is clearly reflexive and transitive, so that the equivalence classes under ~ form a partially ordered set. We sometimes use the notation {A(x)} to denote the equivalence class to which the function given by A(x) belongs (e.g.fx^)). Also, A 5 B means B § A. (2.5) Theorem: The partially ordered sets of equivalence classes of Young's functions, and of Y-functions, are iso¬ morphic . Proof: Let denote the class of Young's functions and suppose A is a Y-function. It is enough to show that {A} fl Id is non-void. Let C(x) = J (A(t)/t)dt. (2.6) 0 C is a Young's function and A(x/2) ^ C(x) % A(x) , (2.7) for A(x/2) * fX (A(t)/t)dt SE C(x) *x • (A(x) /x) .Q.E.D. x/2 C, as defined in (2.6), is a regularization of A to a convex function. The remark which follows discusses a regularization obtained in a natural way: the largest convex function which is pointwise ^ the given function. Given a function C:[0,°°) -> (-»,<»], it can be shown by a geometric argument that the function C*(x) - inf{2lic(u) +£Sc(v):v s x s u} (2.8) 10 (where if u=v=x we mean C(x) by the indicated quantity) is convex. Clearly C*(x) 3 C(x) and C* is the largest such convex function. In case C is a Y-function, C (2x) 5 C(x): if v < %x then 2^C(u) + (v) § ^_C(— £) u-v v ' u-v ' y u-v ' x 2' %x 2u C(x/2) u-v X s C(x/2) , and if v a %x, C(u) = C(v) = C(x/2) , so that ~“”C(u) + ^fc(v) * C(x/2). We give an example to show this estimate best possible: Let 0 £ x £ 1, C(x) = < X 1 < x. 0 3 x £ 1, Then C (x) = 1 < x. and C (2(l+e)) = l+2e for all e>0, while C(l+e) = 1+e. (2.9) Remarks: (i) If A(0) = 0 and A is non-decreasing the extended real-valued function p(f,g) = p(f-g,0) can be defined on the measurable functions on the measure space (X ,S ,|_i) by p(f,0) = A(f) = inf {K>0: f A( | f (x) |/K)dpi(x) 3 K). X 11 p is a metric on the set where it is finite. If A is a Y-function (A need not be left-continuous) and A(f) < %, then if C is defined as in (2.6) 2 (%A(f)) S IIf||c S A(f), are so that the topologies determined by p and by II’IIQ the same. Proof of the inequalities: If % 5 K > A(f) , 1 5 f(l/K)C(|f(x)|/K)dn(x) s fC(|f(x)|/K) dpi(x), so that ||f ||^ ^ A(f) . On the other hand since 1/TK C(X) S C(X4/K) , fA( | f (x) | /2VK)dpi * (7K/7K) /C( | f (x) | /K)dn ^ JK. Jc( | f (x) | /K))d|a < ZjK. 2 Hence ||f||c 5 (%A(f)) . (2.10) Definition: Let A be a Y-function. We define |f|A = inf {K>0:J A( | f (x) |/K)d|_t S 1}. X If A is convex we write ||f||A as usual. ri (2.11) Theorem: Let A be a Y-function; and let C be defined by (2.6). Then I file « mA a 2||f||c. Proof: This is an immediate consequence of (2.7) and the definition (2.10). The following gives a substitute for Jensen's Inequality. (2.12) Lemma: If C is a Y-function then s rc(2f(x))pfa-)dx jp(x)dx > - fp(x)dx (2.13) 12 where p,f are non-negative measurable functions and Jp(x)dx 4- 0. Proof; Let C*(x) = [ (C(t)/t)dt. As in (2.7), C(x/2) 3 C*(x) 5 C(x) so that in (2.12), by the convexity of C* the left-hand side of (2.13) is less than or equal to 2£ p C*(r (x) (x)dx x ^ fC*(2f(x)^p(x)dx v | p (x) dx ' ~ |'p (x) dx which by our first remark is less than or equal to the right- hand side of (2.13). Since Jensen's Inequality is a necessary and sufficient condition for a function to be convex, it is natural to ask whether (2.11) possesses a similar property with respect to functions convex at 0. The following example shows that this is not the case. Let {x 0 5x31,2 x, 1 1 < x < 2. Then C(x) 5x3 C(2x) so that (2.11) holds, although C is not convex at 0. The real-valued functions x v y = max(x,y) , x,y s 0 x A y = min(x,y) , x,y ^ 0 induce the corresponding operations on Y-functions (e.g. (A V B) (x) = A(x) v B(x)) , and both A v B, A A B are Y- functions. The same holds for an arbitrary collection of Y-functions, so that we get (2.14) Theorem: The class of Y-functions is a complete 13 distributive lattice. (2.15) Theorem: The set of equivalence classes of Y-functions is a distributive lattice, with operations fA(x)) v [B(x)} - {A(x)vB(x)} , fA(x)} A {B(X)} = {A(X)AB(X)}. Proof: It is enough to show that the relation ~ has the Substitution Property [2;Ch II, §5] with respect to the lattice of Y-functions. Suppose A~B, C~D. For appropriate positive constants k^, k2, K-^, K2, with k=k^Ak2 and K=K^vK2» A(kx) A C(kx) ^ A(k^x) A C(k2x) 5 B(x) A D(x) ^ A(K-,x) A C(K2X) ^ A(Kx) A C(Kx) , so that A A C ~ B A D. Similarly A v C ~ B v D. Q.E.D. Inverses. Let A:[0,co] -» [0,=°] be non-decreasing and set A“^(y) = inf {x:A(x)>y) (inf 0 = ®) . (2.16) - A ^ is defined on [0,co], A ^(<=o) = “, A ^ is non-decreas¬ ing and right-continuous. (2.17) Lemma: Let A:[0,®) -* [0,o>] be non-decreasing and left-continuous. Suppose A(x) -* <» as x -> ® and A(0) = 0. Then A (A-1 (x) ) 5x = A”1 (A (x) ) , xgO (2.17) (2.18) (2.19) (2.20) We omit the proofs of these. 