Our proof uses a solution of an abstract moment problem (see theorem ...... [16] B. Fuchssteiner and H. König, New versions of the Hahn-Banach the- orem.
a sandwich theorem, the moment problem, finite-simplicial sets and some inequalities CALIN AMBROZIE and OCTAV OLTEANU
We recall the following sandwich type problem. Let X be a convex subset of a vector space E, f, g : X → R two maps, g convex, f concave, g ≤ f . The problem is: in what conditions upon X, for any pair (g, f ) as above, there exists h : X → R affine, such that g ≤ h ≤ f . If X is a compact Hausdorff topological space and S is a convex cone of lower bounded, lower semicontinuous functions on X for which axiom S1 ([5], p.493) is fulfilled, the problem is solved in [5], p.505. In the present work, we show that for any finite-simplicial subset X, the sandwich problem stated above has at least one solution (we say that a convex subset X of a vector space E is finitesimplicial if and only if for any finite dimensional compact convex subset K ⊂ X, there exists a finite dimensional simplex T such that K ⊂ T ⊂ X). The novelty here is the fact that X is not supposed to be bounded in a locally convex topology. The idea of the proof is to reduce the problem to the very particular case when X is a finite dimensional simplex. As applications of our results, we deduce some inequalities for positive numbers and for self-adjoint bounded operators acting on a Hilbert space, pointing out the interdependence between scalar and operator inequalities (see corollary 3.9). In section 4 we solve a concrete moment problem in a space of differentiable functions. 2000 Mathematics Subject Classification: 46A22, 47A57, 52B11, 47A63. Key words: sandwich type theorem for linear operators, the abstract and the classical moment problem, S-affine operators, finite-simplicial sets, the constant β(a, n) = a(n−1)/n (n + a)1/n − a (n ∈ N, n ≥ 2, a > 0) and inequalities for self-adjoint operators.
1
1. INTRODUCTION The results mentioned in the abstract are deduced from theorem 2.3, stated and proved below. A similar result is proved in [17] (theorem 1.5.9), where a characterization for the inserting of an affine map between two arbitrary maps defined on a convex subset X of a convex cone F is given. Our proof uses a solution of an abstract moment problem (see theorem 2.1 below). Theorem 2.3 follows from theorem 2.1 using a natural embedding of an arbitrary set X in a vector space E. 2. A GENERAL SANDWICH-TYPE THEOREM In [32] p.513-515 we proved the following general theorem related to the abstract moment problem. 2.1. THEOREM. Let V be an preordered vector space, let Y be an ordercomplete vector lattice, {vj : j ∈ J} ⊂ V , {yj : j ∈ J} ⊂ Y , F, G ∈ L(V, Y ) two linear operators. We consider the following assertions: (a) There exists H ∈ L(V, Y ) such that H(vj ) = yj ,
j ∈ J,
G(ϕ) ≤ H(ϕ) ≤ F (ϕ),
ϕ ∈ V+ .
(b) For any finite subset J1 ⊂ J and for any {βj : j ∈ J1 } ⊂ R, the following implication holds: X
βj vj = ϕ2 − ϕ1
with
ϕ1 , ϕ2 ∈ V+ ⇒
j∈J1
X
βj yj ≤ F (ϕ2 ) − G(ϕ1 ).
j∈J1
If V is a vector lattice, we also consider the assertion (b0 ): (b0 ) G(ϕ) ≤ F (ϕ), ∀ϕ ∈ V+ and for any finite subset J1 ⊂ J and any {βj : j ∈ J1 } ⊂ R, we have: X j∈J1
βj yj ≤ F
X
+
βj vj − G
X
−
βj vj .