14 Remark; An important Y-function gives an example showing that no inequality of the form A ^(KA(x) £ x holds. Let A(x) = 0 for O^x^l, and +® otherwise. Then A“^(x) ® 1 for Osx<<»; and A ^(co)=co, so that for all e>0, A ^(eA(x))=l for O^x^l, A"^(eA(x))= ® for x>l; thus A ^(eA(x))^ x for all e>0. (2.21) Theorem: If A:[0,®) - [0,®) is non-decreasing, concave at 0, and continuous from the right at 0 then A is absolutely continuous on [0,®). Proof: Let c > 0. We show first that A is absolutely continuous on [c,®). Given e > 0, let 6 = J if A(c) >0; if otherwise there is nothing to prove. Suppose {(xn,yn)} is a system of non-overlapping open intervals with £(yw - x ) < 6 7, and x ^ c, all n. Then since n n rr n A(x)/x is non-increasing, A(y ) A(x ) s(A(yn) - A(xJ) - y„ - xj ir rn n n n' A(0 <: E — (y - x y) x n '•■'n n <; E (y„Ny - x_) < e . c n n Therefore on (0,®) we have the function A' defined almost everywhere and for 0 < x < y A(y) - A(x) = f A'(t)dt . X By the theorem of Beppo Levi 15 A(y) - A(0) = lim (A(y) - A(x)) x-0 = lim f A'(t)dt x->0 x = A'(t)dt . Q.E.D. (2.22) Definition; A:[0,®] - [0,®] is an inverse Y-function or Y~function if A is non-decreasing, concave at 0 (on [0,®)), continuous from the right at 0, and A(®) = ®. A is non-trivial if A 4 0 or 4 08 on [0,®). If A is a non-trivial Y~^-function then in fact 0 < A(x) < ® for 0 < x < ® . As a corollary to (2.21), inverse Y-functions are absolutely continuous on [0,®). (2.23) Lemma; If A;[0,®) -* [0,®] is non-decreasing, left- continuous, tends to +®, and A(0) = 0 then A is a Y-function if and only if A"^ is an inverse Y-function. A is non¬ trivial if and only if A"^ is non-trivial. If A: [0, ®]-»[0, ®] is non-decreasing, and A(®) = ®, then A is an inverse Y-function if and only if its left-continuous inverse is a Y-function, and the same non-triviality assertion holds. Proof; Clearly A-'*' is non-decreasing, continuous from the right at 0, and A"^®) = ®. For y s x > 0 we set Si = {z;A(z) > y} = {z:j A(z) > xj = {z:A(yz) > x} = J{z:A(z) > x} 16 Hence A~^(y) = inf S-, s ^ inf S0 = ^ A"^(x), so that A“^ is concave at 0. The non-triviality of A gives x' > 0 such that A(x') < °°. If A(x') = 0 let x" be the greatest y such that A(y) = 0; x" < “. Then A”'*'(A(x//)) = A"^(0) = inf{y:A(y) > 0} = x" . If A(x') > 0 then A(y) > A(x') for all y > x'. Hence A~^(A(x')) = x'. Thus A-’*' is non-trivial. One point in the converse needs to be given: if for some Xq, 0 < XQ < «>, 0 < A~^(xo) < then A(A”^(y)) £ y < » and A(2A~l(y)) ^ y > 0, so that A is non-trivial. We omit proof of the second part. (2.24) Lemma: If A,B are Y-functions then A s B if and only if B'1(x) £ kA_1(x), x ^ 0 , (2.25) where k is the constant in (2.3). Proof: If A s B, x s A~^(A(x)) s A ^(B(kx)) , so £ B_1(x) £ A_1(B(k B'^x>)) S A-1(x) . If the condition holds x <: B“1(B(x)) <; kA_1(B(x)) so that A(|) £ A(A'1(B(x))) £ B(x) , x * 0 . Q.E.D. Consequently, A,B Y-functions are equivalent if and only if there exist positive constants k,K such that 17 kA-1(x) £ B-1(x) £ KA_1(x), x * 0 . (2.26) Definition: If A,B are inverse Y-functions then A £ B if there is a constant k > 0 such that A(x) £ kB(x), x £ 0. (2.27) Corollary: The partially ordered sets of equivalence classes of inverse Y-functions, and of Y-functions, are dually isomorphic. (2.28) Theorem: The lattice of equivalence classes of non¬ trivial Y-functions has greatest and least elements and one pair of non-trivial complements. Proof: We show that Q(x) = max(0,x-l), x ^ 0, is repre¬ sentative of the least class of non-trivial Y-functions, and that 0 £ x. £ 1 , (2.29) 1 < x represents the greatest. If A is a non-trivial Y-function there is x^ > 0 such o that A(xQ) < * SO that A(x)/x £ A(xo)/xQ , X £ x0 . and thus A(x) £ max(l,A(xo)/xo)x , 0 £ x £ xQ . Hence A £ I . There is also x^ > 0 for which A(x^) >0 so A(x)/x * A(x1)/x1 * min(l,A(x1)/x1) , or A ;> Q . 18 For the complements we set \^(x) *= x , x ^ 0, and 0 0 £ X < 1 , ^(x) - (2.30) +00 1 < X . Then \^(x) v \ro(x) = I(x), x £ 0 and \ 0 0 £ x £ 1 , 4(x) A ^(x) ^ Q(2x) . X 1 < X \ J If now A v B ~ I, A A B ~ Q, non-trivially, choose repre¬ sentatives for {A}, [B] which are finite on [0,1] and positive for x > 1. For some k > 0 I(kx) <: A(x) v B(x) , x £ 0 ; hence A(x) v B(x) = «° for x > 1/k . Suppose A(x) = ® } x > 1/k. For some k > 0 , A(x) A B(x) £ Q(kx) , x s: 0 ; thus A(x) A B(x) = 0 for x s 1/k. If A(x) > 0 for x > 0 B(x) = 0 for x £ 1/k , so A ~ I, contrary to hypothesis. Hence A(x) = 0 on a non-trivial interval, and thus A ~ \ . In a distributive lattice with least, greatest elements com¬ plements are unique so B ~ ^ . Q.E.D. (2.31) Lemma: In any distributive <£ with Q,I, if a and b are complements, then £ is isomorphic to the direct pro¬ duct of [Q,a] and [Q,b], Proof: The desired isomorphism is the mapping x -* (a A x, b A x) . 19 For if x -* (c,d) and y - (c,d) then c = a A x = a Ay and d=bAx=bAy so x = (a A x) V (b A x) = (a A y) v (b A y) = y . If c £ a, d £ b then c v d -* (c,d) since (e.g.) (c v d) A a = (c A a) v (a A d) = c v (a A d) and Q^aAd^aAb=Q. Q.E.D. As a corollary we have (2.32) Proposition: The lattice of equivalence classes of non-trivial Y-functions is isomorphic to the direct product of the intervals [Q,^] and [Q,^] (where Q(x) = max(0,x-l), ^(x) = x; is given by (2.30)). (2.33) Remark: The direct product representation may be made using other intervals; for example [a,I] and [b,I] , since x -♦ x v a is an isomorphism of [Q,b] and [a, I] . (2.34) Theorem: The lattice of equivalence classes of non¬ trivial Y-functions is not a-complete. Proof: Suppose the contrary. Then the set of classes {xp}, 1 < p < <= and rational, has the g.l.b. A, a non-trivial Y-function, such that A(x) <; kpX^, x s 0 , each rational p, 1 < p < °° . Then B(x) = A(x log(x+1)) £ k (x log(x+l))p_e p-e p p e £ s kp_ex [(log(x+l)) ' /x ] p £ kpx where e > 0 is rational and p - e > 1 . 20 Since A is the g.l.b., there is k > 0 for which A(x log(x+l)) <; A(kx), x ^ 0 . For x sufficiently large we must have A ^(A(x)) =* x o since 0 < A(xQ) <£ k2XQ < » for some xQ > 0 and hence 2 0 < XA(Xq) ^ A(x) £ k2X < «, x > xQ . Therefore x log(x+l) <; kx , all sufficiently large x, a contradiction. Q.E.D. (2.35) Theorem: Let (X,S,n) be a totally a-finite atom- free measure space, of infinite measure. As a set of linear subspaces of the space of measurable real- or complex¬ valued functions in X, the set of non-trivial Orlicz spaces over (X,S,|j), ordered by inclusion, 1° is dually isomorphic to the lattice of equivalence classes of non-trivial Y-functions; 2° is a distributive lattice; 3° is a sublattice of the lattice of linear subspaces of the measurable functions; 4° is the direct product of its intervals [L^nL“,L^] 1 and [L nL°°,L“] in the lattice of subspaces; 5° has, as a lattice, the one pair L , L of non¬ trivial complements; 6° is not, as a lattice, cr-complete. Proof of 1° is found in the thesis of W. Luxemburg [4;Th 5,p 52], Then 2°, 5°, 6° follow from preceding propositions. Once we have 3°, 4° follows. 21 Proof of 3°: We must show that as sets of functions L L n and L L + AvB ' A 4 AAB - A 4- Topologies can be given, in natural ways, to the sets fl Lg and LA + L^: + f 1 L - PIIA H HB • A n 4 : f + f ILA+LB - inf tl|f]_llA + l|f2llB l 2 «= f} . The first of these gives to fl Lg the structure of a Banach space, as may easily be seen. The second is clearly a pseudo¬ norm; since the elements of the spaces are functions, it may be easily verified that II •II, ,, is actually a norm making LA+LB LA + Lg into a Banach space. We prove these assertions with two lemmas (2.36) Lemma: LAvB = LA n Lg, as sets, and f 2 f f %(l|f|lA + l|f!lB> = ll llAVB = Proof: If f € LA n Lg then for k > 2(||f||A V ||f||B) , J (AvB) ( |f (x) |/k)dfi s J* A( |f (x) | /k)d(j X X + f B(|f (x) | /k)dja * 1 , X so that the right-hand inequality holds. Similarly we show that the other holds. Q.E.D. (2.37) Lemma: If A,B,C are Young's functions and there is k > 0 such that (AAB) (x/k) ^ C(x) < (AAB) (X) , x s 0 . then 22 1° LA + Lg = Lc (as sets) ; 2° if u € La, v € LB , ||u + v||c S 2(||U||A V ||v||B); 3° if f 6 LQ there exist u € LA , v € Lg such that V V £ k f IMIA H IIB ll He • 4 f s S 2k f ° ll Hc PH^+LB H Hc • Proof: Let u € LA> v € L. . i C (luWI+K(*)l,)du * %c ; + j c (J^J-)du] * M J A(J^|ii)dy + / B . Hence ||u+v||c £ 2(||u||A v ||v||B), and LA + Lg £ LQ . Let f € LQ. Set, for s > 0 Sg - {x € X:A(s|f (x) |)<; B(s|f(x)|)} . We may assume |f (x) | < °=> for all x € X . Let u(x) = f(x)Xg (x) , Sl/k where Xg denotes the characteristic function of S. Define v by u+v = f. Then J A(|u(x) |/K)d|i - J* (AAB) ( |u(x) | /K)d(a s ; c (^)dM * l if K > k||f[|Q, since |u(x)| lf(x)|. Similarly we get 5 k f so L £ L + L and IMIB ll llc c A B 23 V V S k f IMIA II IIB H Hc for this choice of u,v . If l|f|l 0 Chen LA+I_ " £ s 2( u | v u )s 2( u +llv ll ) 0 ll HC H nlA ll JlB H nllA n B ^ so that f is 0 almost everywhere. If f > 0 choose and H ll +L 0 < e < %||f||L +L L^A B ^A^B u,v such that ||f||L +L * HIA + ||v||B - e . A B Then > 2 + V ) PHL.+L.A ij On the other hand, for suitably chosen u,v f U + II lll^+Lg * II IIA IMIB * 2kllfllc * Q.E.D. Order relations among equivalence classes of Young's functions imply the corresponding relations among the Orlicz spaces but not conversely, in general, as the example L^(0,1) £ l5(0,l), p s q, shows. But we may define the equivalence relation {A} = {B} (mod (X,S,M)) if LA(X,S,H) = Lg (X, S, , which has the Substitution Property with respect to the lattice of equivalence classes. This yields 24 (2.38) Corollary: If (X,S,|j) is a measure space the class of Orlicz spaces over (X,S,|j), ordered by inclusion, is a distributive sublattice of the lattice of linear subspaces of measurable functions on X, and is dually homomorphic to the lattice of equivalence classes of Young's functions. (2.39) Definition: If £ is a distributive lattice with Q,I and complements a,b the complementary ordering of £ with respect to a,b is defined by (£,> ) - [Q,a] x dual of ([Q,b]), or x > (w.r.t. a,b) iff xAa ss yAa and xAb s yAb . Example: If a = Q, b ■ I, (£,>* ) is the dual of (£,s). (2.40) Lemma: If £ is a distributive lattice with Q,I and complements a,b the complementary ordinary of £ makes £ into a distributive lattice with greatest element a , least element b and complements Q,I . Proof: Omitted. The new lattice operations are x U y = [(xAa) v (yAa)] v [(xAb) A (yAb)] , x D y ** [ (xAa) A (yAa)] v [(xAb) V (yAb)] . A calculation yields x U y — (xvy) A (avx) A (avy) , (2.41) and dually x fl y = (xvy) A (bvx) A (bVy) . (2.42) (see Birkhoff [2; Ch IX, §3 & Ex 1]) 25 (2.43) Definition: If A,B are Y-functions A passes B if A A L s B H, and A A \ ^ B A A 00 1 1 «> , where \^(x) ® x, x £ 0 and ^(x) = 0, 0 £ x <; 1; +<», x > 1. (2.44) Lemma: If A,B are Y-functions, A ~ B if and only if A passes B and B passes A. A passes B if and only