j∈J1
j∈J1
Then (a) ⇔ (b) holds and, if V is a vector lattice, we have (b0 ) ⇔ (b) ⇔ (a). For the proof and for some applications of theorem 2.1 see [26], [32], [33]. 2
Now, we are going to apply this theorem in order to obtain a sandwich type theorem for maps defined on an arbitrary set X. The idea is to imbed X in an order vector space V . We use the following notations. Let X be an arbitrary set and Y a vector space. We denote F = F(X, Y ) := {f : X → Y }. Let x ∈ X. We denote by εx : F → Y the linear operator defined by εx (f ) = f (x), f ∈ F. Let us also denote ε(X) := {εx : x ∈ X} ⊂ L(F, Y ), V := Sp(ε(X)) ⊂ L(F, Y ), where L(F, Y ) is the space of all linear operators from F into Y and Sp(ε(X)) is the vector subspace of L(F, Y ) generated by ε(X). 2.2. Definition. Let S ⊂ F be a convex cone. We shall say that h ∈ F is S-affine if and only if for any T ∈ V with T (s) = 0, ∀s ∈ S ∩ (−S), we have T (h) = 0. Let us denote S := {T ∈ V : T (s) = 0,
∀s ∈ S ∩ (−S)}.
Obviously, S is a vector subspace of L(F, Y ). In the particular case when Y = R, we consider the dual pair hF, V i, when n n hf,
X
βj εxj i :=
j=1
X
βj f (xj ),
βj ∈ R.
j=1
In this case h : X → R is S-affine if aond only if h ∈ (S ∩ (−S))00 = = S ∩ (−S), where the closure is considered in the weak topology σ(F, V ). 2.3. THEOREM. Let X be a nonempty arbitrary set, Y an order-complete vector lattice, S ⊂ F a convex cone. Let f, g ∈ F be such that g(x) ≤ f (x), x ∈ X. The following assertions are equivalent. (a) There exists a S-affine map h ∈ F such that (1)
g(x) ≤ h(x) ≤ f (x),
x ∈ X.
(b) For any n ∈ N \ {0}, any {x1 , . . . , xn , x01 , . . . , x0n } ⊂ X and any {λ1 , . . . , λn , α1 , . . . , αn } ⊂ R+ such that
(2)
n X
j=1
λj εxj −
n X j=1
αj εx0j (s) = 0 3
∀s ∈ S ∩ (−S),
we must have n X
(3)
αj g(x0j ) ≤
j=1
n X
λj f (xj ).
j=1
Proof. The implication (a) ⇒ (b) is almost obvious. In fact, let x1 , . . . , x0n , λ1 , . . . , αn as at the point (b). We have to show the implication (2) ⇒ (3). From (2) and from the definition of the S-affine map h, we have:
(4)
n X
λj εxj −
j=1
n X j=1
αj εx0j (h) = 0,
which may be rewritten as n X
(40 )
λj h(xj ) =
j=1
n X
αj h(x0j ).
j=1
Now, from (1) and (40 ) we deduce n X
αj g(x0j )
≤
j=1
n X
αj h(x0j )
=
j=1
n X
λj h(xj ) ≤
j=1
n X
λj f (xj ).
j=1
(b) ⇒ (a) Let V := Sp(ε(X)) ⊂ L(F, Y ), V+ :=
X
βj εxj : J1 finite set, βj ∈ R+ ,
∀j ∈ J1
j∈J1
⊃ ε(X).
It is easy to see that V endowed with the order relation v1 ≤ v2 ⇔ v2 − v1 ∈ V+ is a vector lattice and
X
+
γj εxj =
j∈J1
X
γj+ εxj ,
J1 finite , γj ∈ R,
xj ∈ X.