if A v ^ B V and A V <; B v . (2.45) Theorem: Let (X,S,|j) be as in the preceding Theorem. The following are equivalent (we write LA for LA|X,S,u)): 1° A passes B; 1 £ 1 2° LA n L“ = Lg n L“ and L^ fl L Lg (1 L ; 1 1 3° LA + L 2 Lg + L and LA + L°° £ Lg + L°° ; 1 4° LA n Lg « (LA n L ) + (Lg n L") ; 1 5° LA + Lg = (LA + L ) n (LB + L°°) . The proof consists of using Theorem (2.35) to translate 2°-5° into statements about Y-functions, and using the relation A A B = (AAX^) v (BAX^) which is equivalent to 1° . (2.46) Theorem: Let A,B be non-trivial Y-functions. A passes B if and only if there exist positive constants Vko’xl’kl’ suc^ that A(x) s B(kQx) , 0 <; x £ xQ , A(x) ^ B(k^x) , x^ ^ x . Proof: =*: There is x > 0 for which A(x } < ® . O o If k' « max(l,A(xQ)) then A(x) <: k'x , 0 ^ x £ x o 26 A A X^ € {A(x)Ak'x} s {BAX-^} SO that there is a positive constant k o such that A(x) » A(x) A k'(x) <: B(kx) A kx B(kx), for OsxsxQ . Since {AAX^} ^ {BAX^} there is a constant K > 0 such that (AAX^) (X) S (BAX^MKX) , x ^ 0 . If x s 2/K, x ICx A(x) s r A(t) A Xjt) S J B(t) A X (t) & 0 0 Kx j. Kx ~ - r B(t) ^ S r B(t) * B(^§). 1 C Kx C 1 If the inequalities hold A(x) A [min(A(x0) ,1) lx § B(kQx) A [min (B (kQx0) ,1) ] (kQx/k0) , so that [A A X^}S [B A X-^}. Also A(x) A Xoo(x/x1) i B(k-^x) A x^Ckjx/k^) , so that [AAX}5{BAX}. Q.E.D. Remarks: If °° s p s q & i fxp) passes [xq] (where {x“} = (Xj), so that P q P 1 q L n L - (L n L ) + (L n L”), Lp + Lq - (Lp + L1) n (Lq + L°°). (2.48) The sublattice of the lattice of equivalence classes of Y- functions generated by fX^) and fxp), 1 S p < ® consists of all classes of the form {xp A xq) , fxp v xq) ,{xP A Xj ,fxP v X^) Jx^} > 1 = P>q < “• This follows from the reduction of lattice polynomials in these classes to lattice polynomials of positive real numbers, 27 one each for large values and small values, which can be recombined after simplification to one of the indicated form. We shall use later the following (2.49) Proposition: Suppose (X,S,[_L) is such that n(X)<°°, and that there exists XQ>0 for which A(x/k) ^ A^(x) ^ A(k'x) for x ^ XQ. Then |f|A 5 [l+u(X)A(x0/k)]k|f|A and |f|A s [l+n(X)A1(x1)]k'|f|A. Proof: Suppose f € L^. Then rA1(|f(x)|/K)dU = r Ay( | f (x)|/K)dM. 1 X |f(x)|/K ^xQ + J\ A (|f(x)|/K)d|i J 1 |f(x)|/K >xx «WnOO + J* . , . A (k' | f (x) | /K)d|i x xl I f ( ) I /K >XQ S Aj^Cx^nCX) + 1 if K/k' > |f|A, so that f Ax( |f(x)| /K[l+|i(X)A1(x0)])d4 £ 1 and therefore | f |A^ - [1+H(X)A^(XQ) )k'|f |A< A(x) £ A^(kx) for x i xQ/k • Hence the first inequality holds. Q»E.D. 28 (2.50) Proposition: If (X,S,|i) is purely atomic and the measures of the atoms are bounded below by m>0, and if there exist positive constants k,k',XQ such that A(x/k) s A^(x) s A(k'x), 0 ^ x 5 XQ, then 1 |f|A - (1 + (A^ (l/m)/x0))k|f|A> I f lAi S (k' + A~^(l/m) /x0) | f |A* Proof: If SA(|f |/K)n 5 1 then for each a a a a 1 1 A(|fa |/K)na ^ A(|fa|/K)m so that ifj S KA"1(l/m) _1 1 I.e. ||f=§ KA (1/m) , so ||fs* |f |AA~ (l/m) . f f a a" EA(k'(| |/l.)) if l al/x03L, all a. The last sum is 5 1 if L > k'|f|^. Hence if L > (HfH^/xg) v k'|f|A then L 1 |f|, so f v k f f 1 I IA-L - ' I IA - I lACA“ (l/m) /x0 v k') - If+ * For the other inequality, XQ is replaced by Xg/k, k' by k and the argument is the same. Q.E„D. 29 (3.1) Definition: If A is a Y-function, the Young's comple¬ ment , X, of A is defined by A(x) - sup(xy-A(y)) . (3.2) y It is easy to see that A(ax+gy) £ aX(x) + pX(y) , 0 £ a, a+g=l; x,y 5 0, so X is convex. Indeed A is a Young's function. (3.3) Lemma; If A,B are Y-functions and A 5 B then A£B; and X ~ A for all Y-functions A. Thus complementation is an involution (on the lattice of equivalence classes) which reverses order. Proof: If we have B(x) £ A(kx), some k>0, all x50, X(x) » sup(xy-A(y)) = sup(kx(y/k)-A(k(y/k))) £ B(kx) . In particular, y y if A ~ B then X ~ B. If A is a Y-function we define C as in (2.6). C is a Young's function and so C(x) = C(x), xSO. Therefore A ~ C = Cl ~ A. Q.E.D. Remark: The following relation is noteworthy: A(x) £ B(kx) => A(kx) 5 B(x) (xsO). (3.4) We derive immediately the relations X(x/2) £ C(x) £ X(x) £ A(x) £ X(2x) , x 50. (3.5) (3*6) Theorem: Let C be a Y-function, and let C be defined by (2.8). Then C*(x) ■ C(x). Proof: C is the largest convex function which is pointwise £ C, and C(x) £ C(x), xsO. Therefore C(x) £ C*(x) £ C(x) , X50. Thus 30 C(x) ^ C*(x) 5 C(x) , x^O, and C*(x) = C*(x) = C(x) , x^O. Q.E.D. (3.7) Lemma: If X is a lattice and f:£ -* £ is an involu¬ tion which reverses order then f is a dual automorphism. Proof: f(x v y) ^ f(x) , f(y) so f(x v y) s f(x) A f(y). Similarly, f(x A y) S f(x) v f(y). Hence f(f(x) A f(y>) s f(f(x)) v f(f(y)) = x v y. Hence f(x) A f(y) * f(x v y), and so f(x) A f(y) = f(x v y) . Similarly f(x) v f(y) = f(x A y) . Finally since f is an involution f is one-to-one and onto. Q.E.D. Thus complementation is a dual automorphism in the lattice of equivalence classes. In particular A A B ~ A v B. T.Andd [1; Lemma 1] has taken the right-hand side of this relation as the definition of the left-hand side. This yields an immediate proof that the class of Orlicz spaces is a lattice. (3.8) Theorem: If A is a Y-function then x ^ A“^(x)A"^(x) 5 2x, X&0. (3.9) Proof: The right-hand side follows from Young1s Inequality; xy S A(x) + A(y) , x,y&0. (3.10) and the relation (1.15) B(B“*(x)) S X valid for left-continuous functions B. To prove the left-hand inequality for 0 show first that A(A(x)/x) 5 A(x) , x>0 (3.11) (see [3;Ch.I,§2-2]). If not, for some XQ>0 A(A(X0)/Xq) « sup(yA(x) /x-A(y)) > A(xQ) , y so that there is some y $ x for which A(Xq) - A(y) § yA(x0)/xQ - A(y) > A(xQ) , a contradiction. The left-hand side of (3.9) is equivalent, by (2.17) to the inequality A(x/A-1(x)) S x. (3.12) In particular, if for some x, 0 A”1(x) = inf{y:A(x)>y} = inf {y:A(y) >A(t)} = t, so that A(x/A-1(x)) = A(A(t)/t) ^ A(t) = x. -1 — 4. Suppose now that for some x, 0 Then for all t A(t)^t, and A(x/A~^(x))> x. From the first of these, the tending of A to +“ and the left-continuity of A there is XQ>0 such that A(XQ) < x 5 A(XQ+0) . It is clear from the graph of A that A"'*'(x) = XQ. From A(x/A~^(x)) > X we conclude that there is some y such that 32 x < yA"1^) /x - A(y) . If y S A X(x) then x < x-A(y), an absurdity. 1 If y > A (x) = XQ then _1 _1 A(y)/y ^ A(X0+0)/XQ = A(XQ+0)/A (x) g X/A (x) , so that A(y) > xy/A ^(x), from which follows the contradiction x<0. Q.E.D. 2 Remark: The functions x, x /2 are examples which show that both inequalities (3.9) are best possible. (3.13) Corollary: If A is a Y-function A-^ is equivalent to the function x/A-^(x), which is a Y-function. And if B is a Y-function such that for constants k,K>0 kx g A~1(x)B”1(x) £ Kx, xgO, (3.14) Then B ~ A„ In particular A(2x/k) g B(x) g A(x/K) , x 1 0. (3.15) Proof: For the first part we verify that x/A"^(x) is non¬ decreasing. Let 9 § 1. 0x/A (0X) - x/A (X), = (a positive function of x)•(0A (x) ■ A“1(0x)) g 0. For the inequalities, we have B“1(x) * KX/A”1(X) £ KA~1 (x) and B”^(x) g kx/A"^(x) g %kA~^x) . ' Now (3.15) follows from (2.24). Q.E.D. 33 The following results give information on the structure of the orderings on Y-functions. (3.16) Lemma: [Q,X-^] has a dual automorphism. Proof: By (2.33) the mapping A -* A v Xra is an isomorphism of and [X I], Complementation gives a dual isomorphism between [XOT,I] and [Q,X^]. Hence A -* Avx^ -* AvX^ ~ AAX^ is a dual automorphism of [Q,X^]. Two partial orderings s, >- on a set z are isomorphic if (z >=) and (£,>-) are isomorphic (3.17) Theorem: The usual ordering (s) and the complementary ordering ("passing") of classes of Y-functions are isomorphic. Proof: Let g(A) = (A A Xj v (A A xt). (3.18) That is, g(A) is equivalent to A for large values, to A for small values, g is an involution, for g(g(A)) = (g(A) A Xj V (g(A) A X1) , and g(A) = (A v X-^) A (A v \o) is equivalent to A for large values and to A for small values. Thus g(g(A)) ~ A for all values. A A v V If A passes* B,* A X CO§ B X GOand A GOX ^ B X C so that g(A) s (B A Xw) v (A A X-^) ~ (B A XM) v (A v X^)^ g(B) If A S B, g (A) A ~ A A X-j^ ^ B A X^ ~ g(B) A X^ and g(A) A X ~ A A X S B A X ~ g(B) A Xm so g(A) passes g(B). Because g is an involution this 34 proves that g is an isomorphism of the desired type. Remark: Since g(A) ~ A for large values, g(A) ~ A for = x all if and only if AAX-^ ~ AAX^. Using the notation X£ f and the consequence AAA ^ X2 - AvA of (3.10) we obtain A X2 *1 - (AvA)AX-^ = AAXX » (AA^AX-J^ * X2AX^. Finally g(X-^AX2) ® (X^AX^)AX^ = X2AX^ so g(A) ~ A if and only if A ~ X£ for small values. We now establish a connection between "large values" and "small values". (3.19) Definition: If A is a Y-function we define on [0,®) (0 x=0, sup l/A(l/y) x>0, y A(0x)/0x = (l/0x)sup l/A(l/y) = (1/x) sup l/0A(l/0y)sA(x)/x y<0 A(0) ® A(0) =0. If x>0, A(x) « sup l/(sup l/A(l/z)) = sup inf A(z) S A(x) . y For each e>0 there exists YQ A(x) * inf A(z) “ A(y0) > A(x)-e, Z>V0 £d fd so that A(x) 1 A(x), and hence A = A. To prove the last statement suppose for some k>0 A(x) £ B(kx), x § 0. Then for x>0 A(kx) = sup l/A(l/ky) § sup l/B(l/y) = B(x) . y Thus the map reverses order. By (3.7) A -» A is a dual automor¬ phism. In particular x^ = x^ and \m = Xa. (3.21) Lemma; If £ is a distributive lattice with Q,I which has one pair of non-trivial complements a,b and a dual automorphism f then in the complementary ordering of z with respect to a,b f is an (dual) automorphism if (f(a)^a) f (a)=a. Proof: Since f is one-to-one and onto, we need only show that f and f”^ preserve (reverse) the complementary ordering. Since this ordering can be expressed in terms of the original ordering it will be enough to show that f preserves (reverses) the complementary ordering. In the argument we assume f(a) = a. Proof of the dual assertion is obtained by exchanging a and b in the number¬ ed lines. Suppose xAa ^ yAa and xAb £ yAb. Then f(x)va = f(xAa) ^ f(yAa) = f(y)va and (3.22) f(x)vb = f(xAb) ^ f(yAb) = f(y)vb By duality f(x)Aa § f(y)Aa and f(x)Ab ^ f(y)Ab (3.23) so that f preserves the complementary ordering. The argument is reversible. Q.E.D. As a corollary we get the (3.24) Theorem: In the complementary ("passing") order on classes of Y-functions, complementation is a dual automorphism while A -» A is an isomorphism. Proof: T = Xi , X = X . (3.25) Lemma: % ~ S. Moreover X(x) £ A(x) = X(4x) , xsO. Proof: For all Y-functions A, (A)~^(x) = l/A"^(l/x). To show this fix x>0 and suppose A(y) > x. Then