j∈J1
Let f, g ∈ F such that g ≤ f . We define G, F : V → Y by G
X j∈J1
βj εxj :=
X
F
βj g(xj ),
X j∈J1
j∈J1
4
βj εxj :=
X j∈J1
βj f (xj ),
βj ∈ R, J1 finite set. Then G and F are linear operators from V into Y , with the property X
G(ϕ) =
(3)
λj g(xj ) ≤
j∈J1
ϕ ∈ V+ , ϕ =
X
X
λj f (xj ) = F (ϕ),
j∈J1
λj εxj , λj ∈ R+ , J1 finite set. Thus the first condition
j∈J1 (b0 ) of
at the point theorem 2.1. is accomplished (we take in (2) and (3) x0j = xj , αj = λj , j ∈ {1, 2, . . . , n} =: J1 ). Let S := {T ∈ V : T (s) = 0, ∀s ∈ S ∩ (−S)}. In the vector space S we choose a Hamel basis {εxj : j ∈ J}, hence S = Sp{εxj : j ∈ J}. Then we take vj := εxj , yj := 0, j ∈ J. We check the second condition at the point (b0 ) of theorem 2.1. In our case, this condition is
0 ≤ F
X
+
βj εxj − G
j∈J1
We have:
F
=F
X
βj εxj ,
J1 ⊂ J, J1 finite subset.
j∈J1
−
X
X
j∈J1
+
βj εxj − G
βj+ εxj − G
j∈J1
βj− εxj =
j∈J1
−
βj εxj =
j∈J1
X
X
X
βj+ f (xj ) −
j∈J1
X
βj− g(xj ) ≥ 0,
j∈J1
the last inequality being accomplished by (3), where: x0j = xj , λj = βj+ , αj = βj− ,
j ∈ J1 := {j1 , . . . , jn } ⊂ J
(the condition (2) is true since X j∈J1
λj εxj −
X j∈J1
αj εx0j =
X
(βj+ − βj− )εxj =
j∈J1
X
βj εxj ∈ S,
j∈J1
the last relation being true by the choise of the set J ⊃ J1 . Now, the conditions from (b0 ) Theorem 2.1 are fulfilled and, by (b0 ) ⇒ (a) of this theorem, there exists H ∈ L(V, Y ) such that H(εxj ) = 0, j ∈ J
(i.e. H(T ) = 0 ∀T ∈ S) 5
and such that: (5)
G(ϕ) ≤ H(ϕ) ≤ F (ϕ),
ϕ ∈ V+ .
Let h : X → Y , h(x) := H(εx ), x ∈ X. Writing (5) for ϕ = εx , x ∈ X we find: g(x) ≤ h(x) ≤ f (x), x ∈ X, i.e. (1). On the other hand, from H(εxj ) = 0, j ∈ J, it follows
H
X
βj εxj = 0 ∀J1 ⊂ J; J1 finite set,
∀βj ∈ R,
j∈J1
with
X
Thus for any T =
P
βj s(xj ) = 0 ∀s ∈ S ∩ (−S).
j∈J1
βj εxj with T (s) = 0, ∀s ∈ S ∩ (−S), we have
j∈J1
X
0=H
X
βj εxj =
j∈J1
βj h(xj ) = T (h).
j∈J1
The conclusion is that h is S-affine and the proof is complete. 3. AN APPLICATION OF THEOREM 2.3 We start by recalling a well known result. 3.1. LEMMA. Let E be a real vector space and T := co{v1 , . . . , vm } a finite dimensional simplex (v1 , . . . , vm are affine independent). Let Y be an ordered vector space and f˜, g˜ : T → Y , f˜ concave, g˜ convex, with g˜ ≤ f˜. ˜ : T → Y affine such that Then there exists h ˜ ≤ f˜. g˜ ≤ h ˜ may be chosen such that Moreover, if g˜ < f˜, h ˜ < f˜. f˜ < h (If we choose yk ∈ [˜ g (vk ), f˜(vk )], then the function ˜ h
Ãm X k=1
!
λ k vk
:=
m X
λk yk ,
k=1
λk ≥ 0,
m X k=1
6
λk = 1
˜ ≤ f˜). is well define, affine and f˜ ≤ h 3.2. Definition. A convex subset X ⊂ E of the vector space E is said to be finite-simplicial if and only if for any finite-dimensional compact subset K ⊂ X (in a finite dimensional subspace which contains it), there exists a finite dimensional simplex T such that K ⊂ T ⊂ X. 3.3. Exemples. 1◦ Let E = R2 ; then any convex cone X ⊂ E is an (unbounded) finite-simplicial set. 2◦ Let E = Rn , n ≥ 3 and let X be a closed convex cone in E. We suppose that X have a compact base B which is a simplex (the order relation defined by X is latticial). Then X is a finite-simplicial set. 3◦ Let E be an arbitrary real vector space, and let f ∈ E ∗ a linear functional, f 6= 0. Then the sets {x ∈ E : f (x) > α}, {x ∈ E : f (x) ≥ α}, α ∈ R, are finite-simplicial. From here we deduce easily the fact that, in general, the intersection of two finite-simplicial sets is not a finite-simplicial set. 4◦ In E = Rn , n ≥ 3, the cone X = {(x1 , . . . , xn ) ∈ Rn : xn ≥ α(xp1 + . . . + xpn−1 )1/p } (α ∈]0, ∞[, p > 1) has a compact base, but is not finite-simplicial. 3.4. THEOREM. Let E be a real vector space and X ⊂ E a finitesimplicial (convex) subset. Let Y be an order-complete vector lattice, f, g : X → Y , f concave, g convex, such that g ≤ f. Let S be the cone of all concave maps f : X → Y . Then there exists an S-affine map h : X → Y such that g ≤ h ≤ f. Proof. We shall apply Theorem 2.3. We check the condition (b) of this theorem. Let {x1 , . . . , xn , x01 , . . . , x0n } ⊂ X, {λ1 , . . . , λn , α1 , . . . , αn } ⊂ R+ such that (2) holds. On the other hand, for any linear operator U ∈ L(E, Y ) and any y0 ∈ Y , (U + y0 )|X ∈ S ∩ (−S). Hence, by (2), we must have
n X j=1
λj εxj −
n X j=1
αj εx0j ((U + y0 )|X ) = 0. 7
Taking U = 0, y0 6= 0, we get
n X
n X
λj =
j=1
αj . Then we take y0 := 0,
j=1
U ∈ L(E, Y ) arbitrary and we obtain n X
n X
λj xj =
j=1
These yield
n X j=1
j=1
λj n X
αj x0j .
n X
xj =
j=1
λk
k=1
αj n X
x0j .
αk
k=1
Now we check (2) ⇒ (3). To see this, we remark that the set K := co{x1 , . . . , xn , x01 , . . . , x0n } is a finite-dimensional compact subset of X. But X was supposed to be finite-simplicial. Thus, there exists a finite dimensional simplex T = co{v1 , . . . , vm } such that K ⊂ T ⊂ X. ˜ : T → Y affine, Let g˜ := g|T , f˜ := f |T . By lemma 3.1, there exists h ˜ ≤ f˜. Then we get: such that g˜ ≤ h
n X
αj g(x0j )
=
j=1
n X
αj g˜(x0j )
n X
k=1
n X
λk
n X j=1
˜ j) ≤ λj h(x
n X αj ˜ λj h n j=1 X j=1 n X
αk
k=1
n n X X λj xj = λk n j=1 X j=1
λk
k=1 n X
n X
j=1
j=1
λj f˜(xj ) =
k=1
8
λj f (xj ),
≤
n X λj ˜ λj h n j=1 X j=1
=
˜ 0 ) = h(x j
αk
g˜(x0j )
k=1
n X αj αj n j=1 X j=1
≤
n X
αk
=
n X αj αj n j=1 X j=1
=
j=1
i.e. (3).