there is z that l/A(l/z) > x and hence A~^'(l/x)sA”'*'(A(l/z))§l/z>l/y, so that A~^(x) = inf(y:A(y)>x}s 1/A-^(1/x) . Now let 0>1. A(e/A-1(l/x)) = sup , l/A(l/y) s l/A(A_1(l/x)/0^) y<0/A_i(l/x) is 0^/A(A"1(l/x)) § 0^x > x. Therefore 0/A~^(l/x) 6 {y:A(y)>x}, and so 0/A~^(l/x) a A“^X) , all 0>1. Thus the inequality holds even when 0=1, and (A)-1(x) = l/A"1(l/x). A_1(x)A_1(x) = l/A’1(l/x)A “1(1/X) , by what we have shown. The quantity on the right, by (3.8) 37 is between x/2 and x and we apply (3.13) to get the con¬ clusion and the estimate. (3.26) Theorem: The intervals [Q,\^] and [QjX^l are iso¬ morphic . Proof; A - A is an automorphism on the lattice of non-trivial classes , and Q.E.D. In many cases natural constructions based on convex functions preserve convexity. With X this is not true. (3.27) Theorem: Every equivalence class of functions convex at 0 contains a convex function A such that X is not convex Proof: Choose a representative A for the class, which is positive, finite and twice differentiable in the interval (1/3,3) X'(x) = (l/xA(l/x))^ A'(1/x) , which is non-decreasing in (1/3,3) o if and only if (A(x)/x) (1/A1 (x) ) is non-decreasing there, and this is equivalent to 2xA'(x) (xA1 (x)-A(x)) - x^A"(x)A(x) s 0, l/3 This quantity is dominated by 2(A(2x))^ - x^A(x)A"(x) (3.28) Now we modify A by adding to A" the function 0, 05x^1 or 2 This does not affect the equivalence class of A, while the modified X is not convex. Q.E.D. 38 (3.29) Definition: The Young's function A is an N-function if 0 < A(x) < ® for 0 lim A(x)/x = 0, lim A(x)/x = ® (3.30) x-*0 x-*® x With A(x) =f a(t)dt, A(x) $ xa(x) S A(2x) so that A is ' 0 an N-function if and only if 0 < a(x) < ® for 0 (3.31) Theorem: For a non-trivial Y-function A the follow¬ ing are equivalent: 1° 5 is an N-function 2° A is an N-function 3° 0 4° A is not comparable to either of Proof: 1° =» 2°: We may assume that A is a Young's function, so that A is an N-function. For 0 A(x) = sup y(x-A(y)/y) { Q] since for some y A ) > x y0 <3'o % Yi Mv1)/y1 < x If a denotes the left-continuous inverse of a, a has the same properties as a. Hence A is an N-function. 1° and 2° => 3°. 3° => A not comparable to X^; if A were comparable to X^, we would have to have A S X-^ so that A ^ = X0, which contradicts 3°. 4° => 1°: A(x) = 0 for 0 s x £ XQ ® A s x^, so 0 0 A(x)/x ^ C <® => A ^ X-^, so A(x)/x -*® as x -► ®. Similarly A(x)/x -0 as x - 0. Hence A is an N-function. Q.E.D. In [3;§3] there is a construction which shows that to a given N-function there corresponds an N-function which is not compar¬ able to it for large values. By using the mapping A -> A (Definition (3.19)), we can then construct an N-function which is not comparable to the given one, either for large or small values. Let A be the given function, and let B be an N- function not comparable for large values, to A (A is also an N-function), and C such a function with respect to A. It follows that A is not comparable to (5AX1)V(CA\J , either for large or small values. Remark: We sketch a "direct" construction of a non-comparable function (for large values). Again let A be an N-function and consider the family of N-functions A(x); A(x/n), A(nx), n = 2,3,* •*. Let B(x) = A(x) , 0 ^ x § XQ =1. There is a point x^ such that the line joining (x^,A(2x^)) to (1,A(1)) does not lie above the graph of A(2x). Since A is an N- function A(x/2)/x -♦ ® as x -* ®. Thus this supporting line for A(2x) meets the graph of A(x/2) above a point We extend B to be the function having this line segment for its graph, on (XQ,X2]. NOW we let X2 play the role of XQ and 40 construct a supporting line for A(3x) and passing through A(x/3) , and so on. It is evident from the graph obtained that B is not comparable to A for large values. 41 <4. If G is a locally compact unimodular topological group, with Haar measure dt, and f,g are integrable we define their convolution f*g by (f*g)(x) = J- f(t)g(t'1x)dt. G We write LA*Lg to denote the set of all f*g, f € L^. g 6 Lg. In this section our concern is to give conditions on G,A,B and a third function C in order that f*g always exist and that LA*LB S V In order to use Y-functions directly we recall (2.10) that |f|.= inf{K>0:J* A(|f(t)|/K)dt * 1), A G where A is a Y-function. If A is convex we write ||f||A for |f|A. . / (4.1) Theorem; If G is a locally compact unimodular topo¬ logical group and A,B,C are Y-functions for which there exists a constant k>0 such that A“1(x)B'1(x) = kx C’^(x) , x§0 (4.2) then LA*LB = Lc. Moreover If*gIc ^ 4k|f|A|g|B. If C is convex the 4 may be replaced by 2. Proof: To make use of (4.2) we employ the (4.3) Lemma; If (4.2) holds then 42 xy/k * A(x) C_1(B(y)) + B(y)C_1(A(x)) , x,y s 0. (4.4) If (4.4) holds then A’^(x)B"^(x) £ 2kxC"^(x) , xiO. Proof of lemma: Fix x^O, ysO. If A(x) £ B(y) consider xB_1(A(x)) £ A"1(A(x))B"1(A(x)) £ kA(x)C“1(A(x)) (4.5) Since u/B~^(u) does not decrease x/k £(A(x) /B”1 (A(x) ) ) C"1(A (x)) 5 (B(y) /B_ 1 (B (y) ))C~1 (A(x) ) ^ (B (y) /y) C"1 (A (x) ) ; Thus xy/k £ B(y)C"1(A(x)). If A(x)>B(y) we exchange A,B in (4.5) and finally obtain xy/k £ max[A(x)C’1(B(y) ) ,B(y)C-1(A(x)) ] S A(x)C'1(B(y)) + B(y)C'1(A(x)). If (4.4) holds we let x « A”^(u) , y = B“'*'(u). Then A ^(u)B 1(u) £ k[A(A-^(u))C”^(B(B-^(u))) + similar term] £ 2kuC’1(u). This proves the lemma. We now prove (4.1) assuming k=l. Let K > |f|^, L > |g|g. Since 5(x) £ C(x) = C(2x) , we have (f^x) £ C’^(x); now using these, Jensen's Inequality, and (4.3), we get ,fC( | f*g(x)/4KL)dx 2 r5(y(|f(t) | /K) (|g(t-1x) | /L)dt)dx (4.6) 2 JC(%.fA(|f(t) |/K)C”1(B(|g(t"1x) |/K))dt 43 + %fB(|g(t"1x) |/L)C"1(A(|f(t) |/K)dt)dx 1 1 s %rc(rA(|f(t>i/K)r (B(|g(t’ x)i/L)dt))dx + %pS( pB( IgCt"^) I /L)C"1(A( I f (t)|/K))dt)dx = I + J. We use next Jensen's Inequality and the observation that T A( | f (t) | /K)dt =§ 1: I - PA( |f (t) | /K)C“1(B( |g(t-1x) | /L))dt)dx PA( I f (t) |/K)dt a rece^CBduft-1^) I1/iOdt ^ fA(|f(t) |/K)dt S(l/2fA(|f(t) |/K)dt)JTA(|f(t) |/K)B(|g(t_1x) |/L)dt dx 1 -a/2.rA(|f(t)|/K)dt),fA(|£(t)|/K)[,rB(|g(t- x)|/LdXldt S(l/2|A(|f(t)|/K)dt)/A(|f(t)|/K)dt - k. Similarly J £ \ and I + J s 1 so that |f*g| SE 4|f|A|g|B. If C is convex we omit the first step in (4.6) so that the 4 is replaced by 2. Finally if k ^ 1 we observe that kCf^Cx) « Ck"*(x) h k where Ck(x) = C(x/k) , and that Ihl^ “ l / lc " Ih|c/U. Q.E.D. 44 (4.7) Corollary; If G is also totally finite and if only 1 1 1 A" (x)B" (x) ^ kxC" (x) , XSXQ>0, (4.8) then Moreover f 4kM f l *g|c - l lAlg|B» where M depends on the measure of G and XQ. If C is convex, 4 may be replaced by 2. (4.9) Corollary: If G is as in (4.1) and also discrete, and if only 1 1 1 A” (x)B~ (x) S kxC*" (x) , 0 *x Then L^*Lg s L^. Moreover |f*g|c - 4kM|f|A|g|B where M depends on XQ, A,B,C, and the measure M of an atom of G. If C is convex the constant may be replaced by 2. In both cases proof is accomplished by substituting functions A^, C^ for A 1,B~1,C~1. For the new functions (4.2) holds, and where the appropriate inequality (4.8) or (4.10) holds, the substitute functions are equal to the original ones. The constants arise from replacement of the substitute-function "norms” by the equivalent original "norms". 1 For (4.7) we have e.g. A^(x) - A" (XQ)X/XQ, 0^X^Xq. Thus for X 1 1 1 1 2 1 2 1 A^ (x)B^ (x) = (A" (X0)B~ (X0)/X0)X /X0 £ kC" (x0)x /x0 - kxC][ (x). 45 -1 -1 For (4.9) we define A-^ (x) = A (XQ) for x s x , and so on. By (2.50) the first constant M may be taken to be (1+XQ^I(G)) the second by (2.51) to be (1 + A"’^(l/m)/XQ) • (similar terms for B,C). (4.11) Theorem: In order that, over. G, LA*Lg £ 1° (4.2) is necessary if G is the additive group of real numbers with the usual topology. 2° (4.8) is necessary if G is (0,1) with addition modulo 1, and with the usual topology. 3° (4.10) is necessary if G is the additive group of integers, in the discrete topology. Proof of 1°: Suppose (4.2) does not hold, for any k. x x Case 1: There is a sequence ( n3» n i 0, such that (since inverse Y-functions are continuous). Define {in } by x m = 2”n, n^l, m =0. ^ n ^ n n 7 nu Let 0 =i m +• • •+ m_ -i x < m^+*,*+m , nsl; n0 n-i n cp(x) otherwise. J cp(x)dx »1. Let f(x) = A"1(cp(x)), g(x) » B“1(cp(x)). Then |f|A 5 1, |g|B* 1. We set h - f*g. X x h(x) « [ f(x-y)g(y)dy s f(x) f g(y)dy 5 xf(x)g(x) 0 0 since f,g are non-increasing for x>0. ,,, Let e>0 and set un * m^+ +mn, UQ = 0. 46 fC(eh(x) )dx § ftl(exf (x)g(x) )dx co un n _1 = 7, [*J C(ex4 x C (xv ))dx = 7. I . n=l1 ' un-_ 1i n n'' n=l n u <0u Fix 0, O<0<1. If n-i n5 un n _1 n 1 In = ,f C(ex4 xnC (xn))dx 5 (l-e)unC(eeun4 xnC' (xn)). 0u n ,n0-l -n § (l-0)unC(e92 C (xn)) since unxn 5 mnxn= 2 For n so large that e02n>2 we have n 1 1 n In s (l-e)unC/2(%ee)2 C' (xn)) S (l-e)unC(2C- (%e02 xn)> n = (l-9)un(%e0)2 xn S (1-0)(%e0) If un-l ”/®un’ n _ n I 5 mnC(e0un4 xnC ^(xn)) - %e9 for sufficiently large, Thus S I = fC(eh(x))dx = » for all e>0 so h i L . n r n=l Case 2: There is a sequence fxn), t such that A_1(x )B”1(x ) = 4nxC"^(x). s n N n' n n' We define m byJ x m =2 , n^l. Clearly E in < ® n n n ^ ^ i n * n=l so we let u = Em. and define n K k-n un+l < x - un’ cp(x) = elsewhere. Let f(x) = A~^(cp(x)), g(x) = B~ (cp(x)) . As before h(x) ^ xf(x)g(x) 47 and f e L^, g G Lg. For. e>0, oo u fC(eh(x))dx s T. r ” C(ex4nx C_ 1 (x ))dx = 7. I . n= n n=l Vl l Now the proof is exactly the same as that in case 1, except Mu that n_i" replaced by "u^^". Proof of 2°: If (4.8) does not hold for any k, we are in case 2 of the proof of 1° and we restrict cp to (0,1). Proof of 3°: If (4.10) does not hold for any k, we are in m n case 1. Suppose mn S n+i> “ m so Then mn ^ [mn+l] * n(l + 1/ro^) if we write n-1 n cp,k = xn fori=i E [m.+l]1 ^ k < i=lE [m.+l]1 and define f,g as before we have ffn) € L^, Cgn3 € Lfi and {hn3 jt Lc» m n To complete the proof we show that mn 5 n+i> “ 1»2,*** 1 1 0 < A” (x„)B"^(xn) = ^Vcr ^)^ S 1 1 = ix^A (xn+1)B (xn+1)/ 4+1 S ^(xn/xn+l)2A'1(xn)B"1(xn)- lW 1 1 n+1 1 Therefore (2 ‘ mn+1)' - xn+1 5 %xn = (2 mn)' . Q.E.D. We now have the following problems: (4.12) Given non-trivial Y-functions A,C', to find all non-trivial B for which (4.2) holds. Is there a "best possible" such B? 48 (4.13) Given non-trivial Y-functions A,B, to find all non¬ trivial C for which (4.2) holds. Is there a "best possible" such C? (4.14) Definition: Let A,B be non-trivial Y-functions, -1 -l - (A /B ) (X) = x inf A”^(y)/B ^(y) , OSx A"1/B"1(-) = co. - 1 - 1 (4.15) Lemma: A /B is non-deereasing, and (A-1/B”^)(x) /X is a non-increasing function. A'^'/B"*'*' is a non-trivial Y-^-function if and only if A S B for small values. Proof: Let 9§1. A /B~"*■ (0x) = x inf 9A~^(0x)/B~^(9y) ^ x inf A~^(0y) /B”^(y) ySx ygx s (A’1/B‘1)(x) . The second assertion is obvious. Suppose A 1/B’1 is non-trivial: for some XQ, 0 0 < XQinf A-'*'(y)/B-^(y) < ® y^0 -1 -1 Then A (y) l const B (y)/xQ, y S XQ, SO A S B for small x. Conversely, if A(y) S B(ky) , OSySxQ, then A(B”^(x)/k) S x if OS B-'*'(x)/k S XQ, _ and so B '*'(x)/k S A~‘*'(x) , if 0 S B*"^(x)/k S XQ. Hence if B(x) > 0 for x > 0, A"’^'/B”^ is non trivial. 