x0j
=
˜ j ) = h(x
By Theorem 2.3, (b) ⇒ (a), there exists a S-affine map h : X → Y such that (1) holds. The proof is complete. 3.5. COROLLARY. Let E, X be as in theorem 3.4. Let f, g : X → R, f concave, g convex, such that g ≤ f. Then there exists an affine map h : X → R such that g ≤ h ≤ f. Proof. By theorem 3.4, there exists an S-affine map h : X → R such that g ≤ h ≤ f . This implies h ∈ (S ∩ (−S))00 = S ∩ (−S) = S ∩ (−S). (The cone S of all real concave functions on X is closed in the weak topology σ(F, V ), where V := Sp(ε(X)) ⊂ F ∗ ). Thus h is affine on X and the proof is complete. The most simple case when corollary 3.5 may be applied is the case E = R, X = [a, ∞[. For any a > 0, we prove the existence and the unicity of a constant β = β(a) ∈ [0, 1[ which fulfills some interesting inequalities for any x ≥ a. These inequalities stay valid when we replace x by any positive self-adjoint operator A for which SA ⊂ [a, ∞[ ( SA is the spectrum of A). This may be done using the positivity of the spectral measure attached to A (see theorem 3.6 from below). 3.6. THEOREM. Let H be a Hilbert space and let n ∈ N, n ≥ 2. Then for any a ∈ [0, ∞[, there exists a unique β(a, n) ∈ [0, 1] such that for any self-adjoint operator A acting on H, with the spectrum SA ⊂ [a, ∞[, we have (6)
nan−1 I ≤ (A + β(a, n)I)n − An ≤ nAn−1 .
The number β(a, n) is given by the formula (7)
β(a, n) = a(n−1)/n (n + a)1/n − a.
In particular, we have (8)
lim β(a, n) = 0
n→∞
and (9)
lim β(a, n) = 1.
a→∞
9
Proof. Let a ≥ 0. We shall prove that the function g(x) := (xn + nan−1 )1/n is convex on [0, ∞[ and the function f (x) := (xn + nxn−1 )1/n is concave on [0, ∞[. Then, since obviously g(x) ≤ f (x), ∀x ≥ a, we may apply corollary 3.5. A direct computation gives g 0 (x) = (xn + nan−1 )1/n−1 xn−1 , g 00 (x) = n(n − 1)an−1 (xn + nan−1 )1/n−2 xn−2 > 0,
∀x > 0.
Thus, g is strictly convex on [0, ∞[. Computing now the first two derivatives of f , we find f 0 (x) = (xn + nxn−1 )1/n−1 · [xn−1 + (n − 1)xn−2 ], f 00 (x) = −(n − 1)x2n−4 (xn + nxn−1 )(1/n)−2 < 0, µ
(f 00 (x)
=
∀x > 0.
¶
1 − 1 (xn + nxn−1 )1/n−2 [nxn−1 + n(n − 1)xn−2 ][xn−1 + n
+(n − 1)xn−2 ] + (xn + nxn−1 )(1/n)−1 (n − 1)[xn−2 + (n − 2)xn−3 ] = = (n − 1)xn−3 (xn + nxn−1 )(1/n)−2 · {(xn + nxn−1 )(x + n − 2)− −[x2 + (n − 1)x][xn−1 + (n − 1)xn−2 ]} = = (n − 1)xn−3 (xn + nxn−1 )(1/n)−2 [xn+1 + (n − 2)xn + nxn + +n(n − 2)xn−1 − xn+1 − 2(n − 1)xn − −(n − 1)2 xn−1 ] = −(n − 1)x2n−4 (xn + nxn−1 )(1/n)−2 < 0 ∀x > 0). Thus f is concave on [0, ∞[. On the other hand, from their definitions we have g(x) ≤ f (x), ∀x ≥ a. Applying corollary 3.5, there exists an affine function h(x) = αx + β, such that g(x) ≤ h(x) = αx + β ≤ f (x) ∀x ∈ [a, ∞[ , i.e., (10)
(xn + nan−1 )1/n ≤ αx + β ≤ (xn + nxn−1 )1/n ,
Multiplying by x−1 and making x → ∞, we find (11)
α = 1.
10
∀x ∈ [a, ∞[ .
From (10) and (11) we deduce (12) lim [(xn + nan−1 )1/n − x] ≤ β ≤ lim [(xn + nxn−1 )1/n − x], x↑∞
x↑∞
x ≥ a.
It is easy to see that the left hand limit from (12) is 0 and the right one is 1. Thus β ∈ [0, 1]. From the second inequality (10) and by (11) it follows easily that β 6= 1. Thus β = β(a, n) ∈ [0, 1[ . Now we may rewrite (10) as (13)
xn + nan−1 ≤ (x + β)n ≤ xn + nxn−1 ,
x≥a
or, equivalently, (14)
nan−1 ≤ (x + β)n − xn ≤ nxn−1 ,
∀x ∈ [a, ∞[ ,
(β = β(a, n) ∈ [0, 1[ ). Now (6) follows from (14) via the functional calculus. On the other hand, from (14) written for x = a, we obtain immediately (7). From (7) we get (8) easily. To obtain (9), we rewrite (14) for x = a in another form nan−1 = Cn1 β(a, n)an−1 + . . . + Cnk β k (a, n)an−k + . . . + β n (a, n) and now we multiply this relation by a−(n−1) (if a > 0), and we make a → ∞. We get n = Cn1 lim β(a, n) + 0, i.e., (9). a→∞
The proof is complete. 3.7. COROLLARY. For any a ∈ [0, ∞[ we denote β(a, 3) := a2/3 (3 + a)1/3 − a ∈ [0, 1[ . Then for any self-adjoint operator A acting on a Hilbert space H, with the spectrum SA ⊂ [a, ∞[, we have (15)
3β(a, 3)A2 + 3β 2 (a, 3)A + [β 3 (a, 3) − 3a2 ]I ≥ 0
and (16)
3[1 − β(a, 3)]A2 + 3β 2 (a, 3)A − β 3 (a, 3)I ≥ 0;
(for B ∈ A(H) we write B ≥ 0 if hB(h), hi ≥ 0, 11
∀h ∈ H).
Proof. The relation (6) written for n = 3 yields 3a2 I ≤ A3 + 3β(a, 3)A2 + 3β 2 (a, 3)A + β 3 (a, 3)I − A3 ≤ 3A2 . These inequalities are equivalent respectively to (15) and (16). 3.8. COROLLARY. For any self-adjoint operator having the spectrum SA ⊂ [1, ∞[ , we have (17)
3(22/3 − 1)A2 + 3(24/3 + 1 − 25/3 )A + 3(22/3 − 24/3 )I ≥ 0
and (18)
(2 − 22/3 )A2 − (1 + 24/3 − 25/3 )A − (1 + 22/3 − 24/3 )I ≥ 0.
Proof. One applies corollary 3.7 to a = 1 (β(1, 3) = 22/3 − 1). 3.9. COROLLARY. For any ϕ ∈ [0, π/4[ and any a ∈ [0, cos ϕ − sin ϕ], the following inequalities hold 3β(a, 3) + 3 cos ϕβ 2 (a, 3) + β 3 (a, 3) − 3a2 ≥ 0,
(19)
3β(a, 3) + 3 cos ϕβ 2 (a, 3) + β 3 (a, 3) − 3a2 ≥
(20)
≥ 3 sin(2ϕ)β(a, 3) + 3 sin ϕβ 2 (a, 3), 3[1 − β(a, 3)] − 3 cos ϕβ 2 (a, 3) − β 3 (a, 3) ≥ 0,
(21)
{3[1 − β(a, 3)] − 3 cos ϕβ 2 (a, 3) − β 3 (a, 3)}2 ≥
(22)
≥ {3 sin(2ϕ)[1 − β(a, 3)] − 3 sin ϕβ 2 (a, 3)}2 .
Proof. Let (23)
0