49 Suppose B(ky) = 0 for 0 5 y < x^ 2 XQ. Then A(y) = 0, 0 = y < x^, so that A~^(y), B~^(y) 5 const > 0, all y. Hence ® > inf A-^(yyB ^(y) > 0 for all x y^x since A-^, B-^ are positive and finite on (0,°°). Thus A^/B'1 is non-trivial. Q.E.D. The next proposition shows that / operates on equivalence classes. (4.16) Lemma: If A,B,C,D are Y-functions and A^C, B^D then A^/B"1 S C'1^1 and A-1^"1 ^ A^/D-1. ‘ Proof is straightforward. (4.17) Lemma; If A,B,C are Y-functions and A”^ C~^/D“^ 1 1 - - then C’ /A" § D ^. Hence if C ^/D'"^ is non-trivial, so is C"1 /A’1. Proof: A(B~^(x)/k) s x if 0 ^ B“'*'(x)k S XQ, and so B~^(x)/k S A“^(x) , if 0 ^B~^(x)/k 3S XQ. (C“1/A"1)(x) * x inf C‘1(y)/(y inf C“1(z)/D’1(z)) y=x z=y s x inf (C-1(y)/y)/(C-1(y)/D-1(y)) ySx = D- (x) . This proposition shows that the asymmetry we shall see is superficial. 50 (4.18) Theorem: Let A,B,C be non-trivial Y-functions. Then A ^(x)B ^(x) g kxC ^(x), x§0, for some k>0, (4.2) if and only if C -1“/A 1 is a non-trivial Y~-1 -function and B"1 g C"1/A"1. Proof: We may assume k=l. B"1(x) g B_1(x)/x g C“1(x)/A“1(x) , so A § c for small values and thus C’^/Ais a non-trivial Y”^-function. B“1(x)/x = inf B_1(y)/y g inf C_1 (y)/A-1 (y) so that B"1 g C"1/A-1 ygx ygx If: The condition B~^(x) g (C"^/A_^)(x), x§0 forces C"^/A”^ to be a non-trivial Y ^"-function since B-^ is non-trivial. And A”1 (x)B~1 (x) g A‘1(x)(C"1/A"1)(x) g xC_1 (x) . (4.19) The following theorem solves Problem (4.12). (4.20) Theorem: 1° Given non-trivial Y-functions A,C there exists B, a non-trivial Y-function such that (4.2) holds, if and only if C"^/A”^ is a non-trivial Y-function. In this case C 1/A 1 is the best possible such function. 2° For all non-trivial Y-functions A,C there exists a non-trivial Y-function B such that (4.8) holds. In this case there exists a best possible function similar to Cf^/A"*. In 2°, "best possible" refers to large values, and we write, _1 1 c (C /A" )L(X) 1 x inf "^(y)/A"^(y) XSl lgygx 1 1 (C' /A" )L(1) X<1 . 51 (4.21) Definition; A ^*B ^(x) - sup A ^(y)B ^(y)/y y=x The following proposition is similar to (4.15). -1 -1 _ i (4.22) Lemma: A *B is a non-trivial Y -function if and only if A ^(x)B~^(x)/x is bounded near 0, which is the case if and only if A ^ B near 0. However, one can define for large values a non-trivial Y ^-function (A_1*B_1),(x) = sup A"1(y)B"1(y)/y. x 0-y-x The following theorem solves Problem (4.13) (4.23) Theorem: 1° Given non-trivial Y-functions A,B there exists a non-trivial Y-function C such that (4.2) holds if and only if A ^ B near 0. In this case (4.2) holds for all C such that Cf1 s A-1*B~1. 2° For all A,B, non-trivial Y-functions, there exists a non-trivial Y-function C such that (4.8) holds, and C ^ s (A ^*B”^)^, for large values, which is best possible in this sense. A natural question arises: Given A,C, when does there exist B such that A'^B-1 ~ C-1? (4.24) Theorem: Let A, C be non-trivial Y-functions. A"1*(C’1/A_1) ~ C*1 if and only if there exists a Y-function D ~ C for which D~^(x)/A ^(x) is a non-decreasing function. Proof: It is immediate that A-1*(C"1/A“1) =§ C-1. 52 If the reverse inequality holds, denote the left-hand side by D-'*'. Then for a fixed x and each 9 5 1 D-1(0x)/A’1(9x) = sup (A-1(y) inf C_1(z)/A-1(z)) y=0x z=y = sup (A“1/y)d(y))/A_1(0x) y=9x = max (sup (A_1(y)d(y)), sup (A_1(y)d(y)) /A_1(ex)) ygx x=yi0x S max(sup(A’1(y)d(y))/A_1(x) , sup (A 1(y)d(y))/A’1(0x)) . y^x x^y^0x We observe that d is non-increasing. The second quantity is less than or equal to sup A_1(0x)d(y)) /A’1 (0x) = d(x). x^y^0x The first quantity,which is D" ^ (x)/A-^ (x) , is greater than or equal to A" (x) d (x) /A" ^ (x) *= d (x) . Hence D~^(x)/A~^(x) is a non-increasing function. On the other hand if D ~ C and D”^(x)/A” ^(x) is non¬ increasing then 1 1 1 1 1 _1 (D" /A" ) (X) = XD" (X)/A“ (X) = (x/A’ (x))D (x) , which is non-decreasing and non-trivial. Finally 1 -1 -1 -1 -1 1 C' ~ D = A * (D"VA ) ~ A * (C^/A' ). Q.E.D. BIBLIOGRAPHY 1. And6. T. On Products of Orlicz Spaces. Math. Annalen 140, 174-186 (1WT 2. Birkhoff, G. Lattice Theory. A.M.S. Colloquium Publication XXV, New York, 1948 (Revised Edition). 3. Krasnosel'skii, M. A. and Rutickii, Ya.B. Convex Functions and Orlicz Spaces. Trans, from Russian by L. F. Boron. Noordhoff-Groningen, 1961. 4. Luxemburg, W. A. J. Banach Function Spaces. Thesis, Technische Hogeschool te Delft, 1955. 5. Matuszewska, W. and Orlicz, W. A note on the theory of s-normed spaces of co-integrable functions"! Studia Math. 21, 107-115 (1961). 6. O'Neil, R. Fractional Integration in Orlicz Spaces. To appear in Transactions of the A.M.S. 7. Zygmund, A. Triginometric Series, 2 vols. Cambridge Univ. Press, 1959.